List of integrals of rational functions

Template:Short description Template:One source Template:Verifiability The following is a list of integrals (antiderivative functions) of rational functions. Any rational function can be integrated by partial fraction decomposition of the function into a sum of functions of the form: Template:Blockindent which can then be integrated term by term.

For other types of functions that can be integrated, see lists of integrals.

Miscellaneous integrands

• $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
• $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \frac{x}{a}+C$
• $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C=\{\begin{array}{cc}-\frac{1}{a}\mathrm{artanh}\frac{x}{a}+C=\frac{1}{2a}\ln \frac{a-x}{a+x}+C& \text{(for }|x|<|a|\text{)}\\ -\frac{1}{a}\mathrm{arcoth}\frac{x}{a}+C=\frac{1}{2a}\ln \frac{x-a}{x+a}+C& \text{(for }|x|>|a|\text{)}\end{array}$
• $\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C=\{\begin{array}{cc}\frac{1}{a}\mathrm{artanh}\frac{x}{a}+C=\frac{1}{2a}\ln \frac{a+x}{a-x}+C& \text{(for }|x|<|a|\text{)}\\ \frac{1}{a}\mathrm{arcoth}\frac{x}{a}+C=\frac{1}{2a}\ln \frac{x+a}{x-a}+C& \text{(for }|x|>|a|\text{)}\end{array}$
• $\int \frac{1}{x^{2^n}+1}dx=\frac{1}{2^{n-1}}\sum _{k=1}^{2^{n-1}}\sin \left(\frac{2k-1}{2^n}\pi \right)\arctan [(x-\cos \left(\frac{2k-1}{2^n}\pi \right))\csc \left(\frac{2k-1}{2^n}\pi \right)]-\frac{1}{2}\cos \left(\frac{2k-1}{2^n}\pi \right)\ln |x^2-2x\cos \left(\frac{2k-1}{2^n}\pi \right)+1|+C$

Integrands of the form x^m(a x + b)ⁿ

Many of the following antiderivatives have a term of the form ln |ax + b|. Because this is undefined when x = −b / a, the most general form of the antiderivative replaces the constant of integration with a locally constant function. However, it is conventional to omit this from the notation. For example, $$\int \frac{1}{ax+b}dx=\{\begin{array}{cc}\frac{1}{a}\ln (-(ax+b))+C^{-}& ax+b<0\\ \frac{1}{a}\ln (ax+b)+C^{+}& ax+b>0\end{array}$$ is usually abbreviated as $$\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C,$$ where C is to be understood as notation for a locally constant function of x. This convention will be adhered to in the following.

• $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{(for }n\ne -1\text{)}$ (Cavalieri's quadrature formula)
• $\int x^{a-1}(1-x{)}^{b-1}dx=\mathrm{B}(x;a,b)+C\text{(for }\mathrm{Re}(a),\mathrm{Re}(b)>0\text{)}$ (Incomplete beta function)
• $\int \frac{x}{ax+b}dx=\frac{x}{a}-\frac{b}{a^2}\ln |ax+b|+C$
• $\int \frac{mx+n}{ax+b}dx=\frac{m}{a}x+\frac{an-bm}{a^2}\ln |ax+b|+C$
• $\int \frac{x}{(ax+b{)}^2}dx=\frac{b}{a^2(ax+b)}+\frac{1}{a^2}\ln |ax+b|+C$
• $\int \frac{x}{(ax+b{)}^n}dx=\frac{a(1-n)x-b}{a^2(n-1)(n-2)(ax+b{)}^{n-1}}+C\text{(for }n\in ̸\{1,2\}\text{)}$
• $\int x(ax+b{)}^{n}dx=\frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b{)}^{n+1}+C\text{(for }n\in ̸\{-1,-2\}\text{)}$
• $\int \frac{x^2}{ax+b}dx=\frac{b^2\ln (|ax+b|)}{a^3}+\frac{a{x}^2-2bx}{2a^2}+C$
• $\int \frac{x^2}{(ax+b{)}^2}dx=\frac{1}{a^3}(ax-2b\ln |ax+b|-\frac{b^2}{ax+b})+C$
• $\int \frac{x^2}{(ax+b{)}^3}dx=\frac{1}{a^3}(\ln |ax+b|+\frac{2b}{ax+b}-\frac{b^2}{2(ax+b{)}^2})+C$
• $\int \frac{x^2}{(ax+b{)}^n}dx=\frac{1}{a^3}(-\frac{(ax+b{)}^{3-n}}{(n-3)}+\frac{2b(ax+b{)}^{2-n}}{(n-2)}-\frac{b^2(ax+b{)}^{1-n}}{(n-1)})+C\text{(for }n\in ̸\{1,2,3\}\text{)}$
• $\int \frac{1}{x(ax+b)}dx=-\frac{1}{b}\ln \left|\frac{ax+b}{x}\right|+C$
• $\int \frac{1}{x^2(ax+b)}dx=-\frac{1}{bx}+\frac{a}{b^2}\ln \left|\frac{ax+b}{x}\right|+C$
• $\int \frac{1}{x^2(ax+b{)}^2}dx=-a(\frac{1}{b^2(ax+b)}+\frac{1}{a{b}^{2}x}-\frac{2}{b^3}\ln \left|\frac{ax+b}{x}\right|)+C$

Integrands of the form x^m / (a x² + b x + c)ⁿ

For $a\ne 0:$

• $\int \frac{1}{a{x}^2+bx+c}dx=\{\begin{array}{cc}\frac{2}{\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+C& \text{(for }4ac-b^2>0\text{)}\\ \frac{1}{\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|+C=\{\begin{array}{cc}-\frac{2}{\sqrt{b^2-4ac}}\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}}+C& \text{(for }|2ax+b|<\sqrt{b^2-4ac}\text{)}\\ -\frac{2}{\sqrt{b^2-4ac}}\mathrm{arcoth}\frac{2ax+b}{\sqrt{b^2-4ac}}+C& \text{(else)}\end{array}& \text{(for }4ac-b^2<0\text{)}\\ -\frac{2}{2ax+b}+C& \text{(for }4ac-b^2=0\text{)}\end{array}$
• $\int \frac{x}{a{x}^2+bx+c}dx=\frac{1}{2a}\ln |a{x}^2+bx+c|-\frac{b}{2a}\int \frac{dx}{a{x}^2+bx+c}+C$
• $\int \frac{mx+n}{a{x}^2+bx+c}dx=\{\begin{array}{cc}\frac{m}{2a}\ln |a{x}^2+bx+c|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+C& \text{(for }4ac-b^2>0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|+\frac{2an-bm}{2a\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|+C=\{\begin{array}{cc}\frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}}+C& \text{(for }|2ax+b|<\sqrt{b^2-4ac}\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\mathrm{arcoth}\frac{2ax+b}{\sqrt{b^2-4ac}}+C& \text{(else)}\end{array}& \text{(for }4ac-b^2<0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a(2ax+b)}+C=\frac{m}{a}\ln |x+\frac{b}{2a}|-\frac{2an-bm}{a(2ax+b)}+C& \text{(for }4ac-b^2=0\text{)}\end{array}$
• $\int \frac{1}{(a{x}^2+bx+c{)}^n}dx=\frac{2ax+b}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int \frac{1}{(a{x}^2+bx+c{)}^{n-1}}dx+C$
• $\int \frac{x}{(a{x}^2+bx+c{)}^n}dx=-\frac{bx+2c}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int \frac{1}{(a{x}^2+bx+c{)}^{n-1}}dx+C$
• $\int \frac{1}{x(a{x}^2+bx+c)}dx=\frac{1}{2c}\ln \left|\frac{x^2}{a{x}^2+bx+c}\right|-\frac{b}{2c}\int \frac{1}{a{x}^2+bx+c}dx+C$

Integrands of the form x^m (a + b xⁿ)^p

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.

• $\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^p}{m+np+1}+\frac{anp}{m+np+1}\int x^m{(a+b{x}^n)}^{p-1}dx$
• $\int x^m{(a+b{x}^n)}^{p}dx=-\frac{x^{m+1}{(a+b{x}^n)}^{p+1}}{an(p+1)}+\frac{m+n(p+1)+1}{an(p+1)}\int x^m{(a+b{x}^n)}^{p+1}dx$
• $\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^p}{m+1}-\frac{bnp}{m+1}\int x^{m+n}{(a+b{x}^n)}^{p-1}dx$
• $\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m-n+1}{(a+b{x}^n)}^{p+1}}{bn(p+1)}-\frac{m-n+1}{bn(p+1)}\int x^{m-n}{(a+b{x}^n)}^{p+1}dx$
• $\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m-n+1}{(a+b{x}^n)}^{p+1}}{b(m+np+1)}-\frac{a(m-n+1)}{b(m+np+1)}\int x^{m-n}{(a+b{x}^n)}^{p}dx$
• $\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^{p+1}}{a(m+1)}-\frac{b(m+n(p+1)+1)}{a(m+1)}\int x^{m+n}{(a+b{x}^n)}^{p}dx$

Integrands of the form (A + B x) (a + b x)^m (c + d x)ⁿ (e + f x)^p

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, n and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $(a+bx{)}^m(c+dx{)}^n(e+fx{)}^p$ by setting B to 0.

• $\begin{array}{cc}& \int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=-\frac{(Ab-aB)(a+bx{)}^{m+1}(c+dx{)}^n(e+fx{)}^{p+1}}{b(m+1)(af-be)}+\frac{1}{b(m+1)(af-be)}\cdot \\ & \int (bc(m+1)(Af-Be)+(Ab-aB)(nde+cf(p+1))+d(b(m+1)(Af-Be)+f(n+p+1)(Ab-aB))x)(a+bx{)}^{m+1}(c+dx{)}^{n-1}(e+fx{)}^{p}dx\end{array}$
• $\begin{array}{cc}& \int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=\frac{B(a+bx{)}^m(c+dx{)}^{n+1}(e+fx{)}^{p+1}}{df(m+n+p+2)}+\frac{1}{df(m+n+p+2)}\cdot \\ & \int (Aadf(m+n+p+2)-B(bcem+a(de(n+1)+cf(p+1)))+(Abdf(m+n+p+2)+B(adfm-b(de(m+n+1)+cf(m+p+1))))x)(a+bx{)}^{m-1}(c+dx{)}^n(e+fx{)}^{p}dx\end{array}$
• $\begin{array}{cc}& \int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=\frac{(Ab-aB)(a+bx{)}^{m+1}(c+dx{)}^{n+1}(e+fx{)}^{p+1}}{(m+1)(ad-bc)(af-be)}+\frac{1}{(m+1)(ad-bc)(af-be)}\cdot \\ & \int ((m+1)(A(adf-b(cf+de))+Bbce)-(Ab-aB)(de(n+1)+cf(p+1))-df(m+n+p+3)(Ab-aB)x)(a+bx{)}^{m+1}(c+dx{)}^n(e+fx{)}^{p}dx\end{array}$

Integrands of the form x^m (A + B xⁿ) (a + b xⁿ)^p (c + d xⁿ)^q

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form ${(a+b{x}^n)}^p{(c+d{x}^n)}^q$ and $x^m{(a+b{x}^n)}^p{(c+d{x}^n)}^q$ by setting m and/or B to 0.

• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{abn(p+1)}+\frac{1}{abn(p+1)}\cdot \\ & \int x^m(c(Abn(p+1)+(Ab-aB)(m+1))+d(Abn(p+1)+(Ab-aB)(m+nq+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q-1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{b(m+n(p+q+1)+1)}+\frac{1}{b(m+n(p+q+1)+1)}\cdot \\ & \int x^m(c((Ab-aB)(1+m)+Abn(1+p+q))+(d(Ab-aB)(1+m)+Bnq(bc-ad)+Abdn(1+p+q))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{an(bc-ad)(p+1)}+\frac{1}{an(bc-ad)(p+1)}\cdot \\ & \int x^m(c(Ab-aB)(m+1)+An(bc-ad)(p+1)+d(Ab-aB)(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bd(m+n(p+q+1)+1)}-\frac{1}{bd(m+n(p+q+1)+1)}\cdot \\ & \int x^{m-n}(aBc(m-n+1)+(aBd(m+nq+1)-b(-Bc(m+np+1)+Ad(m+n(p+q+1)+1)))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{ac(m+1)}+\frac{1}{ac(m+1)}\cdot \\ & \int x^{m+n}(aBc(m+1)-A(bc+ad)(m+n+1)-An(bcp+adq)-Abd(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{a(m+1)}-\frac{1}{a(m+1)}\cdot \\ & \int x^{m+n}(c(Ab-aB)(m+1)+An(bc(p+1)+adq)+d((Ab-aB)(m+1)+Abn(p+q+1))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{(Ab-aB)x^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bn(bc-ad)(p+1)}-\frac{1}{bn(bc-ad)(p+1)}\cdot \\ & \int x^{m-n}(c(Ab-aB)(m-n+1)+(d(Ab-aB)(m+nq+1)-bn(Bc-Ad)(p+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx\end{array}$

Integrands of the form (d + e x)^m (a + b x + c x²)^p when b² − 4 a c = 0

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form ${(a+bx+c{x}^2)}^p$ when $b^2-4ac=0$ by setting m to 0.

• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^p}{e(m+1)}-\frac{p(d+ex{)}^{m+2}(b+2cx){(a+bx+c{x}^2)}^{p-1}}{e^2(m+1)(m+2p+1)}+\frac{p(2p-1)(2cd-be)}{e^2(m+1)(m+2p+1)}\int (d+ex{)}^{m+1}{(a+bx+c{x}^2)}^{p-1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^p}{e(m+1)}-\frac{p(d+ex{)}^{m+2}(b+2cx){(a+bx+c{x}^2)}^{p-1}}{e^2(m+1)(m+2)}+\frac{2cp(2p-1)}{e^2(m+1)(m+2)}\int (d+ex{)}^{m+2}{(a+bx+c{x}^2)}^{p-1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=-\frac{e(m+2p+2)(d+ex{)}^m{(a+bx+c{x}^2)}^{p+1}}{(p+1)(2p+1)(2cd-be)}+\frac{(d+ex{)}^{m+1}(b+2cx){(a+bx+c{x}^2)}^p}{(2p+1)(2cd-be)}+\frac{e^{2}m(m+2p+2)}{(p+1)(2p+1)(2cd-be)}\int (d+ex{)}^{m-1}{(a+bx+c{x}^2)}^{p+1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=-\frac{em(d+ex{)}^{m-1}{(a+bx+c{x}^2)}^{p+1}}{2c(p+1)(2p+1)}+\frac{(d+ex{)}^m(b+2cx){(a+bx+c{x}^2)}^p}{2c(2p+1)}+\frac{e^{2}m(m-1)}{2c(p+1)(2p+1)}\int (d+ex{)}^{m-2}{(a+bx+c{x}^2)}^{p+1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^p}{e(m+2p+1)}-\frac{p(2cd-be)(d+ex{)}^{m+1}(b+2cx){(a+bx+c{x}^2)}^{p-1}}{2c{e}^2(m+2p)(m+2p+1)}+\frac{p(2p-1)(2cd-be{)}^2}{2c{e}^2(m+2p)(m+2p+1)}\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p-1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=-\frac{2ce(m+2p+2)(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^{p+1}}{(p+1)(2p+1)(2cd-be{)}^2}+\frac{(d+ex{)}^{m+1}(b+2cx){(a+bx+c{x}^2)}^p}{(2p+1)(2cd-be)}+\frac{2c{e}^2(m+2p+2)(m+2p+3)}{(p+1)(2p+1)(2cd-be{)}^2}\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p+1}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^m(b+2cx){(a+bx+c{x}^2)}^p}{2c(m+2p+1)}+\frac{m(2cd-be)}{2c(m+2p+1)}\int (d+ex{)}^{m-1}{(a+bx+c{x}^2)}^{p}dx$
• $\int (d+ex{)}^m{(a+bx+c{x}^2)}^{p}dx=-\frac{(d+ex{)}^{m+1}(b+2cx){(a+bx+c{x}^2)}^p}{(m+1)(2cd-be)}+\frac{2c(m+2p+2)}{(m+1)(2cd-be)}\int (d+ex{)}^{m+1}{(a+bx+c{x}^2)}^{p}dx$

Integrands of the form (d + e x)^m (A + B x) (a + b x + c x²)^p

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form ${(a+bx+c{x}^2)}^p$ and $(d+ex{)}^m{(a+bx+c{x}^2)}^p$ by setting m and/or B to 0.

• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(Ae(m+2p+2)-Bd(2p+1)+eB(m+1)x){(a+bx+c{x}^2)}^p}{e^2(m+1)(m+2p+2)}+\frac{1}{e^2(m+1)(m+2p+2)}p\cdot \\ & \int (d+ex{)}^{m+1}(B(bd+2ae+2aem+2bdp)-Abe(m+2p+2)+(B(2cd+be+bem+4cdp)-2Ace(m+2p+2))x){(a+bx+c{x}^2)}^{p-1}dx\end{array}$
• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^m(Ab-2aB-(bB-2Ac)x){(a+bx+c{x}^2)}^{p+1}}{(p+1)(b^2-4ac)}+\frac{1}{(p+1)(b^2-4ac)}\cdot \\ & \int (d+ex{)}^{m-1}(B(2aem+bd(2p+3))-A(bem+2cd(2p+3))+e(bB-2Ac)(m+2p+3)x){(a+bx+c{x}^2)}^{p+1}dx\end{array}$
• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(Ace(m+2p+2)-B(cd+2cdp-bep)+Bce(m+2p+1)x){(a+bx+c{x}^2)}^p}{c{e}^2(m+2p+1)(m+2p+2)}-\frac{p}{c{e}^2(m+2p+1)(m+2p+2)}\cdot \\ & \int (d+ex{)}^m(Ace(bd-2ae)(m+2p+2)+B(ae(be-2cdm+bem)+bd(bep-cd-2cdp))+\\ & (Ace(2cd-be)(m+2p+2)-B(-b^2{e}^2(m+p+1)+2c^2{d}^2(1+2p)+ce(bd(m-2p)+2ae(m+2p+1))))x){(a+bx+c{x}^2)}^{p-1}dx\end{array}$
• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(A(bcd-b^{2}e+2ace)-aB(2cd-be)+c(A(2cd-be)-B(bd-2ae))x){(a+bx+c{x}^2)}^{p+1}}{(p+1)(b^2-4ac)(c{d}^2-bde+a{e}^2)}+\\ & \frac{1}{(p+1)(b^2-4ac)(c{d}^2-bde+a{e}^2)}\cdot \\ & \int (d+ex{)}^m(A(bcde(2p-m+2)+b^2{e}^2(m+p+2)-2c^2{d}^2(3+2p)-2ac{e}^2(m+2p+3))-\\ & B(ae(be-2cdm+bem)+bd(-3cd+be-2cdp+bep))+ce(B(bd-2ae)-A(2cd-be))(m+2p+4)x){(a+bx+c{x}^2)}^{p+1}dx\end{array}$
• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{B(d+ex{)}^m{(a+bx+c{x}^2)}^{p+1}}{c(m+2p+2)}+\frac{1}{c(m+2p+2)}\cdot \\ & \int (d+ex{)}^{m-1}(m(Acd-aBe)-d(bB-2Ac)(p+1)+((Bcd-bBe+Ace)m-e(bB-2Ac)(p+1))x){(a+bx+c{x}^2)}^{p}dx\end{array}$
• $\begin{array}{cc}& \int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=-\frac{(Bd-Ae)(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^{p+1}}{(m+1)(c{d}^2-bde+a{e}^2)}+\frac{1}{(m+1)(c{d}^2-bde+a{e}^2)}\cdot \\ & \int (d+ex{)}^{m+1}((Acd-Abe+aBe)(m+1)+b(Bd-Ae)(p+1)+c(Bd-Ae)(m+2p+3)x){(a+bx+c{x}^2)}^{p}dx\end{array}$

Integrands of the form x^m (a + b xⁿ + c x²ⁿ)^p when b² − 4 a c = 0

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form ${(a+b{x}^n+c{x}^{2n})}^p$ when $b^2-4ac=0$ by setting m to 0.

• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}{(a+b{x}^n+c{x}^{2n})}^p}{m+2np+1}+\frac{np{x}^{m+1}(2a+b{x}^n){(a+b{x}^n+c{x}^{2n})}^{p-1}}{(m+1)(m+2np+1)}-\frac{b{n}^{2}p(2p-1)}{(m+1)(m+2np+1)}\int x^{m+n}{(a+b{x}^n+c{x}^{2n})}^{p-1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{(m+n(2p-1)+1)x^{m+1}{(a+b{x}^n+c{x}^{2n})}^p}{(m+1)(m+n+1)}+\frac{np{x}^{m+1}(2a+b{x}^n){(a+b{x}^n+c{x}^{2n})}^{p-1}}{(m+1)(m+n+1)}+\frac{2cp{n}^2(2p-1)}{(m+1)(m+n+1)}\int x^{m+2n}{(a+b{x}^n+c{x}^{2n})}^{p-1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{(m+n(2p+1)+1)x^{m-n+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{b{n}^2(p+1)(2p+1)}-\frac{x^{m+1}(b+2c{x}^n){(a+b{x}^n+c{x}^{2n})}^p}{bn(2p+1)}-\frac{(m-n+1)(m+n(2p+1)+1)}{b{n}^2(p+1)(2p+1)}\int x^{m-n}{(a+b{x}^n+c{x}^{2n})}^{p+1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=-\frac{(m-3n-2np+1)x^{m-2n+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{2c{n}^2(p+1)(2p+1)}-\frac{x^{m-2n+1}(2a+b{x}^n){(a+b{x}^n+c{x}^{2n})}^p}{2cn(2p+1)}+\frac{(m-n+1)(m-2n+1)}{2c{n}^2(p+1)(2p+1)}\int x^{m-2n}{(a+b{x}^n+c{x}^{2n})}^{p+1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}{(a+b{x}^n+c{x}^{2n})}^p}{m+2np+1}+\frac{np{x}^{m+1}(2a+b{x}^n){(a+b{x}^n+c{x}^{2n})}^{p-1}}{(m+2np+1)(m+n(2p-1)+1)}+\frac{2a{n}^{2}p(2p-1)}{(m+2np+1)(m+n(2p-1)+1)}\int x^m{(a+b{x}^n+c{x}^{2n})}^{p-1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=-\frac{(m+n+2np+1)x^{m+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{2a{n}^2(p+1)(2p+1)}-\frac{x^{m+1}(2a+b{x}^n){(a+b{x}^n+c{x}^{2n})}^p}{2an(2p+1)}+\frac{(m+n(2p+1)+1)(m+2n(p+1)+1)}{2a{n}^2(p+1)(2p+1)}\int x^m{(a+b{x}^n+c{x}^{2n})}^{p+1}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m-n+1}(b+2c{x}^n){(a+b{x}^n+c{x}^{2n})}^p}{2c(m+2np+1)}-\frac{b(m-n+1)}{2c(m+2np+1)}\int x^{m-n}{(a+b{x}^n+c{x}^{2n})}^{p}dx$
• $\int x^m{(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}(b+2c{x}^n){(a+b{x}^n+c{x}^{2n})}^p}{b(m+1)}-\frac{2c(m+n(2p+1)+1)}{b(m+1)}\int x^{m+n}{(a+b{x}^n+c{x}^{2n})}^{p}dx$

Integrands of the form x^m (A + B xⁿ) (a + b xⁿ + c x²ⁿ)^p

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form ${(a+b{x}^n+c{x}^{2n})}^p$ and $x^m{(a+b{x}^n+c{x}^{2n})}^p$ by setting m and/or B to 0.

• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}(A(m+n(2p+1)+1)+B(m+1)x^n){(a+b{x}^n+c{x}^{2n})}^p}{(m+1)(m+n(2p+1)+1)}+\frac{np}{(m+1)(m+n(2p+1)+1)}\cdot \\ & \int x^{m+n}(2aB(m+1)-Ab(m+n(2p+1)+1)+(bB(m+1)-2Ac(m+n(2p+1)+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p-1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m-n+1}(Ab-2aB-(bB-2Ac)x^n){(a+b{x}^n+c{x}^{2n})}^{p+1}}{n(p+1)(b^2-4ac)}+\frac{1}{n(p+1)(b^2-4ac)}\cdot \\ & \int x^{m-n}((m-n+1)(2aB-Ab)+(m+2n(p+1)+1)(bB-2Ac)x^n){(a+b{x}^n+c{x}^{2n})}^{p+1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}(bBnp+Ac(m+n(2p+1)+1)+Bc(m+2np+1)x^n){(a+b{x}^n+c{x}^{2n})}^p}{c(m+2np+1)(m+n(2p+1)+1)}+\frac{np}{c(m+2np+1)(m+n(2p+1)+1)}\cdot \\ & \int x^m(2aAc(m+n(2p+1)+1)-abB(m+1)+(2aBc(m+2np+1)+Abc(m+n(2p+1)+1)-b^{2}B(m+np+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p-1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=-\frac{x^{m+1}(A{b}^2-abB-2aAc+(Ab-2aB)c{x}^n){(a+b{x}^n+c{x}^{2n})}^{p+1}}{an(p+1)(b^2-4ac)}+\frac{1}{an(p+1)(b^2-4ac)}\cdot \\ & \int x^m((m+n(p+1)+1)A{b}^2-abB(m+1)-2(m+2n(p+1)+1)aAc+(m+n(2p+3)+1)(Ab-2aB)c{x}^n){(a+b{x}^n+c{x}^{2n})}^{p+1}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{B{x}^{m-n+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{c(m+n(2p+1)+1)}-\frac{1}{c(m+n(2p+1)+1)}\cdot \\ & \int x^{m-n}(aB(m-n+1)+(bB(m+np+1)-Ac(m+n(2p+1)+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p}dx\end{array}$
• $\begin{array}{cc}& \int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{A{x}^{m+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{a(m+1)}+\frac{1}{a(m+1)}\cdot \\ & \int x^{m+n}(aB(m+1)-Ab(m+n(p+1)+1)-Ac(m+2n(p+1)+1)x^n){(a+b{x}^n+c{x}^{2n})}^{p}dx\end{array}$