Automorphic number

The following is a list of integrals (antiderivative functions) of rational functions. For other types of functions, see lists of integrals.

Miscellaneous integrands

$\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \frac{x}{a}+C$

$\int \frac{1}{x^2-a^2}dx=\{\begin{array}{ll}-\frac{1}{a}\mathrm{arctanh}\frac{x}{a}=\frac{1}{2a}\ln \frac{a-x}{a+x}+C& \text{(for }\left|x\right|<\left|a\right|\text{)}\\ -\frac{1}{a}\mathrm{arccoth}\frac{x}{a}=\frac{1}{2a}\ln \frac{x-a}{x+a}+C& \text{(for }\left|x\right|>\left|a\right|\text{)}\end{array}$

$\int \frac{dx}{x^{2^n}+1}=\sum _{k=1}^{2^{n-1}}\{\frac{1}{2^{n-1}}[\sin \left(\frac{(2k-1)\pi }{2^n}\right)\arctan [(x-\cos \left(\frac{(2k-1)\pi }{2^n}\right))\csc \left(\frac{(2k-1)\pi }{2^n}\right)]]-\frac{1}{2^n}[\cos \left(\frac{(2k-1)\pi }{2^n}\right)\ln |x^2-2x\cos \left(\frac{(2k-1)\pi }{2^n}\right)+1|]\}+C$

Any rational function can be integrated using partial fractions in integration, by decomposing the rational function into a sum of functions of the form:

$\frac{a}{(x-b{)}^n}$, and $\frac{ax+b}{{((x-c{)}^2+d^2)}^n}.$

Integrands of the form x^m(a x + b)ⁿ

$\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$

More generally,

$\int \frac{1}{ax+b}dx=\{\begin{array}{ll}\frac{1}{a}\ln |ax+b|+C^{-}& x<-b/a\\ \frac{1}{a}\ln |ax+b|+C^{+}& x>-b/a\end{array}$

$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{(for }n\ne -1\text{)}$ (Cavalieri's quadrature formula)

$\int \frac{x}{ax+b}dx=\frac{x}{a}-\frac{b}{a^2}\ln |ax+b|+C$

$\int \frac{x}{(ax+b{)}^2}dx=\frac{b}{a^2(ax+b)}+\frac{1}{a^2}\ln |ax+b|+C$

$\int \frac{x}{(ax+b{)}^n}dx=\frac{a(1-n)x-b}{a^2(n-1)(n-2)(ax+b{)}^{n-1}}+C\text{(for }n\in ̸\{1,2\}\text{)}$

$\int x(ax+b{)}^{n}dx=\frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b{)}^{n+1}+C\text{(for }n\in ̸\{-1,-2\}\text{)}$

$\int \frac{x^2}{ax+b}dx=\frac{b^2\ln (|ax+b|)}{a^3}+\frac{a{x}^2-2bx}{2a^2}+C$

$\int \frac{x^2}{(ax+b{)}^2}dx=\frac{1}{a^3}(ax-2b\ln |ax+b|-\frac{b^2}{ax+b})+C$

$\int \frac{x^2}{(ax+b{)}^3}dx=\frac{1}{a^3}(\ln |ax+b|+\frac{2b}{ax+b}-\frac{b^2}{2(ax+b{)}^2})+C$

$\int \frac{x^2}{(ax+b{)}^n}dx=\frac{1}{a^3}(-\frac{(ax+b{)}^{3-n}}{(n-3)}+\frac{2b(ax+b{)}^{2-n}}{(n-2)}-\frac{b^2(ax+b{)}^{1-n}}{(n-1)})+C\text{(for }n\in ̸\{1,2,3\}\text{)}$

$\int \frac{1}{x(ax+b)}dx=-\frac{1}{b}\ln \left|\frac{ax+b}{x}\right|+C$

$\int \frac{1}{x^2(ax+b)}dx=-\frac{1}{bx}+\frac{a}{b^2}\ln \left|\frac{ax+b}{x}\right|+C$

$\int \frac{1}{x^2(ax+b{)}^2}dx=-a(\frac{1}{b^2(ax+b)}+\frac{1}{a{b}^{2}x}-\frac{2}{b^3}\ln \left|\frac{ax+b}{x}\right|)+C$

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

For $a\ne 0:$

$\int \frac{1}{a{x}^2+bx+c}dx=\{\begin{array}{ll}\frac{2}{\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+C& \text{(for }4ac-b^2>0\text{)}\\ -\frac{2}{\sqrt{b^2-4ac}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\frac{2ax+b}{\sqrt{b^2-4ac}}+C=\frac{1}{\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|+C& \text{(for }4ac-b^2<0\text{)}\\ -\frac{2}{2ax+b}+C& \text{(for }4ac-b^2=0\text{)}\end{array}$

$\int \frac{x}{a{x}^2+bx+c}dx=\frac{1}{2a}\ln |a{x}^2+bx+c|-\frac{b}{2a}\int \frac{dx}{a{x}^2+bx+c}+C$

$\int \frac{mx+n}{a{x}^2+bx+c}dx=\{\begin{array}{ll}\frac{m}{2a}\ln |a{x}^2+bx+c|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+C& \text{(for }4ac-b^2>0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\frac{2ax+b}{\sqrt{b^2-4ac}}+C& \text{(for }4ac-b^2<0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a(2ax+b)}+C& \text{(for }4ac-b^2=0\text{)}\end{array}$

$\int \frac{1}{(a{x}^2+bx+c{)}^n}dx=\frac{2ax+b}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int \frac{1}{(a{x}^2+bx+c{)}^{n-1}}dx+C$

$\int \frac{x}{(a{x}^2+bx+c{)}^n}dx=-\frac{bx+2c}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int \frac{1}{(a{x}^2+bx+c{)}^{n-1}}dx+C$

$\int \frac{1}{x(a{x}^2+bx+c)}dx=\frac{1}{2c}\ln \left|\frac{x^2}{a{x}^2+bx+c}\right|-\frac{b}{2c}\int \frac{1}{a{x}^2+bx+c}dx+C$

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.

$\int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=-\frac{(Ab-aB)(a+bx{)}^{m+1}(c+dx{)}^n(e+fx{)}^{p+1}}{b(m+1)(af-be)}+\frac{1}{b(m+1)(af-be)}\cdot$

$\int (bc(m+1)(Af-Be)+(Ab-aB)(nde+cf(p+1))+d(b(m+1)(Af-Be)+f(n+p+1)(Ab-aB))x)(a+bx{)}^{m+1}(c+dx{)}^{n-1}(e+fx{)}^{p}dx$

$\int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=\frac{B(a+bx{)}^m(c+dx{)}^{n+1}(e+fx{)}^{p+1}}{df(m+n+p+2)}+\frac{1}{df(m+n+p+2)}\cdot$

$\int (Aadf(m+n+p+2)-B(bcem+a(de(n+1)+cf(p+1)))+(Abdf(m+n+p+2)+B(adfm-b(de(m+n+1)+cf(m+p+1))))x)(a+bx{)}^{m-1}(c+dx{)}^n(e+fx{)}^{p}dx$

$\int (A+Bx)(a+bx{)}^m(c+dx{)}^n(e+fx{)}^{p}dx=\frac{(Ab-aB)(a+bx{)}^{m+1}(c+dx{)}^{n+1}(e+fx{)}^{p+1}}{(m+1)(ad-bc)(af-be)}+\frac{1}{(m+1)(ad-bc)(af-be)}\cdot$

$\int ((m+1)(A(adf-b(cf+de))+Bbce)-(Ab-aB)(de(n+1)+cf(p+1))-df(m+n+p+3)(Ab-aB)x)(a+bx{)}^{m+1}(c+dx{)}^n(e+fx{)}^{p}dx$

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.

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{abn(p+1)}+\frac{1}{abn(p+1)}\cdot$

$\int x^m(c(Abn(p+1)+(Ab-aB)(m+1))+d(Abn(p+1)+(Ab-aB)(m+nq+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{b(m+n(p+q+1)+1)}+\frac{1}{b(m+n(p+q+1)+1)}\cdot$

$\int x^m(c((Ab-aB)(1+m)+Abn(1+p+q))+(d(Ab-aB)(1+m)+Bnq(bc-ad)+Abdn(1+p+q))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{an(bc-ad)(p+1)}+\frac{1}{an(bc-ad)(p+1)}\cdot$

$\int x^m(c(Ab-aB)(m+1)+An(bc-ad)(p+1)+d(Ab-aB)(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bd(m+n(p+q+1)+1)}-\frac{1}{bd(m+n(p+q+1)+1)}\cdot$

$\int x^{m-n}(aBc(m-n+1)+(aBd(m+nq+1)-b(-Bc(m+np+1)+Ad(m+n(p+q+1)+1)))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{ac(m+1)}+\frac{1}{ac(m+1)}\cdot$

$\int x^{m+n}(aBc(m+1)-A(bc+ad)(m+n+1)-An(bcp+adq)-Abd(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{a(m+1)}-\frac{1}{a(m+1)}\cdot$

$\int x^{m+n}(c(Ab-aB)(m+1)+An(bc(p+1)+adq)+d((Ab-aB)(m+1)+Abn(p+q+1))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{(Ab-aB)x^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bn(bc-ad)(p+1)}-\frac{1}{bn(bc-ad)(p+1)}\cdot$

$\int x^{m-n}(c(Ab-aB)(m-n+1)+(d(Ab-aB)(m+nq+1)-bn(Bc-Ad)(p+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx$

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.

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(Ae(m+2p+2)-Bd(2p+1)+eB(m+1)x){(a+bx+c{x}^2)}^p}{e^2(m+1)(m+2p+2)}+\frac{1}{e^2(m+1)(m+2p+2)}p\cdot$

$\int (d+ex{)}^{m+1}(B(bd+2ae+2aem+2bdp)-Abe(m+2p+2)+(B(2cd+be+bem+4cdp)-2Ace(m+2p+2))x){(a+bx+c{x}^2)}^{p-1}dx$

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^m(Ab-2aB-(bB-2Ac)x){(a+bx+c{x}^2)}^{p+1}}{(p+1)(b^2-4ac)}+\frac{1}{(p+1)(b^2-4ac)}\cdot$

$\int (d+ex{)}^{m-1}(B(2aem+bd(2p+3))-A(bem+2cd(2p+3))+e(bB-2Ac)(m+2p+3)x){(a+bx+c{x}^2)}^{p+1}dx$

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(Ace(m+2p+2)-B(cd+2cdp-bep)+Bce(m+2p+1)x){(a+bx+c{x}^2)}^p}{c{e}^2(m+2p+1)(m+2p+2)}-\frac{p}{c{e}^2(m+2p+1)(m+2p+2)}\cdot$

$\int (d+ex{)}^m(Ace(bd-2ae)(m+2p+2)+B(ae(be-2cdm+bem)+bd(bep-cd-2cdp))+$

$(Ace(2cd-be)(m+2p+2)-B(-b^2{e}^2(m+p+1)+2c^2{d}^2(1+2p)+ce(bd(m-2p)+2ae(m+2p+1))))x){(a+bx+c{x}^2)}^{p-1}dx$

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{(d+ex{)}^{m+1}(A(bcd-b^{2}e+2ace)-aB(2cd-be)+c(A(2cd-be)-B(bd-2ae))x){(a+bx+c{x}^2)}^{p+1}}{(p+1)(b^2-4ac)(c{d}^2-bde+a{e}^2)}+$

$\frac{1}{(p+1)(b^2-4ac)(c{d}^2-bde+a{e}^2)}\cdot$

$\int (d+ex{)}^m(A(bcde(2p-m+2)+b^2{e}^2(m+p+2)-2c^2{d}^2(3+2p)-2ac{e}^2(m+2p+3))-$

$B(ae(be-2cdm+bem)+bd(-3cd+be-2cdp+bep))+ce(B(bd-2ae)-A(2cd-be))(m+2p+4)x){(a+bx+c{x}^2)}^{p+1}dx$

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=\frac{B(d+ex{)}^m{(a+bx+c{x}^2)}^{p+1}}{c(m+2p+2)}+\frac{1}{c(m+2p+2)}\cdot$

$\int (d+ex{)}^{m-1}(m(Acd-aBe)-d(bB-2Ac)(p+1)+((Bcd-bBe+Ace)m-e(bB-2Ac)(p+1))x){(a+bx+c{x}^2)}^{p}dx$

$\int (d+ex{)}^m(A+Bx){(a+bx+c{x}^2)}^{p}dx=-\frac{(Bd-Ae)(d+ex{)}^{m+1}{(a+bx+c{x}^2)}^{p+1}}{(m+1)(c{d}^2-bde+a{e}^2)}+\frac{1}{(m+1)(c{d}^2-bde+a{e}^2)}\cdot$

$\int (d+ex{)}^{m+1}((Acd-Abe+aBe)(m+1)+b(Bd-Ae)(p+1)+c(Bd-Ae)(m+2p+3)x){(a+bx+c{x}^2)}^{p}dx$

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.

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}(A(m+n(2p+1)+1)+B(m+1)x^n){(a+b{x}^n+c{x}^{2n})}^p}{(m+1)(m+n(2p+1)+1)}+\frac{np}{(m+1)(m+n(2p+1)+1)}\cdot$

$\int x^{m+n}(2aB(m+1)-Ab(m+n(2p+1)+1)+(bB(m+1)-2Ac(m+n(2p+1)+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m-n+1}(Ab-2aB-(bB-2Ac)x^n){(a+b{x}^n+c{x}^{2n})}^{p+1}}{n(p+1)(b^2-4ac)}+\frac{1}{n(p+1)(b^2-4ac)}\cdot$

$\int x^{m-n}((m-n+1)(2aB-Ab)+(m+2n(p+1)+1)(bB-2Ac)x^n){(a+b{x}^n+c{x}^{2n})}^{p+1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{x^{m+1}(bBnp+Ac(m+n(2p+1)+1)+Bc(m+2np+1)x^n){(a+b{x}^n+c{x}^{2n})}^p}{c(m+2np+1)(m+n(2p+1)+1)}+\frac{np}{c(m+2np+1)(m+n(2p+1)+1)}\cdot$

$\int x^m(2aAc(m+n(2p+1)+1)-abB(m+1)+(2aBc(m+2np+1)+Abc(m+n(2p+1)+1)-b^{2}B(m+np+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=-\frac{x^{m+1}(A{b}^2-abB-2aAc+(Ab-2aB)c{x}^n){(a+b{x}^n+c{x}^{2n})}^{p+1}}{an(p+1)(b^2-4ac)}+\frac{1}{an(p+1)(b^2-4ac)}\cdot$

$\int x^m((m+n(p+1)+1)A{b}^2-abB(m+1)-2(m+2n(p+1)+1)aAc+(m+n(2p+3)+1)(Ab-2aB)c{x}^n){(a+b{x}^n+c{x}^{2n})}^{p+1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{B{x}^{m-n+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{c(m+n(2p+1)+1)}-\frac{1}{c(m+n(2p+1)+1)}\cdot$

$\int x^{m-n}(aB(m-n+1)+(bB(m+np+1)-Ac(m+n(2p+1)+1))x^n){(a+b{x}^n+c{x}^{2n})}^{p}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n+c{x}^{2n})}^{p}dx=\frac{A{x}^{m+1}{(a+b{x}^n+c{x}^{2n})}^{p+1}}{a(m+1)}+\frac{1}{a(m+1)}\cdot$

$\int x^{m+n}(aB(m+1)-Ab(m+n(p+1)+1)-Ac(m+2n(p+1)+1)x^n){(a+b{x}^n+c{x}^{2n})}^{p}dx$

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.