3.221 \(\int \frac{(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log (e (\frac{a+b x}{c+d x})^n)} \, dx\)

Optimal. Leaf size=128 \[ \frac{(a+b x) e^{\frac{A (m+1)}{B n}} (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{m+1}{n}} \text{Ei}\left (-\frac{(m+1) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B i^2 n (c+d x) (b c-a d)} \]

[Out]

(E^((A*(1 + m))/(B*n))*(a + b*x)*(g*(a + b*x))^(-2 - m)*(e*((a + b*x)/(c + d*x))^n)^((1 + m)/n)*(i*(c + d*x))^
(2 + m)*ExpIntegralEi[-(((1 + m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n))])/(B*(b*c - a*d)*i^2*n*(c + d*
x))

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Rubi [F]  time = 0.694768, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[((a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m)/(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

Defer[Int][((a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m)/(A + B*Log[e*((a + b*x)/(c + d*x))^n]), x]

Rubi steps

\begin{align*} \int \frac{(221 c+221 d x)^m (a g+b g x)^{-2-m}}{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )} \, dx &=\int \frac{(221 c+221 d x)^m (a g+b g x)^{-2-m}}{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )} \, dx\\ \end{align*}

Mathematica [F]  time = 0.234732, size = 0, normalized size = 0. \[ \int \frac{(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m)/(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

Integrate[((a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m)/(A + B*Log[e*((a + b*x)/(c + d*x))^n]), x]

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Maple [F]  time = 3.039, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bgx+ag \right ) ^{-2-m} \left ( dix+ci \right ) ^{m} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

int((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b g x + a g\right )}^{-m - 2}{\left (d i x + c i\right )}^{m}}{B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="maxima")

[Out]

integrate((b*g*x + a*g)^(-m - 2)*(d*i*x + c*i)^m/(B*log(e*((b*x + a)/(d*x + c))^n) + A), x)

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Fricas [A]  time = 0.528145, size = 216, normalized size = 1.69 \begin{align*} \frac{{\rm Ei}\left (-\frac{{\left (B m + B\right )} n \log \left (\frac{b x + a}{d x + c}\right ) + A m +{\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right ) e^{\left (\frac{B m n \log \left (\frac{i}{g}\right ) + A m +{\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right )}}{{\left (B b c - B a d\right )} g^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="fricas")

[Out]

Ei(-((B*m + B)*n*log((b*x + a)/(d*x + c)) + A*m + (B*m + B)*log(e) + A)/(B*n))*e^((B*m*n*log(i/g) + A*m + (B*m
+ B)*log(e) + A)/(B*n))/((B*b*c - B*a*d)*g^2*n)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)**(-2-m)*(d*i*x+c*i)**m/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b g x + a g\right )}^{-m - 2}{\left (d i x + c i\right )}^{m}}{B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="giac")

[Out]

integrate((b*g*x + a*g)^(-m - 2)*(d*i*x + c*i)^m/(B*log(e*((b*x + a)/(d*x + c))^n) + A), x)