3.22 \(\int \frac{1}{(a g+b g x)^2 (A+B \log (e (\frac{a+b x}{c+d x})^n))} \, dx\)

Optimal. Leaf size=94 \[ \frac{e^{\frac{A}{B n}} (c+d x) \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{1}{n}} \text{Ei}\left (-\frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (a+b x) (b c-a d)} \]

[Out]

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

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Rubi [F]  time = 0.0990798, 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 = {} \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[1/((a*g + b*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

Defer[Int][1/((a*g + b*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])), x]

Rubi steps

\begin{align*} \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx &=\int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx\\ \end{align*}

Mathematica [A]  time = 0.131033, size = 94, normalized size = 1. \[ \frac{e^{\frac{A}{B n}} (c+d x) \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{1}{n}} \text{Ei}\left (-\frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a*g + b*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

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

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Maple [F]  time = 0.444, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{2}} \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(1/(b*g*x+a*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

int(1/(b*g*x+a*g)^2/(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{1}{{\left (b g x + a g\right )}^{2}{\left (B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="maxima")

[Out]

integrate(1/((b*g*x + a*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)

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Fricas [A]  time = 0.867845, size = 149, normalized size = 1.59 \begin{align*} \frac{e^{\left (\frac{B \log \left (e\right ) + A}{B n}\right )} \logintegral \left (\frac{{\left (d x + c\right )} e^{\left (-\frac{B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\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(1/(b*g*x+a*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="fricas")

[Out]

e^((B*log(e) + A)/(B*n))*log_integral((d*x + c)*e^(-(B*log(e) + A)/(B*n))/(b*x + a))/((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(1/(b*g*x+a*g)**2/(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{1}{{\left (b g x + a g\right )}^{2}{\left (B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="giac")

[Out]

integrate(1/((b*g*x + a*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)