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

Optimal. Leaf size=37 \[ \text{Unintegrable}\left (\frac{1}{(a g+b g x) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.0908582, 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 = {} \[ \int \frac{1}{(a g+b g x) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.530636, size = 0, normalized size = 0. \[ \int \frac{1}{(a g+b g x) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.437, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{bgx+ag} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) ^{-2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} d \int \frac{1}{{\left (b c g n - a d g n\right )} B^{2} \log \left ({\left (b x + a\right )}^{n}\right ) -{\left (b c g n - a d g n\right )} B^{2} \log \left ({\left (d x + c\right )}^{n}\right ) +{\left (b c g n - a d g n\right )} A B +{\left (b c g n \log \left (e\right ) - a d g n \log \left (e\right )\right )} B^{2}}\,{d x} - \frac{d x + c}{{\left (b c g n - a d g n\right )} B^{2} \log \left ({\left (b x + a\right )}^{n}\right ) -{\left (b c g n - a d g n\right )} B^{2} \log \left ({\left (d x + c\right )}^{n}\right ) +{\left (b c g n - a d g n\right )} A B +{\left (b c g n \log \left (e\right ) - a d g n \log \left (e\right )\right )} B^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{A^{2} b g x + A^{2} a g +{\left (B^{2} b g x + B^{2} a g\right )} \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} + 2 \,{\left (A B b g x + A B a g\right )} \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(A+B*ln(e*((b*x+a)/(d*x+c))**n))**2,x)

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b g x + a g\right )}{\left (B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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