Optimal. Leaf size=160 \[ -\frac{\text{PolyLog}\left (2,1-\frac{b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}+\frac{2 n \text{PolyLog}\left (3,1-\frac{b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}-\frac{2 n^2 \text{PolyLog}\left (4,1-\frac{b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]
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Rubi [A] time = 0.250151, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 58, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.052, Rules used = {2506, 2508, 6610} \[ -\frac{\text{PolyLog}\left (2,1-\frac{b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}+\frac{2 n \text{PolyLog}\left (3,1-\frac{b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}-\frac{2 n^2 \text{PolyLog}\left (4,1-\frac{b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]
Antiderivative was successfully verified.
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Rule 2506
Rule 2508
Rule 6610
Rubi steps
\begin{align*} \int \frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \log \left (\frac{b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{(2 n) \int \frac{\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g}\\ &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{2 n \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac{\left (2 n^2\right ) \int \frac{\text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g}\\ &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{2 n \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac{2 n^2 \text{Li}_4\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}\\ \end{align*}
Mathematica [B] time = 0.455601, size = 559, normalized size = 3.49 \[ \frac{3 n \left (-2 \text{PolyLog}\left (3,\frac{d (a+b x)}{b (c+d x)}\right )+2 \log \left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )+\log \left (\frac{b c-a d}{b c+b d x}\right ) \log ^2\left (\frac{a+b x}{c+d x}\right )\right ) \left (n \log \left (\frac{a+b x}{c+d x}\right )-\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )+\frac{3}{2} \left (2 \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )-2 \log \left (\frac{a}{b}+x\right ) \log (c+d x)+2 \log (c+d x) \log \left (\frac{a+b x}{c+d x}\right )+2 \log \left (\frac{a}{b}+x\right ) \log \left (\frac{b (c+d x)}{b c-a d}\right )-\log ^2\left (\frac{c}{d}+x\right )+2 \log (c+d x) \log \left (\frac{c}{d}+x\right )\right ) \left (\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-n \log \left (\frac{a+b x}{c+d x}\right )\right )^2+n^2 \left (-\left (6 \text{PolyLog}\left (4,\frac{d (a+b x)}{b (c+d x)}\right )+3 \log ^2\left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )-6 \log \left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (3,\frac{d (a+b x)}{b (c+d x)}\right )+\log \left (\frac{b c-a d}{b c+b d x}\right ) \log ^3\left (\frac{a+b x}{c+d x}\right )\right )\right )+\log \left (\frac{a+b x}{c+d x}\right ) \log \left (\frac{b c-a d}{b c+b d x}\right ) \left (3 \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-3 n \log \left (\frac{a+b x}{c+d x}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+n^2 \log ^2\left (\frac{a+b x}{c+d x}\right )\right )}{3 g (b c-a d)} \]
Antiderivative was successfully verified.
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Maple [F] time = 5.443, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( dx+c \right ) \left ( bgx+ag \right ) } \left ( \ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) ^{2}\ln \left ({\frac{-ad+bc}{b \left ( dx+c \right ) }} \right ) }\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )}{\left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{b d x + b c}\right )}{b d g x^{2} + a c g +{\left (b c + a d\right )} g x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )}{\left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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