Optimal. Leaf size=80 \[ \frac{B \text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right )}{b g}-\frac{\log \left (-\frac{b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b g} \]
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Rubi [A] time = 0.217414, antiderivative size = 120, normalized size of antiderivative = 1.5, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2524, 12, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac{B \text{PolyLog}\left (2,-\frac{d (a+b x)}{b c-a d}\right )}{b g}+\frac{\log (a g+b g x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b g}+\frac{B \log (a g+b g x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b g}-\frac{B \log ^2(g (a+b x))}{2 b g} \]
Antiderivative was successfully verified.
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Rule 2524
Rule 12
Rule 2418
Rule 2390
Rule 2301
Rule 2394
Rule 2393
Rule 2391
Rubi steps
\begin{align*} \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{a g+b g x} \, dx &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (a g+b g x)}{e (a+b x)} \, dx}{b g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (a g+b g x)}{a+b x} \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \left (\frac{b e \log (a g+b g x)}{a+b x}-\frac{d e \log (a g+b g x)}{c+d x}\right ) \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{\log (a g+b g x)}{a+b x} \, dx}{g}+\frac{(B d) \int \frac{\log (a g+b g x)}{c+d x} \, dx}{b g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}+\frac{B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}-B \int \frac{\log \left (\frac{b g (c+d x)}{b c g-a d g}\right )}{a g+b g x} \, dx-\frac{B \operatorname{Subst}\left (\int \frac{g \log (x)}{x} \, dx,x,a g+b g x\right )}{b g^2}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}+\frac{B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}-\frac{B \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a g+b g x\right )}{b g}-\frac{B \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{d x}{b c g-a d g}\right )}{x} \, dx,x,a g+b g x\right )}{b g}\\ &=-\frac{B \log ^2(g (a+b x))}{2 b g}+\frac{\left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b g}+\frac{B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac{B \text{Li}_2\left (-\frac{d (a+b x)}{b c-a d}\right )}{b g}\\ \end{align*}
Mathematica [A] time = 0.0420439, size = 95, normalized size = 1.19 \[ \frac{2 B \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )+\log (g (a+b x)) \left (2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+B \log \left (\frac{b (c+d x)}{b c-a d}\right )+A\right )-B \log (g (a+b x))\right )}{2 b g} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.089, size = 602, normalized size = 7.5 \begin{align*}{\frac{dAa}{g \left ( ad-bc \right ) b}\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) }-{\frac{Ac}{g \left ( ad-bc \right ) }\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) }-{\frac{dAa}{g \left ( ad-bc \right ) b}\ln \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) }+{\frac{Ac}{g \left ( ad-bc \right ) }\ln \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) }+{\frac{Bad}{2\,g \left ( ad-bc \right ) b} \left ( \ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \right ) ^{2}}-{\frac{Bc}{2\,g \left ( ad-bc \right ) } \left ( \ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \right ) ^{2}}-{\frac{Bad}{g \left ( ad-bc \right ) b}{\it dilog} \left ( -{\frac{1}{be} \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) } \right ) }+{\frac{Bc}{g \left ( ad-bc \right ) }{\it dilog} \left ( -{\frac{1}{be} \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) } \right ) }-{\frac{Bad}{g \left ( ad-bc \right ) b}\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \ln \left ( -{\frac{1}{be} \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) } \right ) }+{\frac{Bc}{g \left ( ad-bc \right ) }\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \ln \left ( -{\frac{1}{be} \left ( d \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) -be \right ) } \right ) } \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*} -B{\left (\frac{\log \left (b x + a\right ) \log \left (d x + c\right )}{b g} - \int \frac{b d x \log \left (e\right ) + b c \log \left (e\right ) +{\left (2 \, b d x + b c + a d\right )} \log \left (b x + a\right )}{b^{2} d g x^{2} + a b c g +{\left (b^{2} c g + a b d g\right )} x}\,{d x}\right )} + \frac{A \log \left (b g x + a g\right )}{b g} \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{B \log \left (\frac{b e x + a e}{d x + c}\right ) + A}{b g x + a g}, 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{B \log \left (\frac{{\left (b x + a\right )} e}{d x + c}\right ) + A}{b g x + a g}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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