Optimal. Leaf size=83 \[ -\frac{2 B \text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right )}{b g}-\frac{\log \left (-\frac{b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g} \]
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Rubi [A] time = 0.291895, antiderivative size = 121, normalized size of antiderivative = 1.46, number of steps used = 10, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2524, 12, 2418, 2390, 2301, 2394, 2393, 2391} \[ -\frac{2 B \text{PolyLog}\left (2,-\frac{d (a+b x)}{b c-a d}\right )}{b g}+\frac{\log (a g+b g x) \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}-\frac{2 B \log (a g+b g x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b g}+\frac{B \log ^2(g (a+b x))}{b g} \]
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 (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx &=\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(a+b x)^2 \left (\frac{2 d e (c+d x)}{(a+b x)^2}-\frac{2 b e (c+d x)^2}{(a+b x)^3}\right ) \log (a g+b g x)}{e (c+d x)^2} \, dx}{b g}\\ &=\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(a+b x)^2 \left (\frac{2 d e (c+d x)}{(a+b x)^2}-\frac{2 b e (c+d x)^2}{(a+b x)^3}\right ) \log (a g+b g x)}{(c+d x)^2} \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \left (-\frac{2 b e \log (a g+b g x)}{a+b x}+\frac{2 d e \log (a g+b g x)}{c+d x}\right ) \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}+\frac{(2 B) \int \frac{\log (a g+b g x)}{a+b x} \, dx}{g}-\frac{(2 B d) \int \frac{\log (a g+b g x)}{c+d x} \, dx}{b g}\\ &=-\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}+(2 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{(2 B) \operatorname{Subst}\left (\int \frac{g \log (x)}{x} \, dx,x,a g+b g x\right )}{b g^2}\\ &=-\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a g+b g x\right )}{b g}+\frac{(2 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))}{b g}-\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac{\left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{2 B \text{Li}_2\left (-\frac{d (a+b x)}{b c-a d}\right )}{b g}\\ \end{align*}
Mathematica [A] time = 0.0366503, size = 87, normalized size = 1.05 \[ \frac{\log (a+b x) \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )-2 B \log \left (\frac{b (c+d x)}{b c-a d}\right )+B \log (a+b x)+A\right )-2 B \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )}{b g} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.335, size = 265, normalized size = 3.2 \begin{align*} -{\frac{A\ln \left ( \left ( bx+a \right ) ^{-1} \right ) }{bg}}-{\frac{B\ln \left ( \left ( bx+a \right ) ^{-1} \right ) }{bg}\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ) }+2\,{\frac{Bad}{bg \left ( ad-bc \right ) }{\it dilog} \left ( -{\frac{1}{d} \left ({\frac{ad-bc}{bx+a}}-d \right ) } \right ) }-2\,{\frac{Bc}{g \left ( ad-bc \right ) }{\it dilog} \left ( -{\frac{1}{d} \left ({\frac{ad-bc}{bx+a}}-d \right ) } \right ) }+2\,{\frac{B\ln \left ( \left ( bx+a \right ) ^{-1} \right ) ad}{bg \left ( ad-bc \right ) }\ln \left ( -{\frac{1}{d} \left ({\frac{ad-bc}{bx+a}}-d \right ) } \right ) }-2\,{\frac{B\ln \left ( \left ( bx+a \right ) ^{-1} \right ) c}{g \left ( ad-bc \right ) }\ln \left ( -{\frac{1}{d} \left ({\frac{ad-bc}{bx+a}}-d \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{2 \, \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 ) - 2 \,{\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{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\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 (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\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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