Optimal. Leaf size=173 \[ \frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 a^2}+\frac{b^2 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{2 a^2}-\frac{\text{PolyLog}(2,c (a+b x))}{2 x^2}+\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}+\frac{b^2 c \log (x)}{2 a (1-a c)}-\frac{b^2 c \log (-a c-b c x+1)}{2 a (1-a c)}+\frac{b \log (-a c-b c x+1)}{2 a x} \]
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Rubi [A] time = 0.177859, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.846, Rules used = {6598, 44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ \frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 a^2}+\frac{b^2 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{2 a^2}-\frac{\text{PolyLog}(2,c (a+b x))}{2 x^2}+\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}+\frac{b^2 c \log (x)}{2 a (1-a c)}-\frac{b^2 c \log (-a c-b c x+1)}{2 a (1-a c)}+\frac{b \log (-a c-b c x+1)}{2 a x} \]
Antiderivative was successfully verified.
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Rule 6598
Rule 44
Rule 2416
Rule 2395
Rule 36
Rule 29
Rule 31
Rule 2394
Rule 2315
Rule 2393
Rule 2391
Rubi steps
\begin{align*} \int \frac{\text{Li}_2(c (a+b x))}{x^3} \, dx &=-\frac{\text{Li}_2(c (a+b x))}{2 x^2}-\frac{1}{2} b \int \frac{\log (1-a c-b c x)}{x^2 (a+b x)} \, dx\\ &=-\frac{\text{Li}_2(c (a+b x))}{2 x^2}-\frac{1}{2} b \int \left (\frac{\log (1-a c-b c x)}{a x^2}-\frac{b \log (1-a c-b c x)}{a^2 x}+\frac{b^2 \log (1-a c-b c x)}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac{\text{Li}_2(c (a+b x))}{2 x^2}-\frac{b \int \frac{\log (1-a c-b c x)}{x^2} \, dx}{2 a}+\frac{b^2 \int \frac{\log (1-a c-b c x)}{x} \, dx}{2 a^2}-\frac{b^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=\frac{b \log (1-a c-b c x)}{2 a x}+\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac{\text{Li}_2(c (a+b x))}{2 x^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 a^2}+\frac{\left (b^2 c\right ) \int \frac{1}{x (1-a c-b c x)} \, dx}{2 a}+\frac{\left (b^3 c\right ) \int \frac{\log \left (-\frac{b c x}{-1+a c}\right )}{1-a c-b c x} \, dx}{2 a^2}\\ &=\frac{b \log (1-a c-b c x)}{2 a x}+\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}+\frac{b^2 \text{Li}_2(c (a+b x))}{2 a^2}-\frac{\text{Li}_2(c (a+b x))}{2 x^2}+\frac{b^2 \text{Li}_2\left (1-\frac{b c x}{1-a c}\right )}{2 a^2}+\frac{\left (b^2 c\right ) \int \frac{1}{x} \, dx}{2 a (1-a c)}+\frac{\left (b^3 c^2\right ) \int \frac{1}{1-a c-b c x} \, dx}{2 a (1-a c)}\\ &=\frac{b^2 c \log (x)}{2 a (1-a c)}-\frac{b^2 c \log (1-a c-b c x)}{2 a (1-a c)}+\frac{b \log (1-a c-b c x)}{2 a x}+\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}+\frac{b^2 \text{Li}_2(c (a+b x))}{2 a^2}-\frac{\text{Li}_2(c (a+b x))}{2 x^2}+\frac{b^2 \text{Li}_2\left (1-\frac{b c x}{1-a c}\right )}{2 a^2}\\ \end{align*}
Mathematica [A] time = 0.180152, size = 131, normalized size = 0.76 \[ \frac{b x \left (b x (a c-1) \text{PolyLog}\left (2,\frac{a c+b c x-1}{a c-1}\right )-a b c x \log (x)+\left (a (a c+b c x-1)+b x (a c-1) \log \left (\frac{b c x}{1-a c}\right )\right ) \log (-a c-b c x+1)\right )-(a c-1) \left (a^2-b^2 x^2\right ) \text{PolyLog}(2,c (a+b x))}{2 a^2 x^2 (a c-1)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.227, size = 195, normalized size = 1.1 \begin{align*} -{\frac{{\it polylog} \left ( 2,xbc+ac \right ) }{2\,{x}^{2}}}+{\frac{{b}^{2}\ln \left ( -xbc-ac+1 \right ) }{2\,{a}^{2}}\ln \left ( -{\frac{xbc}{ac-1}} \right ) }+{\frac{{b}^{2}}{2\,{a}^{2}}{\it dilog} \left ( -{\frac{xbc}{ac-1}} \right ) }+{\frac{{b}^{2}{\it dilog} \left ( -xbc-ac+1 \right ) }{2\,{a}^{2}}}-{\frac{{b}^{2}c\ln \left ( -xbc \right ) }{2\,a \left ( ac-1 \right ) }}+{\frac{{b}^{2}c\ln \left ( -xbc-ac+1 \right ) }{2\,a \left ( ac-1 \right ) }}+{\frac{bc\ln \left ( -xbc-ac+1 \right ) }{ \left ( 2\,ac-2 \right ) x}}-{\frac{b\ln \left ( -xbc-ac+1 \right ) }{2\,a \left ( ac-1 \right ) x}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.994426, size = 261, normalized size = 1.51 \begin{align*} -\frac{b^{2} c \log \left (x\right )}{2 \,{\left (a^{2} c - a\right )}} - \frac{{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) +{\rm Li}_2\left (-b c x - a c + 1\right )\right )} b^{2}}{2 \, a^{2}} + \frac{{\left (\log \left (-b c x - a c + 1\right ) \log \left (-\frac{b c x + a c - 1}{a c - 1} + 1\right ) +{\rm Li}_2\left (\frac{b c x + a c - 1}{a c - 1}\right )\right )} b^{2}}{2 \, a^{2}} - \frac{{\left (a^{2} c - a\right )}{\rm Li}_2\left (b c x + a c\right ) -{\left (b^{2} c x^{2} +{\left (a b c - b\right )} x\right )} \log \left (-b c x - a c + 1\right )}{2 \,{\left (a^{2} c - a\right )} x^{2}} \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{{\rm Li}_2\left (b c x + a c\right )}{x^{3}}, 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{{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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