Optimal. Leaf size=138 \[ \frac{b \text{PolyLog}(2,c (a+b x))}{e (b d-a e)}-\frac{\text{PolyLog}(2,c (a+b x))}{e (d+e x)}+\frac{b \text{PolyLog}\left (2,\frac{e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{e (b d-a e)}+\frac{b \log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )}{e (b d-a e)} \]
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Rubi [A] time = 0.18397, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6598, 2418, 2393, 2391, 2394} \[ \frac{b \text{PolyLog}(2,c (a+b x))}{e (b d-a e)}-\frac{\text{PolyLog}(2,c (a+b x))}{e (d+e x)}+\frac{b \text{PolyLog}\left (2,\frac{e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{e (b d-a e)}+\frac{b \log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )}{e (b d-a e)} \]
Antiderivative was successfully verified.
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Rule 6598
Rule 2418
Rule 2393
Rule 2391
Rule 2394
Rubi steps
\begin{align*} \int \frac{\text{Li}_2(c (a+b x))}{(d+e x)^2} \, dx &=-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}-\frac{b \int \frac{\log (1-a c-b c x)}{(a+b x) (d+e x)} \, dx}{e}\\ &=-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}-\frac{b \int \left (\frac{b \log (1-a c-b c x)}{(b d-a e) (a+b x)}-\frac{e \log (1-a c-b c x)}{(b d-a e) (d+e x)}\right ) \, dx}{e}\\ &=-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}+\frac{b \int \frac{\log (1-a c-b c x)}{d+e x} \, dx}{b d-a e}-\frac{b^2 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{e (b d-a e)}\\ &=\frac{b \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}-\frac{b \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{e (b d-a e)}+\frac{\left (b^2 c\right ) \int \frac{\log \left (-\frac{b c (d+e x)}{-b c d-(1-a c) e}\right )}{1-a c-b c x} \, dx}{e (b d-a e)}\\ &=\frac{b \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}+\frac{b \text{Li}_2(c (a+b x))}{e (b d-a e)}-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}-\frac{b \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e x}{-b c d-(1-a c) e}\right )}{x} \, dx,x,1-a c-b c x\right )}{e (b d-a e)}\\ &=\frac{b \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}+\frac{b \text{Li}_2(c (a+b x))}{e (b d-a e)}-\frac{\text{Li}_2(c (a+b x))}{e (d+e x)}+\frac{b \text{Li}_2\left (\frac{e (1-a c-b c x)}{b c d+e-a c e}\right )}{e (b d-a e)}\\ \end{align*}
Mathematica [A] time = 0.140869, size = 108, normalized size = 0.78 \[ \frac{\frac{b \left (\text{PolyLog}\left (2,\frac{e (a c+b c x-1)}{a c e-b c d-e}\right )+\text{PolyLog}(2,c (a+b x))+\log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )\right )}{b d-a e}-\frac{\text{PolyLog}(2,c (a+b x))}{d+e x}}{e} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.38, size = 189, normalized size = 1.4 \begin{align*} -{\frac{bc{\it polylog} \left ( 2,xbc+ac \right ) }{ \left ( bcex+bcd \right ) e}}-{\frac{b}{e \left ( ae-bd \right ) }{\it dilog} \left ({\frac{ace-bcd+ \left ( -xbc-ac+1 \right ) e-e}{ace-bcd-e}} \right ) }-{\frac{b\ln \left ( -xbc-ac+1 \right ) }{e \left ( ae-bd \right ) }\ln \left ({\frac{ace-bcd+ \left ( -xbc-ac+1 \right ) e-e}{ace-bcd-e}} \right ) }-{\frac{b{\it dilog} \left ( -xbc-ac+1 \right ) }{e \left ( ae-bd \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03118, size = 224, normalized size = 1.62 \begin{align*} -\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}{b d e - a e^{2}} + \frac{{\left (\log \left (-b c x - a c + 1\right ) \log \left (\frac{b c e x +{\left (a c - 1\right )} e}{b c d -{\left (a c - 1\right )} e} + 1\right ) +{\rm Li}_2\left (-\frac{b c e x +{\left (a c - 1\right )} e}{b c d -{\left (a c - 1\right )} e}\right )\right )} b}{b d e - a e^{2}} - \frac{{\rm Li}_2\left (b c x + a c\right )}{e^{2} x + d e} \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 )}{e^{2} x^{2} + 2 \, d e x + d^{2}}, 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 )}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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