Optimal. Leaf size=287 \[ \frac{c^2 \text{PolyLog}(2,c x)}{8 x^2}-\frac{1}{8} c^4 \text{PolyLog}(2,c x)-\frac{1}{4} c^4 \text{PolyLog}(3,c x)-\frac{1}{2} c^4 \text{PolyLog}(3,1-c x)+\frac{c^3 \text{PolyLog}(2,c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{PolyLog}(2,c x)+\frac{1}{2} c^4 \log (1-c x) \text{PolyLog}(2,1-c x)+\frac{c \text{PolyLog}(2,c x)}{12 x^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{4 x^4}+\frac{5 c^2}{144 x^2}-\frac{c^2 \log (1-c x)}{8 x^2}+\frac{7 c^3}{36 x}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{16} c^4 \log ^2(1-c x)-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}-\frac{5 c \log (1-c x)}{72 x^3} \]
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Rubi [A] time = 0.449191, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 37, number of rules used = 17, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.062, Rules used = {6591, 2395, 44, 6603, 2398, 2410, 36, 29, 31, 2391, 2390, 2301, 6589, 6596, 2396, 2433, 2374} \[ \frac{c^2 \text{PolyLog}(2,c x)}{8 x^2}-\frac{1}{8} c^4 \text{PolyLog}(2,c x)-\frac{1}{4} c^4 \text{PolyLog}(3,c x)-\frac{1}{2} c^4 \text{PolyLog}(3,1-c x)+\frac{c^3 \text{PolyLog}(2,c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{PolyLog}(2,c x)+\frac{1}{2} c^4 \log (1-c x) \text{PolyLog}(2,1-c x)+\frac{c \text{PolyLog}(2,c x)}{12 x^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{4 x^4}+\frac{5 c^2}{144 x^2}-\frac{c^2 \log (1-c x)}{8 x^2}+\frac{7 c^3}{36 x}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{16} c^4 \log ^2(1-c x)-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}-\frac{5 c \log (1-c x)}{72 x^3} \]
Antiderivative was successfully verified.
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Rule 6591
Rule 2395
Rule 44
Rule 6603
Rule 2398
Rule 2410
Rule 36
Rule 29
Rule 31
Rule 2391
Rule 2390
Rule 2301
Rule 6589
Rule 6596
Rule 2396
Rule 2433
Rule 2374
Rubi steps
\begin{align*} \int \frac{\log (1-c x) \text{Li}_2(c x)}{x^5} \, dx &=-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} \int \frac{\log ^2(1-c x)}{x^5} \, dx-\frac{1}{4} c \int \left (\frac{\text{Li}_2(c x)}{x^4}+\frac{c \text{Li}_2(c x)}{x^3}+\frac{c^2 \text{Li}_2(c x)}{x^2}+\frac{c^3 \text{Li}_2(c x)}{x}-\frac{c^4 \text{Li}_2(c x)}{-1+c x}\right ) \, dx\\ &=\frac{\log ^2(1-c x)}{16 x^4}-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{8} c \int \frac{\log (1-c x)}{x^4 (1-c x)} \, dx-\frac{1}{4} c \int \frac{\text{Li}_2(c x)}{x^4} \, dx-\frac{1}{4} c^2 \int \frac{\text{Li}_2(c x)}{x^3} \, dx-\frac{1}{4} c^3 \int \frac{\text{Li}_2(c x)}{x^2} \, dx-\frac{1}{4} c^4 \int \frac{\text{Li}_2(c x)}{x} \, dx+\frac{1}{4} c^5 \int \frac{\text{Li}_2(c x)}{-1+c x} \, dx\\ &=\frac{\log ^2(1-c x)}{16 x^4}+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)+\frac{1}{12} c \int \frac{\log (1-c x)}{x^4} \, dx+\frac{1}{8} c \int \left (\frac{\log (1-c x)}{x^4}+\frac{c \log (1-c x)}{x^3}+\frac{c^2 \log (1-c x)}{x^2}+\frac{c^3 \log (1-c x)}{x}-\frac{c^4 \log (1-c x)}{-1+c x}\right ) \, dx+\frac{1}{8} c^2 \int \frac{\log (1-c x)}{x^3} \, dx+\frac{1}{4} c^3 \int \frac{\log (1-c x)}{x^2} \, dx+\frac{1}{4} c^4 \int \frac{\log ^2(1-c x)}{x} \, dx\\ &=-\frac{c \log (1-c x)}{36 x^3}-\frac{c^2 \log (1-c x)}{16 x^2}-\frac{c^3 \log (1-c x)}{4 x}+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)+\frac{1}{8} c \int \frac{\log (1-c x)}{x^4} \, dx-\frac{1}{36} c^2 \int \frac{1}{x^3 (1-c x)} \, dx+\frac{1}{8} c^2 \int \frac{\log (1-c x)}{x^3} \, dx-\frac{1}{16} c^3 \int \frac{1}{x^2 (1-c x)} \, dx+\frac{1}{8} c^3 \int \frac{\log (1-c x)}{x^2} \, dx+\frac{1}{8} c^4 \int \frac{\log (1-c x)}{x} \, dx-\frac{1}{4} c^4 \int \frac{1}{x (1-c x)} \, dx-\frac{1}{8} c^5 \int \frac{\log (1-c x)}{-1+c x} \, dx+\frac{1}{2} c^5 \int \frac{\log (c x) \log (1-c x)}{1-c x} \, dx\\ &=-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{36} c^2 \int \left (\frac{1}{x^3}+\frac{c}{x^2}+\frac{c^2}{x}-\frac{c^3}{-1+c x}\right ) \, dx-\frac{1}{24} c^2 \int \frac{1}{x^3 (1-c x)} \, dx-\frac{1}{16} c^3 \int \frac{1}{x^2 (1-c x)} \, dx-\frac{1}{16} c^3 \int \left (\frac{1}{x^2}+\frac{c}{x}-\frac{c^2}{-1+c x}\right ) \, dx-\frac{1}{8} c^4 \int \frac{1}{x (1-c x)} \, dx-\frac{1}{8} c^4 \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1-c x\right )-\frac{1}{4} c^4 \int \frac{1}{x} \, dx-\frac{1}{2} c^4 \operatorname{Subst}\left (\int \frac{\log (x) \log \left (c \left (\frac{1}{c}-\frac{x}{c}\right )\right )}{x} \, dx,x,1-c x\right )-\frac{1}{4} c^5 \int \frac{1}{1-c x} \, dx\\ &=\frac{c^2}{72 x^2}+\frac{13 c^3}{144 x}-\frac{49}{144} c^4 \log (x)+\frac{49}{144} c^4 \log (1-c x)-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}-\frac{1}{16} c^4 \log ^2(1-c x)+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{2} c^4 \log (1-c x) \text{Li}_2(1-c x)-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{24} c^2 \int \left (\frac{1}{x^3}+\frac{c}{x^2}+\frac{c^2}{x}-\frac{c^3}{-1+c x}\right ) \, dx-\frac{1}{16} c^3 \int \left (\frac{1}{x^2}+\frac{c}{x}-\frac{c^2}{-1+c x}\right ) \, dx-\frac{1}{8} c^4 \int \frac{1}{x} \, dx-\frac{1}{2} c^4 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-c x\right )-\frac{1}{8} c^5 \int \frac{1}{1-c x} \, dx\\ &=\frac{5 c^2}{144 x^2}+\frac{7 c^3}{36 x}-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}-\frac{1}{16} c^4 \log ^2(1-c x)+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{2} c^4 \log (1-c x) \text{Li}_2(1-c x)-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{2} c^4 \text{Li}_3(1-c x)\\ \end{align*}
Mathematica [A] time = 0.235598, size = 277, normalized size = 0.97 \[ \frac{-36 c^4 x^4 \text{PolyLog}(3,c x)-72 c^4 x^4 \text{PolyLog}(3,1-c x)+18 c^4 x^4 (4 \log (1-c x)+1) \text{PolyLog}(2,1-c x)+6 \left (c x \left (6 c^2 x^2+3 c x+2\right )+6 \left (c^4 x^4-1\right ) \log (1-c x)\right ) \text{PolyLog}(2,c x)-18 c^4 x^4+28 c^3 x^3+5 c^2 x^2-9 c^4 x^4 \log ^2(1-c x)+36 c^4 x^4 \log (c x) \log ^2(1-c x)-49 c^4 x^4 \log (x)-33 c^4 x^4 \log (c x)+82 c^4 x^4 \log (1-c x)+18 c^4 x^4 \log (c x) \log (1-c x)-54 c^3 x^3 \log (1-c x)-18 c^2 x^2 \log (1-c x)+9 \log ^2(1-c x)-10 c x \log (1-c x)}{144 x^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( -cx+1 \right ){\it polylog} \left ( 2,cx \right ) }{{x}^{5}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.73984, size = 289, normalized size = 1.01 \begin{align*} \frac{1}{4} \,{\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \,{\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \,{\rm Li}_{3}(-c x + 1)\right )} c^{4} + \frac{1}{8} \,{\left (\log \left (c x\right ) \log \left (-c x + 1\right ) +{\rm Li}_2\left (-c x + 1\right )\right )} c^{4} - \frac{41}{72} \, c^{4} \log \left (x\right ) - \frac{1}{4} \, c^{4}{\rm Li}_{3}(c x) + \frac{28 \, c^{3} x^{3} + 5 \, c^{2} x^{2} - 9 \,{\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )^{2} + 6 \,{\left (6 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + 2 \, c x + 6 \,{\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) + 2 \,{\left (41 \, c^{4} x^{4} - 27 \, c^{3} x^{3} - 9 \, c^{2} x^{2} - 5 \, c x\right )} \log \left (-c x + 1\right )}{144 \, x^{4}} \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 (c x\right ) \log \left (-c x + 1\right )}{x^{5}}, 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 (c x\right ) \log \left (-c x + 1\right )}{x^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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