Optimal. Leaf size=132 \[ -x \text {Li}_2(c x)+\frac {2 \text {Li}_3(1-c x)}{c}+x \text {Li}_2(c x) \log (1-c x)-\frac {\text {Li}_2(c x) \log (1-c x)}{c}-\frac {2 \text {Li}_2(1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}+\frac {3 (1-c x) \log (1-c x)}{c}+3 x \]
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Rubi [A] time = 0.21, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {6586, 2389, 2295, 6600, 2296, 6688, 6742, 6596, 2396, 2433, 2374, 6589} \[ -x \text {PolyLog}(2,c x)+\frac {2 \text {PolyLog}(3,1-c x)}{c}+x \log (1-c x) \text {PolyLog}(2,c x)-\frac {\log (1-c x) \text {PolyLog}(2,c x)}{c}-\frac {2 \log (1-c x) \text {PolyLog}(2,1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}+\frac {3 (1-c x) \log (1-c x)}{c}+3 x \]
Antiderivative was successfully verified.
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Rule 2295
Rule 2296
Rule 2374
Rule 2389
Rule 2396
Rule 2433
Rule 6586
Rule 6589
Rule 6596
Rule 6600
Rule 6688
Rule 6742
Rubi steps
\begin {align*} \int \log (1-c x) \text {Li}_2(c x) \, dx &=x \log (1-c x) \text {Li}_2(c x)+c \int \left (-\frac {1}{c}-\frac {1}{c (-1+c x)}\right ) \text {Li}_2(c x) \, dx+\int \log ^2(1-c x) \, dx\\ &=x \log (1-c x) \text {Li}_2(c x)-\frac {\operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{c}+c \int \frac {x \text {Li}_2(c x)}{1-c x} \, dx\\ &=-\frac {(1-c x) \log ^2(1-c x)}{c}+x \log (1-c x) \text {Li}_2(c x)+\frac {2 \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c}+c \int \left (-\frac {\text {Li}_2(c x)}{c}-\frac {\text {Li}_2(c x)}{c (-1+c x)}\right ) \, dx\\ &=2 x+\frac {2 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}+x \log (1-c x) \text {Li}_2(c x)-\int \text {Li}_2(c x) \, dx-\int \frac {\text {Li}_2(c x)}{-1+c x} \, dx\\ &=2 x+\frac {2 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-x \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{c}+x \log (1-c x) \text {Li}_2(c x)-\frac {\int \frac {\log ^2(1-c x)}{x} \, dx}{c}-\int \log (1-c x) \, dx\\ &=2 x+\frac {2 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}-x \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{c}+x \log (1-c x) \text {Li}_2(c x)-2 \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx+\frac {\operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c}\\ &=3 x+\frac {3 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}-x \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{c}+x \log (1-c x) \text {Li}_2(c x)+\frac {2 \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 )}{c}\\ &=3 x+\frac {3 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}-x \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{c}+x \log (1-c x) \text {Li}_2(c x)-\frac {2 \log (1-c x) \text {Li}_2(1-c x)}{c}+\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{c}\\ &=3 x+\frac {3 (1-c x) \log (1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}-\frac {\log (c x) \log ^2(1-c x)}{c}-x \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{c}+x \log (1-c x) \text {Li}_2(c x)-\frac {2 \log (1-c x) \text {Li}_2(1-c x)}{c}+\frac {2 \text {Li}_3(1-c x)}{c}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 119, normalized size = 0.90 \[ \frac {2 \text {Li}_3(1-c x)-2 \text {Li}_2(1-c x) \log (1-c x)+\text {Li}_2(c x) ((c x-1) \log (1-c x)-c x)+3 c x+c x \log ^2(1-c x)-\log (c x) \log ^2(1-c x)-\log ^2(1-c x)-3 c x \log (1-c x)+3 \log (1-c x)-2}{c} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ), x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int \ln \left (-c x +1\right ) \polylog \left (2, c x \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 141, normalized size = 1.07 \[ c {\left (\frac {x}{c} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} + \frac {{\left (c x {\rm Li}_2\left (c x\right ) - c x + {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{c} - \frac {\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)}{c} + \frac {2 \, c x - {\left (c x + \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 2 \, {\left (c x - 1\right )} \log \left (-c x + 1\right )}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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