Optimal. Leaf size=153 \[ -\frac {1}{2} a \text {Li}_2(c x){}^2-b x \text {Li}_2(c x)+\frac {2 b \text {Li}_3(1-c x)}{c}+b x \text {Li}_2(c x) \log (1-c x)-\frac {b \text {Li}_2(c x) \log (1-c x)}{c}-\frac {2 b \text {Li}_2(1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}+\frac {3 b (1-c x) \log (1-c x)}{c}+3 b x \]
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Rubi [A] time = 0.34, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6742, 6586, 2389, 2295, 6589, 6605, 6601, 12, 6600, 2296, 6688, 6596, 2396, 2433, 2374} \[ -\frac {1}{2} a \text {PolyLog}(2,c x)^2-b x \text {PolyLog}(2,c x)+\frac {2 b \text {PolyLog}(3,1-c x)}{c}+b x \log (1-c x) \text {PolyLog}(2,c x)-\frac {b \log (1-c x) \text {PolyLog}(2,c x)}{c}-\frac {2 b \log (1-c x) \text {PolyLog}(2,1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}+\frac {3 b (1-c x) \log (1-c x)}{c}+3 b x \]
Antiderivative was successfully verified.
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Rule 12
Rule 2295
Rule 2296
Rule 2374
Rule 2389
Rule 2396
Rule 2433
Rule 6586
Rule 6589
Rule 6596
Rule 6600
Rule 6601
Rule 6605
Rule 6688
Rule 6742
Rubi steps
\begin {align*} \int \frac {(a+b x) \log (1-c x) \text {Li}_2(c x)}{x} \, dx &=a \int \frac {\log (1-c x) \text {Li}_2(c x)}{x} \, dx+\int b \log (1-c x) \text {Li}_2(c x) \, dx\\ &=-\frac {1}{2} a \text {Li}_2(c x){}^2+b \int \log (1-c x) \text {Li}_2(c x) \, dx\\ &=b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2+b \int \log ^2(1-c x) \, dx+(b c) \int \left (-\frac {1}{c}-\frac {1}{c (-1+c x)}\right ) \text {Li}_2(c x) \, dx\\ &=b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-\frac {b \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{c}+(b c) \int \frac {x \text {Li}_2(c x)}{1-c x} \, dx\\ &=-\frac {b (1-c x) \log ^2(1-c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2+\frac {(2 b) \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c}+(b c) \int \left (-\frac {\text {Li}_2(c x)}{c}-\frac {\text {Li}_2(c x)}{c (-1+c x)}\right ) \, dx\\ &=2 b x+\frac {2 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-b \int \text {Li}_2(c x) \, dx-b \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx\\ &=2 b x+\frac {2 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-b x \text {Li}_2(c x)-\frac {b \log (1-c x) \text {Li}_2(c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-b \int \log (1-c x) \, dx-\frac {b \int \frac {\log ^2(1-c x)}{x} \, dx}{c}\\ &=2 b x+\frac {2 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}-b x \text {Li}_2(c x)-\frac {b \log (1-c x) \text {Li}_2(c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-(2 b) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx+\frac {b \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c}\\ &=3 b x+\frac {3 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}-b x \text {Li}_2(c x)-\frac {b \log (1-c x) \text {Li}_2(c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2+\frac {(2 b) \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 b x+\frac {3 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}-b x \text {Li}_2(c x)-\frac {b \log (1-c x) \text {Li}_2(c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-\frac {2 b \log (1-c x) \text {Li}_2(1-c x)}{c}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{c}\\ &=3 b x+\frac {3 b (1-c x) \log (1-c x)}{c}-\frac {b (1-c x) \log ^2(1-c x)}{c}-\frac {b \log (c x) \log ^2(1-c x)}{c}-b x \text {Li}_2(c x)-\frac {b \log (1-c x) \text {Li}_2(c x)}{c}+b x \log (1-c x) \text {Li}_2(c x)-\frac {1}{2} a \text {Li}_2(c x){}^2-\frac {2 b \log (1-c x) \text {Li}_2(1-c x)}{c}+\frac {2 b \text {Li}_3(1-c x)}{c}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 137, normalized size = 0.90 \[ -\frac {1}{2} a \text {Li}_2(c x){}^2+\frac {b \left (2 \text {Li}_3(1-c x)-2 \text {Li}_2(1-c x) \log (1-c x)+3 c x+c x \log ^2(1-c x)-\log (c x) \log ^2(1-c x)-\log ^2(1-c x)-3 c x \log (1-c x)+3 \log (1-c x)-2\right )}{c}+\frac {b \text {Li}_2(c x) ((c x-1) \log (1-c x)-c x)}{c} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x}, x\right ) \]
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 \[ \int \frac {{\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x}\,{d x} \]
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 \[ \int \frac {\left (b x +a \right ) \ln \left (-c x +1\right ) \polylog \left (2, c x \right )}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x}\,{d x} \]
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 \[ \int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right )}{x} \,d x \]
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 \[ \int \frac {\left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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