Optimal. Leaf size=390 \[ \frac {(2 a c+b) \text {Li}_3(1-c x)}{c^2}-\frac {(2 a c+b) \text {Li}_2(c x) \log (1-c x)}{2 c^2}-\frac {(2 a c+b) \text {Li}_2(1-c x) \log (1-c x)}{c^2}-\frac {(2 a c+b) \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {(1-c x) (2 a c+b) \log (1-c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \text {Li}_2(c x) \log (1-c x)-\frac {x (2 a c+b) \text {Li}_2(c x)}{2 c}+\frac {x (2 a c+b)}{2 c}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {2 a (1-c x) \log (1-c x)}{c}+2 a x+\frac {b (1-c x)^2}{8 c^2}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {1}{8} b x^2 \log (1-c x)+\frac {9 b x}{8 c}+\frac {b x^2}{16} \]
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Rubi [A] time = 0.45, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 20, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.111, Rules used = {6598, 43, 2416, 2389, 2295, 2391, 2395, 6604, 2296, 2401, 2390, 2305, 2304, 6586, 6591, 6596, 2396, 2433, 2374, 6589} \[ \frac {(2 a c+b) \text {PolyLog}(3,1-c x)}{c^2}-\frac {(2 a c+b) \log (1-c x) \text {PolyLog}(2,c x)}{2 c^2}-\frac {(2 a c+b) \log (1-c x) \text {PolyLog}(2,1-c x)}{c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {PolyLog}(2,c x)-\frac {x (2 a c+b) \text {PolyLog}(2,c x)}{2 c}-\frac {1}{4} b x^2 \text {PolyLog}(2,c x)-\frac {(2 a c+b) \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {(1-c x) (2 a c+b) \log (1-c x)}{2 c^2}+\frac {x (2 a c+b)}{2 c}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {2 a (1-c x) \log (1-c x)}{c}+2 a x+\frac {b (1-c x)^2}{8 c^2}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {9 b x}{8 c}+\frac {b x^2}{16} \]
Antiderivative was successfully verified.
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Rule 43
Rule 2295
Rule 2296
Rule 2304
Rule 2305
Rule 2374
Rule 2389
Rule 2390
Rule 2391
Rule 2395
Rule 2396
Rule 2401
Rule 2416
Rule 2433
Rule 6586
Rule 6589
Rule 6591
Rule 6596
Rule 6598
Rule 6604
Rubi steps
\begin {align*} \int (a+b x) \log (1-c x) \text {Li}_2(c x) \, dx &=\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+c \int \left (\frac {(-b-2 a c) \text {Li}_2(c x)}{2 c^2}-\frac {b x \text {Li}_2(c x)}{2 c}+\frac {(-b-2 a c) \text {Li}_2(c x)}{2 c^2 (-1+c x)}\right ) \, dx+\int \left (a \log ^2(1-c x)+\frac {1}{2} b x \log ^2(1-c x)\right ) \, dx\\ &=\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+a \int \log ^2(1-c x) \, dx+\frac {1}{2} b \int x \log ^2(1-c x) \, dx-\frac {1}{2} b \int x \text {Li}_2(c x) \, dx-\frac {(b+2 a c) \int \text {Li}_2(c x) \, dx}{2 c}-\frac {(b+2 a c) \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{2 c}\\ &=-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {1}{4} b \int x \log (1-c x) \, dx+\frac {1}{2} b \int \left (\frac {\log ^2(1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}\right ) \, dx-\frac {a \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{c}-\frac {(b+2 a c) \int \frac {\log ^2(1-c x)}{x} \, dx}{2 c^2}-\frac {(b+2 a c) \int \log (1-c x) \, dx}{2 c}\\ &=-\frac {1}{8} b x^2 \log (1-c x)-\frac {a (1-c x) \log ^2(1-c x)}{c}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+\frac {(2 a) \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c}+\frac {b \int \log ^2(1-c x) \, dx}{2 c}-\frac {b \int (1-c x) \log ^2(1-c x) \, dx}{2 c}-\frac {1}{8} (b c) \int \frac {x^2}{1-c x} \, dx+\frac {(b+2 a c) \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{2 c^2}-\frac {(b+2 a c) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{c}\\ &=2 a x+\frac {(b+2 a c) x}{2 c}-\frac {1}{8} b x^2 \log (1-c x)+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {b \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}+\frac {b \operatorname {Subst}\left (\int x \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}-\frac {1}{8} (b c) \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx+\frac {(b+2 a c) \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^2}\\ &=2 a x+\frac {b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(1-c x)}{c^2}-\frac {b \operatorname {Subst}(\int x \log (x) \, dx,x,1-c x)}{2 c^2}+\frac {b \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{c^2}+\frac {(b+2 a c) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{c^2}\\ &=2 a x+\frac {9 b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b (1-c x)^2}{8 c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(1-c x)}{c^2}+\frac {(b+2 a c) \text {Li}_3(1-c x)}{c^2}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 285, normalized size = 0.73 \[ \frac {4 \text {Li}_2(c x) (2 (c x-1) \log (1-c x) (2 a c+b c x+b)-c x (4 a c+b c x+2 b))-16 (2 a c+b) \text {Li}_2(1-c x) \log (1-c x)+48 a c^2 x+16 a c^2 x \log ^2(1-c x)-48 a c^2 x \log (1-c x)+32 a c \text {Li}_3(1-c x)-16 a c \log ^2(1-c x)-16 a c \log (c x) \log ^2(1-c x)+48 a c \log (1-c x)-32 a c+3 b c^2 x^2+4 b c^2 x^2 \log ^2(1-c x)-6 b c^2 x^2 \log (1-c x)+16 b \text {Li}_3(1-c x)+22 b c x-4 b \log ^2(1-c x)-8 b \log (c x) \log ^2(1-c x)-16 b c x \log (1-c x)+22 b \log (1-c x)-14 b}{16 c^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ), x\right ) \]
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 \[ \int {\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )\,{d x} \]
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 \[ \int \left (b x +a \right ) \ln \left (-c x +1\right ) \polylog \left (2, c x \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 258, normalized size = 0.66 \[ -\frac {1}{16} \, c {\left (\frac {8 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} {\left (2 \, a c + b\right )}}{c^{3}} - \frac {3 \, b c^{2} x^{2} + 2 \, {\left (24 \, a c^{2} + 11 \, b c\right )} x - 4 \, {\left (b c^{2} x^{2} + 2 \, {\left (2 \, a c^{2} + b c\right )} x + 2 \, {\left (2 \, a c + b\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 2 \, {\left (2 \, b c^{2} x^{2} - 24 \, a c + 2 \, {\left (8 \, a c^{2} + 3 \, b c\right )} x - 11 \, b\right )} \log \left (-c x + 1\right )}{c^{3}}\right )} + \frac {1}{8} \, {\left (\frac {8 \, {\left (c x {\rm Li}_2\left (c x\right ) - c x + {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} a}{c} + \frac {{\left (4 \, c^{2} x^{2} {\rm Li}_2\left (c x\right ) - c^{2} x^{2} - 2 \, c x + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{2}}\right )} \log \left (-c x + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right ) \,d x \]
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 \[ \int \left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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