Optimal. Leaf size=152 \[ -\frac {a^2 \text {Li}_2(c (a+b x))}{2 b^2}-\frac {(1-a c)^2 \log (-a c-b c x+1)}{4 b^2 c^2}+\frac {a (-a c-b c x+1) \log (-a c-b c x+1)}{2 b^2 c}+\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {1}{4} x^2 \log (-a c-b c x+1)-\frac {x (1-a c)}{4 b c}+\frac {a x}{2 b}-\frac {x^2}{8} \]
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Rubi [A] time = 0.17, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {6598, 43, 2416, 2389, 2295, 2395, 2393, 2391} \[ -\frac {a^2 \text {PolyLog}(2,c (a+b x))}{2 b^2}+\frac {1}{2} x^2 \text {PolyLog}(2,c (a+b x))-\frac {(1-a c)^2 \log (-a c-b c x+1)}{4 b^2 c^2}+\frac {a (-a c-b c x+1) \log (-a c-b c x+1)}{2 b^2 c}+\frac {1}{4} x^2 \log (-a c-b c x+1)-\frac {x (1-a c)}{4 b c}+\frac {a x}{2 b}-\frac {x^2}{8} \]
Antiderivative was successfully verified.
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Rule 43
Rule 2295
Rule 2389
Rule 2391
Rule 2393
Rule 2395
Rule 2416
Rule 6598
Rubi steps
\begin {align*} \int x \text {Li}_2(c (a+b x)) \, dx &=\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {1}{2} b \int \frac {x^2 \log (1-a c-b c x)}{a+b x} \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {1}{2} b \int \left (-\frac {a \log (1-a c-b c x)}{b^2}+\frac {x \log (1-a c-b c x)}{b}+\frac {a^2 \log (1-a c-b c x)}{b^2 (a+b x)}\right ) \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {1}{2} \int x \log (1-a c-b c x) \, dx-\frac {a \int \log (1-a c-b c x) \, dx}{2 b}+\frac {a^2 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 b}\\ &=\frac {1}{4} x^2 \log (1-a c-b c x)+\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {a^2 \operatorname {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 b^2}+\frac {a \operatorname {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{2 b^2 c}+\frac {1}{4} (b c) \int \frac {x^2}{1-a c-b c x} \, dx\\ &=\frac {a x}{2 b}+\frac {1}{4} x^2 \log (1-a c-b c x)+\frac {a (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}-\frac {a^2 \text {Li}_2(c (a+b x))}{2 b^2}+\frac {1}{2} x^2 \text {Li}_2(c (a+b x))+\frac {1}{4} (b c) \int \left (\frac {-1+a c}{b^2 c^2}-\frac {x}{b c}-\frac {(-1+a c)^2}{b^2 c^2 (-1+a c+b c x)}\right ) \, dx\\ &=\frac {a x}{2 b}-\frac {(1-a c) x}{4 b c}-\frac {x^2}{8}-\frac {(1-a c)^2 \log (1-a c-b c x)}{4 b^2 c^2}+\frac {1}{4} x^2 \log (1-a c-b c x)+\frac {a (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}-\frac {a^2 \text {Li}_2(c (a+b x))}{2 b^2}+\frac {1}{2} x^2 \text {Li}_2(c (a+b x))\\ \end {align*}
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Mathematica [A] time = 0.09, size = 96, normalized size = 0.63 \[ \frac {-4 c^2 \left (a^2-b^2 x^2\right ) \text {Li}_2(c (a+b x))+\left (-6 a^2 c^2-4 a c (b c x-2)+2 b^2 c^2 x^2-2\right ) \log (-a c-b c x+1)-b c x (-6 a c+b c x+2)}{8 b^2 c^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.36, size = 110, normalized size = 0.72 \[ -\frac {b^{2} c^{2} x^{2} - 2 \, {\left (3 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]
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 \[ \int x {\rm Li}_2\left ({\left (b x + a\right )} c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 177, normalized size = 1.16 \[ -\frac {\polylog \left (2, b c x +a c \right ) a^{2}}{2 b^{2}}+\frac {\polylog \left (2, b c x +a c \right ) x^{2}}{2}-\frac {\ln \left (-b c x -a c +1\right ) x a}{2 b}-\frac {3 \ln \left (-b c x -a c +1\right ) a^{2}}{4 b^{2}}+\frac {3 a x}{4 b}+\frac {7 a^{2}}{8 b^{2}}+\frac {\ln \left (-b c x -a c +1\right ) a}{b^{2} c}-\frac {5 a}{4 b^{2} c}+\frac {x^{2} \ln \left (-b c x -a c +1\right )}{4}-\frac {\ln \left (-b c x -a c +1\right )}{4 b^{2} c^{2}}-\frac {x^{2}}{8}-\frac {x}{4 b c}+\frac {3}{8 b^{2} c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 145, normalized size = 0.95 \[ \frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{2}}{2 \, b^{2}} + \frac {4 \, b^{2} c^{2} x^{2} {\rm Li}_2\left (b c x + a c\right ) - b^{2} c^{2} x^{2} + 2 \, {\left (3 \, a b c^{2} - b c\right )} x + 2 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]
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 \[ \int x\,\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.04, size = 153, normalized size = 1.01 \[ \begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\\frac {x^{2} \operatorname {Li}_{2}\left (a c\right )}{2} & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {3 a^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2}} - \frac {a^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2 b^{2}} + \frac {a x \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b} + \frac {3 a x}{4 b} - \frac {a \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{2} c} - \frac {x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4} + \frac {x^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2} - \frac {x^{2}}{8} - \frac {x}{4 b c} + \frac {\operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2} c^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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