Optimal. Leaf size=152 \[ \frac {a x}{2 b}-\frac {(1-a c) x}{4 b c}-\frac {x^2}{8}-\frac {(1-a c)^2 \log (1-a c-b c x)}{4 b^2 c^2}+\frac {1}{4} x^2 \log (1-a c-b c x)+\frac {a (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}-\frac {a^2 \text {PolyLog}(2,c (a+b x))}{2 b^2}+\frac {1}{2} x^2 \text {PolyLog}(2,c (a+b x)) \]
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Rubi [A]
time = 0.12, 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 = {6733, 45,
2463, 2436, 2332, 2442, 2440, 2438} \begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 2332
Rule 2436
Rule 2438
Rule 2440
Rule 2442
Rule 2463
Rule 6733
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 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 b^2}+\frac {a \text {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 = 0.07, size = 96, normalized size = 0.63 \begin {gather*} \frac {-b c x (2-6 a c+b c x)+\left (-2-6 a^2 c^2+2 b^2 c^2 x^2-4 a c (-2+b c x)\right ) \log (1-a c-b c x)-4 c^2 \left (a^2-b^2 x^2\right ) \text {PolyLog}(2,c (a+b x))}{8 b^2 c^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.57, size = 166, normalized size = 1.09
method | result | size |
derivativedivides | \(\frac {-\polylog \left (2, x b c +a c \right ) a c \left (x b c +a c \right )+\frac {\polylog \left (2, x b c +a c \right ) \left (x b c +a c \right )^{2}}{2}+\left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right ) a c +\frac {\left (-x b c -a c +1\right )^{2} \ln \left (-x b c -a c +1\right )}{4}-\frac {\left (-x b c -a c +1\right )^{2}}{8}-\frac {\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )}{2}+\frac {1}{2}-\frac {x b c}{2}-\frac {a c}{2}}{c^{2} b^{2}}\) | \(166\) |
default | \(\frac {-\polylog \left (2, x b c +a c \right ) a c \left (x b c +a c \right )+\frac {\polylog \left (2, x b c +a c \right ) \left (x b c +a c \right )^{2}}{2}+\left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right ) a c +\frac {\left (-x b c -a c +1\right )^{2} \ln \left (-x b c -a c +1\right )}{4}-\frac {\left (-x b c -a c +1\right )^{2}}{8}-\frac {\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )}{2}+\frac {1}{2}-\frac {x b c}{2}-\frac {a c}{2}}{c^{2} b^{2}}\) | \(166\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 145, normalized size = 0.95 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 110, normalized size = 0.72 \begin {gather*} -\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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 2.47, size = 153, normalized size = 1.01 \begin {gather*} \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} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x\,\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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