Optimal. Leaf size=347 \[ \frac {11 a^2 x}{18 b^2}-\frac {5 a (1-a c) x}{36 b^2 c}+\frac {(1-a c)^2 x}{27 b^2 c^2}-\frac {5 a x^2}{72 b}+\frac {(1-a c) x^2}{54 b c}+\frac {x^3}{81}-\frac {5 a (1-a c)^2 \log (1-a c-b c x)}{36 b^3 c^2}+\frac {(1-a c)^3 \log (1-a c-b c x)}{27 b^3 c^3}+\frac {5 a x^2 \log (1-a c-b c x)}{36 b}-\frac {1}{27} x^3 \log (1-a c-b c x)+\frac {11 a^2 (1-a c-b c x) \log (1-a c-b c x)}{18 b^3 c}-\frac {11 a^3 \text {PolyLog}(2,c (a+b x))}{18 b^3}-\frac {a^2 x \text {PolyLog}(2,c (a+b x))}{3 b^2}+\frac {a x^2 \text {PolyLog}(2,c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {PolyLog}(2,c (a+b x))+\frac {2 a^3 \text {PolyLog}(3,c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {PolyLog}(3,c (a+b x))}{3 b^3} \]
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Rubi [A]
time = 0.45, antiderivative size = 347, normalized size of antiderivative = 1.00, number
of steps used = 33, number of rules used = 13, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules
used = {6734, 6730, 2494, 2436, 2332, 2468, 2440, 2438, 6733, 45, 2463, 2442, 6724}
\begin {gather*} -\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}-\frac {11 a^3 \text {Li}_2(c (a+b x))}{18 b^3}+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}+\frac {11 a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{18 b^3 c}-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {11 a^2 x}{18 b^2}+\frac {(1-a c)^3 \log (-a c-b c x+1)}{27 b^3 c^3}-\frac {5 a (1-a c)^2 \log (-a c-b c x+1)}{36 b^3 c^2}+\frac {x (1-a c)^2}{27 b^2 c^2}-\frac {5 a x (1-a c)}{36 b^2 c}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{27} x^3 \log (-a c-b c x+1)+\frac {x^2 (1-a c)}{54 b c}+\frac {5 a x^2 \log (-a c-b c x+1)}{36 b}-\frac {5 a x^2}{72 b}+\frac {x^3}{81} \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 2468
Rule 2494
Rule 6724
Rule 6730
Rule 6733
Rule 6734
Rubi steps
\begin {align*} \int x^2 \text {Li}_3(c (a+b x)) \, dx &=-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}+\frac {\int \left (-a^2 \text {Li}_2(c (a+b x))+a b x \text {Li}_2(c (a+b x))-b^2 x^2 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_2(c (a+b x))}{a+b x}\right ) \, dx}{3 b^2}\\ &=-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}-\frac {1}{3} \int x^2 \text {Li}_2(c (a+b x)) \, dx-\frac {a^2 \int \text {Li}_2(c (a+b x)) \, dx}{3 b^2}+\frac {\left (2 a^3\right ) \int \frac {\text {Li}_2(c (a+b x))}{a+b x} \, dx}{3 b^2}+\frac {a \int x \text {Li}_2(c (a+b x)) \, dx}{3 b}\\ &=-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}+\frac {1}{6} a \int \frac {x^2 \log (1-a c-b c x)}{a+b x} \, dx-\frac {a^2 \int \log (1-c (a+b x)) \, dx}{3 b^2}+\frac {a^3 \int \frac {\log (1-c (a+b x))}{a+b x} \, dx}{3 b^2}-\frac {1}{9} b \int \frac {x^3 \log (1-a c-b c x)}{a+b x} \, dx\\ &=-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}+\frac {1}{6} a \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 {a^2 \int \log (1-a c-b c x) \, dx}{3 b^2}+\frac {a^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{3 b^2}-\frac {1}{9} b \int \left (\frac {a^2 \log (1-a c-b c x)}{b^3}-\frac {a x \log (1-a c-b c x)}{b^2}+\frac {x^2 \log (1-a c-b c x)}{b}-\frac {a^3 \log (1-a c-b c x)}{b^3 (a+b x)}\right ) \, dx\\ &=-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}-\frac {1}{9} \int x^2 \log (1-a c-b c x) \, dx+\frac {a^3 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{3 b^3}-\frac {a^2 \int \log (1-a c-b c x) \, dx}{9 b^2}-\frac {a^2 \int \log (1-a c-b c x) \, dx}{6 b^2}+\frac {a^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{9 b^2}+\frac {a^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{6 b^2}+\frac {a \int x \log (1-a c-b c x) \, dx}{9 b}+\frac {a \int x \log (1-a c-b c x) \, dx}{6 b}+\frac {a^2 \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{3 b^3 c}\\ &=\frac {a^2 x}{3 b^2}+\frac {5 a x^2 \log (1-a c-b c x)}{36 b}-\frac {1}{27} x^3 \log (1-a c-b c x)+\frac {a^2 (1-a c-b c x) \log (1-a c-b c x)}{3 b^3 c}-\frac {a^3 \text {Li}_2(c (a+b x))}{3 b^3}-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}+\frac {a^3 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{9 b^3}+\frac {a^3 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{6 b^3}+\frac {a^2 \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{9 b^3 c}+\frac {a^2 \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{6 b^3 c}+\frac {1}{18} (a c) \int \frac {x^2}{1-a c-b c x} \, dx+\frac {1}{12} (a c) \int \frac {x^2}{1-a c-b c x} \, dx-\frac {1}{27} (b c) \int \frac {x^3}{1-a c-b c x} \, dx\\ &=\frac {11 a^2 x}{18 b^2}+\frac {5 a x^2 \log (1-a c-b c x)}{36 b}-\frac {1}{27} x^3 \log (1-a c-b c x)+\frac {11 a^2 (1-a c-b c x) \log (1-a c-b c x)}{18 b^3 c}-\frac {11 a^3 \text {Li}_2(c (a+b x))}{18 b^3}-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}+\frac {1}{18} (a 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 {1}{12} (a 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 {1}{27} (b c) \int \left (-\frac {(-1+a c)^2}{b^3 c^3}+\frac {(-1+a c) x}{b^2 c^2}-\frac {x^2}{b c}+\frac {(-1+a c)^3}{b^3 c^3 (-1+a c+b c x)}\right ) \, dx\\ &=\frac {11 a^2 x}{18 b^2}-\frac {5 a (1-a c) x}{36 b^2 c}+\frac {(1-a c)^2 x}{27 b^2 c^2}-\frac {5 a x^2}{72 b}+\frac {(1-a c) x^2}{54 b c}+\frac {x^3}{81}-\frac {5 a (1-a c)^2 \log (1-a c-b c x)}{36 b^3 c^2}+\frac {(1-a c)^3 \log (1-a c-b c x)}{27 b^3 c^3}+\frac {5 a x^2 \log (1-a c-b c x)}{36 b}-\frac {1}{27} x^3 \log (1-a c-b c x)+\frac {11 a^2 (1-a c-b c x) \log (1-a c-b c x)}{18 b^3 c}-\frac {11 a^3 \text {Li}_2(c (a+b x))}{18 b^3}-\frac {a^2 x \text {Li}_2(c (a+b x))}{3 b^2}+\frac {a x^2 \text {Li}_2(c (a+b x))}{6 b}-\frac {1}{9} x^3 \text {Li}_2(c (a+b x))+\frac {2 a^3 \text {Li}_3(c (a+b x))}{3 b^3}-\frac {\left (a^3-b^3 x^3\right ) \text {Li}_3(c (a+b x))}{3 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 296, normalized size = 0.85 \begin {gather*} \frac {24 a c-150 a^2 c^2+575 a^3 c^3+24 b c x-138 a b c^2 x+510 a^2 b c^3 x+12 b^2 c^2 x^2-57 a b^2 c^3 x^2+8 b^3 c^3 x^3+24 \log (1-a c-b c x)-162 a c \log (1-a c-b c x)+648 a^2 c^2 \log (1-a c-b c x)-510 a^3 c^3 \log (1-a c-b c x)-396 a^2 b c^3 x \log (1-a c-b c x)+90 a b^2 c^3 x^2 \log (1-a c-b c x)-24 b^3 c^3 x^3 \log (1-a c-b c x)-36 c^3 \left (11 a^3+6 a^2 b x-3 a b^2 x^2+2 b^3 x^3\right ) \text {PolyLog}(2,c (a+b x))+216 c^3 \left (a^3+b^3 x^3\right ) \text {PolyLog}(3,c (a+b x))}{648 b^3 c^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int x^{2} \polylog \left (3, c \left (b x +a \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 264, normalized size = 0.76 \begin {gather*} \frac {11 \, {\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^{3}}{18 \, b^{3}} + \frac {a^{3} {\rm Li}_{3}(b c x + a c)}{3 \, b^{3}} + \frac {216 \, b^{3} c^{3} x^{3} {\rm Li}_{3}(b c x + a c) + 8 \, b^{3} c^{3} x^{3} - 3 \, {\left (19 \, a b^{2} c^{3} - 4 \, b^{2} c^{2}\right )} x^{2} + 6 \, {\left (85 \, a^{2} b c^{3} - 23 \, a b c^{2} + 4 \, b c\right )} x - 36 \, {\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x\right )} {\rm Li}_2\left (b c x + a c\right ) - 6 \, {\left (4 \, b^{3} c^{3} x^{3} - 15 \, a b^{2} c^{3} x^{2} + 66 \, a^{2} b c^{3} x + 85 \, a^{3} c^{3} - 108 \, a^{2} c^{2} + 27 \, a c - 4\right )} \log \left (-b c x - a c + 1\right )}{648 \, b^{3} c^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 219, normalized size = 0.63 \begin {gather*} \frac {8 \, b^{3} c^{3} x^{3} - 3 \, {\left (19 \, a b^{2} c^{3} - 4 \, b^{2} c^{2}\right )} x^{2} + 6 \, {\left (85 \, a^{2} b c^{3} - 23 \, a b c^{2} + 4 \, b c\right )} x - 36 \, {\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3}\right )} {\rm Li}_2\left (b c x + a c\right ) - 6 \, {\left (4 \, b^{3} c^{3} x^{3} - 15 \, a b^{2} c^{3} x^{2} + 66 \, a^{2} b c^{3} x + 85 \, a^{3} c^{3} - 108 \, a^{2} c^{2} + 27 \, a c - 4\right )} \log \left (-b c x - a c + 1\right ) + 216 \, {\left (b^{3} c^{3} x^{3} + a^{3} c^{3}\right )} {\rm polylog}\left (3, b c x + a c\right )}{648 \, b^{3} c^{3}} \end {gather*}
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 \begin {gather*} \int x^{2} \operatorname {Li}_{3}\left (a c + b c x\right )\, dx \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 = -0.00 \begin {gather*} \int x^2\,\mathrm {polylog}\left (3,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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