Optimal. Leaf size=661 \[ \frac {(4 a c+3 b) \text {Li}_3(1-c x)}{6 c^4}-\frac {(4 a c+3 b) \text {Li}_2(c x) \log (1-c x)}{12 c^4}-\frac {(4 a c+3 b) \text {Li}_2(1-c x) \log (1-c x)}{6 c^4}-\frac {(4 a c+3 b) \log (c x) \log ^2(1-c x)}{12 c^4}+\frac {13 (4 a c+3 b) \log (1-c x)}{432 c^4}+\frac {(1-c x) (4 a c+3 b) \log (1-c x)}{12 c^4}-\frac {x (4 a c+3 b) \text {Li}_2(c x)}{12 c^3}+\frac {49 x (4 a c+3 b)}{432 c^3}-\frac {x^2 (4 a c+3 b) \text {Li}_2(c x)}{24 c^2}+\frac {13 x^2 (4 a c+3 b)}{864 c^2}-\frac {x^2 (4 a c+3 b) \log (1-c x)}{48 c^2}-\frac {x^3 (4 a c+3 b) \text {Li}_2(c x)}{36 c}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \text {Li}_2(c x) \log (1-c x)+\frac {x^3 (4 a c+3 b)}{324 c}-\frac {x^3 (4 a c+3 b) \log (1-c x)}{108 c}-\frac {a \log ^2(1-c x)}{9 c^3}+\frac {2 a (1-c x) \log (1-c x)}{9 c^3}+\frac {5 a \log (1-c x)}{27 c^3}+\frac {11 a x}{27 c^2}+\frac {1}{9} a x^3 \log ^2(1-c x)-\frac {2}{27} a x^3 \log (1-c x)+\frac {5 a x^2}{54 c}-\frac {a x^2 \log (1-c x)}{9 c}+\frac {2 a x^3}{81}-\frac {b \log ^2(1-c x)}{16 c^4}+\frac {b (1-c x) \log (1-c x)}{8 c^4}+\frac {29 b \log (1-c x)}{192 c^4}+\frac {53 b x}{192 c^3}+\frac {29 b x^2}{384 c^2}-\frac {b x^2 \log (1-c x)}{16 c^2}-\frac {1}{16} b x^4 \text {Li}_2(c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {3}{64} b x^4 \log (1-c x)+\frac {17 b x^3}{576 c}-\frac {b x^3 \log (1-c x)}{24 c}+\frac {3 b x^4}{256} \]
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Rubi [A] time = 0.98, antiderivative size = 661, normalized size of antiderivative = 1.00, number of steps used = 52, number of rules used = 17, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.810, Rules used = {6742, 6591, 2395, 43, 6604, 2398, 2410, 2389, 2295, 2390, 2301, 6586, 6596, 2396, 2433, 2374, 6589} \[ -\frac {x^2 (4 a c+3 b) \text {PolyLog}(2,c x)}{24 c^2}-\frac {x (4 a c+3 b) \text {PolyLog}(2,c x)}{12 c^3}+\frac {(4 a c+3 b) \text {PolyLog}(3,1-c x)}{6 c^4}-\frac {(4 a c+3 b) \log (1-c x) \text {PolyLog}(2,c x)}{12 c^4}-\frac {(4 a c+3 b) \log (1-c x) \text {PolyLog}(2,1-c x)}{6 c^4}-\frac {x^3 (4 a c+3 b) \text {PolyLog}(2,c x)}{36 c}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {PolyLog}(2,c x)-\frac {1}{16} b x^4 \text {PolyLog}(2,c x)+\frac {13 x^2 (4 a c+3 b)}{864 c^2}-\frac {x^2 (4 a c+3 b) \log (1-c x)}{48 c^2}+\frac {49 x (4 a c+3 b)}{432 c^3}-\frac {(4 a c+3 b) \log (c x) \log ^2(1-c x)}{12 c^4}+\frac {13 (4 a c+3 b) \log (1-c x)}{432 c^4}+\frac {(1-c x) (4 a c+3 b) \log (1-c x)}{12 c^4}+\frac {x^3 (4 a c+3 b)}{324 c}-\frac {x^3 (4 a c+3 b) \log (1-c x)}{108 c}+\frac {11 a x}{27 c^2}-\frac {a \log ^2(1-c x)}{9 c^3}+\frac {2 a (1-c x) \log (1-c x)}{9 c^3}+\frac {5 a \log (1-c x)}{27 c^3}+\frac {5 a x^2}{54 c}+\frac {1}{9} a x^3 \log ^2(1-c x)-\frac {2}{27} a x^3 \log (1-c x)-\frac {a x^2 \log (1-c x)}{9 c}+\frac {2 a x^3}{81}+\frac {29 b x^2}{384 c^2}-\frac {b x^2 \log (1-c x)}{16 c^2}+\frac {53 b x}{192 c^3}-\frac {b \log ^2(1-c x)}{16 c^4}+\frac {b (1-c x) \log (1-c x)}{8 c^4}+\frac {29 b \log (1-c x)}{192 c^4}+\frac {17 b x^3}{576 c}+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {3}{64} b x^4 \log (1-c x)-\frac {b x^3 \log (1-c x)}{24 c}+\frac {3 b x^4}{256} \]
Antiderivative was successfully verified.
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Rule 43
Rule 2295
Rule 2301
Rule 2374
Rule 2389
Rule 2390
Rule 2395
Rule 2396
Rule 2398
Rule 2410
Rule 2433
Rule 6586
Rule 6589
Rule 6591
Rule 6596
Rule 6604
Rule 6742
Rubi steps
\begin {align*} \int x^2 (a+b x) \log (1-c x) \text {Li}_2(c x) \, dx &=\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)+c \int \left (\frac {(-3 b-4 a c) \text {Li}_2(c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{12 c^2}-\frac {b x^3 \text {Li}_2(c x)}{4 c}+\frac {(-3 b-4 a c) \text {Li}_2(c x)}{12 c^4 (-1+c x)}\right ) \, dx+\int \left (\frac {1}{3} a x^2 \log ^2(1-c x)+\frac {1}{4} b x^3 \log ^2(1-c x)\right ) \, dx\\ &=\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)+\frac {1}{3} a \int x^2 \log ^2(1-c x) \, dx+\frac {1}{4} b \int x^3 \log ^2(1-c x) \, dx-\frac {1}{4} b \int x^3 \text {Li}_2(c x) \, dx-\frac {(3 b+4 a c) \int \text {Li}_2(c x) \, dx}{12 c^3}-\frac {(3 b+4 a c) \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{12 c^3}-\frac {(3 b+4 a c) \int x \text {Li}_2(c x) \, dx}{12 c^2}-\frac {(3 b+4 a c) \int x^2 \text {Li}_2(c x) \, dx}{12 c}\\ &=\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)-\frac {1}{16} b \int x^3 \log (1-c x) \, dx+\frac {1}{9} (2 a c) \int \frac {x^3 \log (1-c x)}{1-c x} \, dx+\frac {1}{8} (b c) \int \frac {x^4 \log (1-c x)}{1-c x} \, dx-\frac {(3 b+4 a c) \int \frac {\log ^2(1-c x)}{x} \, dx}{12 c^4}-\frac {(3 b+4 a c) \int \log (1-c x) \, dx}{12 c^3}-\frac {(3 b+4 a c) \int x \log (1-c x) \, dx}{24 c^2}-\frac {(3 b+4 a c) \int x^2 \log (1-c x) \, dx}{36 c}\\ &=-\frac {(3 b+4 a c) x^2 \log (1-c x)}{48 c^2}-\frac {(3 b+4 a c) x^3 \log (1-c x)}{108 c}-\frac {1}{64} b x^4 \log (1-c x)+\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) \log (c x) \log ^2(1-c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)+\frac {1}{9} (2 a c) \int \left (-\frac {\log (1-c x)}{c^3}-\frac {x \log (1-c x)}{c^2}-\frac {x^2 \log (1-c x)}{c}-\frac {\log (1-c x)}{c^3 (-1+c x)}\right ) \, dx-\frac {1}{64} (b c) \int \frac {x^4}{1-c x} \, dx+\frac {1}{8} (b c) \int \left (-\frac {\log (1-c x)}{c^4}-\frac {x \log (1-c x)}{c^3}-\frac {x^2 \log (1-c x)}{c^2}-\frac {x^3 \log (1-c x)}{c}-\frac {\log (1-c x)}{c^4 (-1+c x)}\right ) \, dx-\frac {1}{108} (3 b+4 a c) \int \frac {x^3}{1-c x} \, dx+\frac {(3 b+4 a c) \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{12 c^4}-\frac {(3 b+4 a c) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{6 c^3}-\frac {(3 b+4 a c) \int \frac {x^2}{1-c x} \, dx}{48 c}\\ &=\frac {(3 b+4 a c) x}{12 c^3}-\frac {(3 b+4 a c) x^2 \log (1-c x)}{48 c^2}-\frac {(3 b+4 a c) x^3 \log (1-c x)}{108 c}-\frac {1}{64} b x^4 \log (1-c x)+\frac {(3 b+4 a c) (1-c x) \log (1-c x)}{12 c^4}+\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) \log (c x) \log ^2(1-c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)-\frac {1}{9} (2 a) \int x^2 \log (1-c x) \, dx-\frac {1}{8} b \int x^3 \log (1-c x) \, dx-\frac {b \int \log (1-c x) \, dx}{8 c^3}-\frac {b \int \frac {\log (1-c x)}{-1+c x} \, dx}{8 c^3}-\frac {(2 a) \int \log (1-c x) \, dx}{9 c^2}-\frac {(2 a) \int \frac {\log (1-c x)}{-1+c x} \, dx}{9 c^2}-\frac {b \int x \log (1-c x) \, dx}{8 c^2}-\frac {(2 a) \int x \log (1-c x) \, dx}{9 c}-\frac {b \int x^2 \log (1-c x) \, dx}{8 c}-\frac {1}{64} (b c) \int \left (-\frac {1}{c^4}-\frac {x}{c^3}-\frac {x^2}{c^2}-\frac {x^3}{c}-\frac {1}{c^4 (-1+c x)}\right ) \, dx-\frac {1}{108} (3 b+4 a c) \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx+\frac {(3 b+4 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 )}{6 c^4}-\frac {(3 b+4 a c) \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx}{48 c}\\ &=\frac {b x}{64 c^3}+\frac {49 (3 b+4 a c) x}{432 c^3}+\frac {b x^2}{128 c^2}+\frac {13 (3 b+4 a c) x^2}{864 c^2}+\frac {b x^3}{192 c}+\frac {(3 b+4 a c) x^3}{324 c}+\frac {b x^4}{256}+\frac {b \log (1-c x)}{64 c^4}+\frac {13 (3 b+4 a c) \log (1-c x)}{432 c^4}-\frac {b x^2 \log (1-c x)}{16 c^2}-\frac {a x^2 \log (1-c x)}{9 c}-\frac {(3 b+4 a c) x^2 \log (1-c x)}{48 c^2}-\frac {2}{27} a x^3 \log (1-c x)-\frac {b x^3 \log (1-c x)}{24 c}-\frac {(3 b+4 a c) x^3 \log (1-c x)}{108 c}-\frac {3}{64} b x^4 \log (1-c x)+\frac {(3 b+4 a c) (1-c x) \log (1-c x)}{12 c^4}+\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) \log (c x) \log ^2(1-c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(1-c x)}{6 c^4}-\frac {1}{9} a \int \frac {x^2}{1-c x} \, dx-\frac {1}{24} b \int \frac {x^3}{1-c x} \, dx+\frac {b \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{8 c^4}-\frac {b \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{8 c^4}+\frac {(2 a) \operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{9 c^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{9 c^3}-\frac {b \int \frac {x^2}{1-c x} \, dx}{16 c}-\frac {1}{27} (2 a c) \int \frac {x^3}{1-c x} \, dx-\frac {1}{32} (b c) \int \frac {x^4}{1-c x} \, dx+\frac {(3 b+4 a c) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{6 c^4}\\ &=\frac {9 b x}{64 c^3}+\frac {2 a x}{9 c^2}+\frac {49 (3 b+4 a c) x}{432 c^3}+\frac {b x^2}{128 c^2}+\frac {13 (3 b+4 a c) x^2}{864 c^2}+\frac {b x^3}{192 c}+\frac {(3 b+4 a c) x^3}{324 c}+\frac {b x^4}{256}+\frac {b \log (1-c x)}{64 c^4}+\frac {13 (3 b+4 a c) \log (1-c x)}{432 c^4}-\frac {b x^2 \log (1-c x)}{16 c^2}-\frac {a x^2 \log (1-c x)}{9 c}-\frac {(3 b+4 a c) x^2 \log (1-c x)}{48 c^2}-\frac {2}{27} a x^3 \log (1-c x)-\frac {b x^3 \log (1-c x)}{24 c}-\frac {(3 b+4 a c) x^3 \log (1-c x)}{108 c}-\frac {3}{64} b x^4 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{8 c^4}+\frac {2 a (1-c x) \log (1-c x)}{9 c^3}+\frac {(3 b+4 a c) (1-c x) \log (1-c x)}{12 c^4}-\frac {b \log ^2(1-c x)}{16 c^4}-\frac {a \log ^2(1-c x)}{9 c^3}+\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) \log (c x) \log ^2(1-c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(1-c x)}{6 c^4}+\frac {(3 b+4 a c) \text {Li}_3(1-c x)}{6 c^4}-\frac {1}{9} a \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx-\frac {1}{24} b \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx-\frac {b \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx}{16 c}-\frac {1}{27} (2 a c) \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx-\frac {1}{32} (b c) \int \left (-\frac {1}{c^4}-\frac {x}{c^3}-\frac {x^2}{c^2}-\frac {x^3}{c}-\frac {1}{c^4 (-1+c x)}\right ) \, dx\\ &=\frac {53 b x}{192 c^3}+\frac {11 a x}{27 c^2}+\frac {49 (3 b+4 a c) x}{432 c^3}+\frac {29 b x^2}{384 c^2}+\frac {5 a x^2}{54 c}+\frac {13 (3 b+4 a c) x^2}{864 c^2}+\frac {2 a x^3}{81}+\frac {17 b x^3}{576 c}+\frac {(3 b+4 a c) x^3}{324 c}+\frac {3 b x^4}{256}+\frac {29 b \log (1-c x)}{192 c^4}+\frac {5 a \log (1-c x)}{27 c^3}+\frac {13 (3 b+4 a c) \log (1-c x)}{432 c^4}-\frac {b x^2 \log (1-c x)}{16 c^2}-\frac {a x^2 \log (1-c x)}{9 c}-\frac {(3 b+4 a c) x^2 \log (1-c x)}{48 c^2}-\frac {2}{27} a x^3 \log (1-c x)-\frac {b x^3 \log (1-c x)}{24 c}-\frac {(3 b+4 a c) x^3 \log (1-c x)}{108 c}-\frac {3}{64} b x^4 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{8 c^4}+\frac {2 a (1-c x) \log (1-c x)}{9 c^3}+\frac {(3 b+4 a c) (1-c x) \log (1-c x)}{12 c^4}-\frac {b \log ^2(1-c x)}{16 c^4}-\frac {a \log ^2(1-c x)}{9 c^3}+\frac {1}{9} a x^3 \log ^2(1-c x)+\frac {1}{16} b x^4 \log ^2(1-c x)-\frac {(3 b+4 a c) \log (c x) \log ^2(1-c x)}{12 c^4}-\frac {(3 b+4 a c) x \text {Li}_2(c x)}{12 c^3}-\frac {(3 b+4 a c) x^2 \text {Li}_2(c x)}{24 c^2}-\frac {(3 b+4 a c) x^3 \text {Li}_2(c x)}{36 c}-\frac {1}{16} b x^4 \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(c x)}{12 c^4}+\frac {1}{12} \left (4 a x^3+3 b x^4\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(3 b+4 a c) \log (1-c x) \text {Li}_2(1-c x)}{6 c^4}+\frac {(3 b+4 a c) \text {Li}_3(1-c x)}{6 c^4}\\ \end {align*}
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Mathematica [A] time = 0.71, size = 425, normalized size = 0.64 \[ \frac {48 \text {Li}_2(c x) \left (12 \log (1-c x) \left (4 a c \left (c^3 x^3-1\right )+3 b \left (c^4 x^4-1\right )\right )-c x \left (8 a c \left (2 c^2 x^2+3 c x+6\right )+3 b \left (3 c^3 x^3+4 c^2 x^2+6 c x+12\right )\right )\right )-1152 (4 a c+3 b) \text {Li}_2(1-c x) \log (1-c x)+256 a c^4 x^3+768 a c^4 x^3 \log ^2(1-c x)-768 a c^4 x^3 \log (1-c x)+1056 a c^3 x^2-1344 a c^3 x^2 \log (1-c x)+5952 a c^2 x-3840 a c^2 x \log (1-c x)+4608 a c \text {Li}_3(1-c x)-768 a c \log ^2(1-c x)-2304 a c \log (c x) \log ^2(1-c x)+5952 a c \log (1-c x)+81 b c^4 x^4+432 b c^4 x^4 \log ^2(1-c x)-324 b c^4 x^4 \log (1-c x)+268 b c^3 x^3-480 b c^3 x^3 \log (1-c x)+834 b c^2 x^2-864 b c^2 x^2 \log (1-c x)+3456 b \text {Li}_3(1-c x)+4260 b c x-432 b \log ^2(1-c x)-1728 b \log (c x) \log ^2(1-c x)-2592 b c x \log (1-c x)+4260 b \log (1-c x)}{6912 c^4} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{3} + a x^{2}\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ), x\right ) \]
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 \[ \int {\left (b x + a\right )} x^{2} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )\,{d x} \]
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 \[ \int x^{2} \left (b x +a \right ) \ln \left (-c x +1\right ) \polylog \left (2, c x \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 415, normalized size = 0.63 \[ -\frac {1}{6912} \, c {\left (\frac {576 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} {\left (4 \, a c + 3 \, b\right )}}{c^{5}} - \frac {81 \, b c^{4} x^{4} + 4 \, {\left (64 \, a c^{4} + 67 \, b c^{3}\right )} x^{3} + 6 \, {\left (176 \, a c^{3} + 139 \, b c^{2}\right )} x^{2} + 12 \, {\left (496 \, a c^{2} + 355 \, b c\right )} x - 48 \, {\left (9 \, b c^{4} x^{4} + 4 \, {\left (4 \, a c^{4} + 3 \, b c^{3}\right )} x^{3} + 6 \, {\left (4 \, a c^{3} + 3 \, b c^{2}\right )} x^{2} + 12 \, {\left (4 \, a c^{2} + 3 \, b c\right )} x + 12 \, {\left (4 \, a c + 3 \, b\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 4 \, {\left (54 \, b c^{4} x^{4} + 4 \, {\left (32 \, a c^{4} + 21 \, b c^{3}\right )} x^{3} + 6 \, {\left (40 \, a c^{3} + 27 \, b c^{2}\right )} x^{2} - 1488 \, a c + 12 \, {\left (64 \, a c^{2} + 45 \, b c\right )} x - 1065 \, b\right )} \log \left (-c x + 1\right )}{c^{5}}\right )} + \frac {1}{1728} \, {\left (\frac {32 \, {\left (18 \, c^{3} x^{3} {\rm Li}_2\left (c x\right ) - 2 \, c^{3} x^{3} - 3 \, c^{2} x^{2} - 6 \, c x + 6 \, {\left (c^{3} x^{3} - 1\right )} \log \left (-c x + 1\right )\right )} a}{c^{3}} + \frac {9 \, {\left (48 \, c^{4} x^{4} {\rm Li}_2\left (c x\right ) - 3 \, c^{4} x^{4} - 4 \, c^{3} x^{3} - 6 \, c^{2} x^{2} - 12 \, c x + 12 \, {\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{4}}\right )} \log \left (-c x + 1\right ) \]
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 \[ \int x^2\,\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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