Optimal. Leaf size=221 \[ \frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 e^2}-\frac {1}{4} x^2 \text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} x^2 \text {Li}_3(e x) \left (a+b \log \left (c x^n\right )\right )+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {1}{8} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}-\frac {3 b n \log (1-e x)}{16 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {3}{16} b n x^2 \log (1-e x)-\frac {5 b n x}{16 e}-\frac {1}{8} b n x^2 \]
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Rubi [A] time = 0.19, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2385, 2395, 43, 2376, 2391, 6591} \[ -\frac {1}{4} x^2 \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} x^2 \text {PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \text {PolyLog}(2,e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {PolyLog}(2,e x)-\frac {1}{4} b n x^2 \text {PolyLog}(3,e x)+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 e^2}-\frac {1}{8} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {3 b n \log (1-e x)}{16 e^2}+\frac {3}{16} b n x^2 \log (1-e x)-\frac {5 b n x}{16 e}-\frac {1}{8} b n x^2 \]
Antiderivative was successfully verified.
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Rule 43
Rule 2376
Rule 2385
Rule 2391
Rule 2395
Rule 6591
Rubi steps
\begin {align*} \int x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x) \, dx &=-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)-\frac {1}{2} \int x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \, dx+\frac {1}{4} (b n) \int x \text {Li}_2(e x) \, dx\\ &=\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)-\frac {1}{4} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx+2 \left (\frac {1}{8} (b n) \int x \log (1-e x) \, dx\right )\\ &=\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{4} (b n) \int \left (-\frac {1}{2 e}-\frac {x}{4}-\frac {\log (1-e x)}{2 e^2 x}+\frac {1}{2} x \log (1-e x)\right ) \, dx+2 \left (\frac {1}{16} b n x^2 \log (1-e x)+\frac {1}{16} (b e n) \int \frac {x^2}{1-e x} \, dx\right )\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{8} (b n) \int x \log (1-e x) \, dx-\frac {(b n) \int \frac {\log (1-e x)}{x} \, dx}{8 e^2}+2 \left (\frac {1}{16} b n x^2 \log (1-e x)+\frac {1}{16} (b e n) \int \left (-\frac {1}{e^2}-\frac {x}{e}-\frac {1}{e^2 (-1+e x)}\right ) \, dx\right )\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{16} (b e n) \int \frac {x^2}{1-e x} \, dx\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{16} (b e n) \int \left (-\frac {1}{e^2}-\frac {x}{e}-\frac {1}{e^2 (-1+e x)}\right ) \, dx\\ &=-\frac {3 b n x}{16 e}-\frac {1}{16} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)\\ \end {align*}
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Mathematica [F] time = 0.12, size = 0, normalized size = 0.00 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x) \, dx \]
Verification is Not applicable to the result.
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fricas [C] time = 0.88, size = 257, normalized size = 1.16 \[ -\frac {{\left (2 \, b e^{2} n - a e^{2}\right )} x^{2} + {\left (5 \, b e n - 2 \, a e\right )} x - 2 \, {\left (2 \, {\left (b e^{2} n - a e^{2}\right )} x^{2} + b n\right )} {\rm Li}_2\left (e x\right ) - {\left ({\left (3 \, b e^{2} n - 2 \, a e^{2}\right )} x^{2} - 3 \, b n + 2 \, a\right )} \log \left (-e x + 1\right ) + {\left (4 \, b e^{2} x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} x^{2} - 2 \, b e x + 2 \, {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + {\left (4 \, b e^{2} n x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} n x^{2} - 2 \, b e n x + 2 \, {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x) - 4 \, {\left (2 \, b e^{2} n x^{2} \log \relax (x) + 2 \, b e^{2} x^{2} \log \relax (c) - {\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2}\right )} {\rm polylog}\left (3, e x\right )}{16 \, e^{2}} \]
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 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} x {\rm Li}_{3}(e x)\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.42, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) x \polylog \left (3, e x \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{16} \, b {\left (\frac {4 \, {\left (e^{2} x^{2} \log \left (x^{n}\right ) - {\left (e^{2} n - e^{2} \log \relax (c)\right )} x^{2}\right )} {\rm Li}_2\left (e x\right ) - {\left ({\left (3 \, e^{2} n - 2 \, e^{2} \log \relax (c)\right )} x^{2} - 2 \, n \log \relax (x)\right )} \log \left (-e x + 1\right ) - {\left (e^{2} x^{2} + 2 \, e x - 2 \, {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right ) - 4 \, {\left (2 \, e^{2} x^{2} \log \left (x^{n}\right ) - {\left (e^{2} n - 2 \, e^{2} \log \relax (c)\right )} x^{2}\right )} {\rm Li}_{3}(e x)}{e^{2}} - 16 \, \int -\frac {e n x + 2 \, {\left (2 \, e^{2} n - e^{2} \log \relax (c)\right )} x^{2} - 2 \, n \log \relax (x) - 2 \, n}{16 \, {\left (e^{2} x - e\right )}}\,{d x}\right )} - \frac {{\left (4 \, e^{2} x^{2} {\rm Li}_2\left (e x\right ) - 8 \, e^{2} x^{2} {\rm Li}_{3}(e x) - e^{2} x^{2} - 2 \, e x + 2 \, {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} a}{16 \, e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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 \[ \int x \left (a + b \log {\left (c x^{n} \right )}\right ) \operatorname {Li}_{3}\left (e x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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