Optimal. Leaf size=221 \[ -\frac{1}{4} x^2 \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} x^2 \text{PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )+\frac{b n \text{PolyLog}(2,e x)}{8 e^2}+\frac{1}{4} b n x^2 \text{PolyLog}(2,e x)-\frac{1}{4} b n x^2 \text{PolyLog}(3,e x)+\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 e^2}-\frac{1}{8} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )+\frac{x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac{1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{3 b n \log (1-e x)}{16 e^2}+\frac{3}{16} b n x^2 \log (1-e x)-\frac{5 b n x}{16 e}-\frac{1}{8} b n x^2 \]
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Rubi [A] time = 0.189606, antiderivative size = 221, normalized size of antiderivative = 1., 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 2385
Rule 2395
Rule 43
Rule 2376
Rule 2391
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*}
Mathematica [F] time = 0.116756, size = 0, normalized size = 0. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \text{PolyLog}(3,e x) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.336, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\it polylog} \left ( 3,ex \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, b{\left (\frac{4 \,{\left (e^{2} x^{2} \log \left (x^{n}\right ) -{\left (e^{2} n - e^{2} \log \left (c\right )\right )} x^{2}\right )}{\rm Li}_2\left (e x\right ) -{\left ({\left (3 \, e^{2} n - 2 \, e^{2} \log \left (c\right )\right )} x^{2} - 2 \, n \log \left (x\right )\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 \left (c\right )\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 \left (c\right )\right )} x^{2} - 2 \, n \log \left (x\right ) - 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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 0.866368, size = 639, normalized size = 2.89 \begin{align*} -\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 \, b e^{2} n x^{2} \log \left (x\right ) + 2 \, b e^{2} x^{2} \log \left (c\right ) - 2 \,{\left (b e^{2} n - a e^{2}\right )} x^{2} - b n\right )}{\rm \%iint}\left (e, x, -\frac{\log \left (-e x + 1\right )}{e}, -\frac{\log \left (-e x + 1\right )}{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 (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 \left (c\right ) -{\left (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 \left (x\right ) - 4 \,{\left (2 \, b e^{2} n x^{2} \log \left (x\right ) + 2 \, b e^{2} x^{2} \log \left (c\right ) -{\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2}\right )}{\rm polylog}\left (3, e x\right )}{16 \, e^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \log{\left (c x^{n} \right )}\right ) \operatorname{Li}_{3}\left (e x\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} x{\rm Li}_{3}(e x)\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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