Optimal. Leaf size=238 \[ -\frac{\text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac{\text{PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{1}{8} b e^2 n \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{4 x^2}-\frac{b n \text{PolyLog}(3,e x)}{4 x^2}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 x^2}-\frac{1}{16} b e^2 n \log ^2(x)+\frac{3}{16} b e^2 n \log (x)-\frac{3}{16} b e^2 n \log (1-e x)+\frac{3 b n \log (1-e x)}{16 x^2}-\frac{5 b e n}{16 x} \]
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Rubi [A] time = 0.217687, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {2385, 2395, 44, 2376, 2301, 2391, 6591} \[ -\frac{\text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac{\text{PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{1}{8} b e^2 n \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{4 x^2}-\frac{b n \text{PolyLog}(3,e x)}{4 x^2}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 x^2}-\frac{1}{16} b e^2 n \log ^2(x)+\frac{3}{16} b e^2 n \log (x)-\frac{3}{16} b e^2 n \log (1-e x)+\frac{3 b n \log (1-e x)}{16 x^2}-\frac{5 b e n}{16 x} \]
Antiderivative was successfully verified.
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Rule 2385
Rule 2395
Rule 44
Rule 2376
Rule 2301
Rule 2391
Rule 6591
Rubi steps
\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{x^3} \, dx &=-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}+\frac{1}{2} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{x^3} \, dx+\frac{1}{4} (b n) \int \frac{\text{Li}_2(e x)}{x^3} \, dx\\ &=-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}-\frac{1}{4} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3} \, dx-2 \left (\frac{1}{8} (b n) \int \frac{\log (1-e x)}{x^3} \, dx\right )\\ &=-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}+\frac{1}{4} (b n) \int \left (\frac{e}{2 x^2}-\frac{e^2 \log (x)}{2 x}-\frac{\log (1-e x)}{2 x^3}+\frac{e^2 \log (1-e x)}{2 x}\right ) \, dx-2 \left (-\frac{b n \log (1-e x)}{16 x^2}-\frac{1}{16} (b e n) \int \frac{1}{x^2 (1-e x)} \, dx\right )\\ &=-\frac{b e n}{8 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}-\frac{1}{8} (b n) \int \frac{\log (1-e x)}{x^3} \, dx-2 \left (-\frac{b n \log (1-e x)}{16 x^2}-\frac{1}{16} (b e n) \int \left (\frac{1}{x^2}+\frac{e}{x}-\frac{e^2}{-1+e x}\right ) \, dx\right )-\frac{1}{8} \left (b e^2 n\right ) \int \frac{\log (x)}{x} \, dx+\frac{1}{8} \left (b e^2 n\right ) \int \frac{\log (1-e x)}{x} \, dx\\ &=-\frac{b e n}{8 x}-\frac{1}{16} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{16 x^2}-\frac{1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac{b e n}{16 x}-\frac{1}{16} b e^2 n \log (x)+\frac{1}{16} b e^2 n \log (1-e x)-\frac{b n \log (1-e x)}{16 x^2}\right )-\frac{1}{8} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}+\frac{1}{16} (b e n) \int \frac{1}{x^2 (1-e x)} \, dx\\ &=-\frac{b e n}{8 x}-\frac{1}{16} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{16 x^2}-\frac{1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac{b e n}{16 x}-\frac{1}{16} b e^2 n \log (x)+\frac{1}{16} b e^2 n \log (1-e x)-\frac{b n \log (1-e x)}{16 x^2}\right )-\frac{1}{8} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}+\frac{1}{16} (b e n) \int \left (\frac{1}{x^2}+\frac{e}{x}-\frac{e^2}{-1+e x}\right ) \, dx\\ &=-\frac{3 b e n}{16 x}+\frac{1}{16} b e^2 n \log (x)-\frac{1}{16} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac{1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{16} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{16 x^2}-\frac{1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac{b e n}{16 x}-\frac{1}{16} b e^2 n \log (x)+\frac{1}{16} b e^2 n \log (1-e x)-\frac{b n \log (1-e x)}{16 x^2}\right )-\frac{1}{8} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{4 x^2}-\frac{b n \text{Li}_3(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(e x)}{2 x^2}\\ \end{align*}
Mathematica [F] time = 0.122723, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{PolyLog}(3,e x)}{x^3} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.309, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\it polylog} \left ( 3,ex \right ) }{{x}^{3}}}\, 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}{8} \,{\left (e^{2} \log \left (x\right ) - \frac{e x +{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right ) + 2 \,{\rm Li}_2\left (e x\right ) + 4 \,{\rm Li}_{3}(e x)}{x^{2}}\right )} a - \frac{1}{16} \, b{\left (\frac{4 \,{\left (n + \log \left (c\right ) + \log \left (x^{n}\right )\right )}{\rm Li}_2\left (e x\right ) -{\left (2 \, e^{2} n x^{2} \log \left (x\right ) + 3 \, n + 2 \, \log \left (c\right )\right )} \log \left (-e x + 1\right ) - 2 \,{\left (e^{2} x^{2} \log \left (x\right ) - e x -{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right ) + 4 \,{\left (n + 2 \, \log \left (c\right ) + 2 \, \log \left (x^{n}\right )\right )}{\rm Li}_{3}(e x)}{x^{2}} + 16 \, \int -\frac{2 \, e^{2} n x - 5 \, e n - 2 \, e \log \left (c\right ) - 2 \,{\left (2 \, e^{3} n x^{2} - e^{2} n x\right )} \log \left (x\right )}{16 \,{\left (e x^{3} - x^{2}\right )}}\,{d x}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 0.958351, size = 594, normalized size = 2.5 \begin{align*} \frac{b e^{2} n x^{2} \log \left (x\right )^{2} -{\left (5 \, b e n + 2 \, a e\right )} x - 2 \,{\left (b e^{2} n x^{2} + 2 \, b n \log \left (x\right ) + 2 \, b n + 2 \, b \log \left (c\right ) + 2 \, a\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 ) - 2 \,{\left (b e x +{\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) +{\left (2 \, b e^{2} x^{2} \log \left (c\right ) - 2 \, b e n x +{\left (3 \, b e^{2} n + 2 \, a e^{2}\right )} x^{2} - 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 n \log \left (x\right ) + b n + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\rm polylog}\left (3, e x\right )}{16 \, x^{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 \frac{\left (a + b \log{\left (c x^{n} \right )}\right ) \operatorname{Li}_{3}\left (e x\right )}{x^{3}}\, 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 \frac{{\left (b \log \left (c x^{n}\right ) + a\right )}{\rm Li}_{3}(e x)}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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