Optimal. Leaf size=202 \[ -\frac {b e n}{2 x}+\frac {1}{4} b e^2 n \log (x)-\frac {1}{8} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac {1}{4} b e^2 n \text {Li}_2(e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{2 x^2} \]
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Rubi [A]
time = 0.11, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2432, 2442,
46, 2423, 2338, 2438} \begin {gather*} -\frac {\text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {1}{4} b e^2 n \text {PolyLog}(2,e x)-\frac {b n \text {PolyLog}(2,e x)}{4 x^2}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} 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 )}{4 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac {1}{8} b e^2 n \log ^2(x)+\frac {1}{4} b e^2 n \log (x)-\frac {1}{4} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {b e n}{2 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 46
Rule 2338
Rule 2423
Rule 2432
Rule 2438
Rule 2442
Rubi steps
\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \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)}{2 x^2}-\frac {1}{2} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3} \, dx-\frac {1}{4} (b n) \int \frac {\log (1-e x)}{x^3} \, dx\\ &=-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{8 x^2}-\frac {1}{4} 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)}{4 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)}{2 x^2}+\frac {1}{2} (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+\frac {1}{8} (b e n) \int \frac {1}{x^2 (1-e x)} \, dx\\ &=-\frac {b e n}{4 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{8 x^2}-\frac {1}{4} 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)}{4 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)}{2 x^2}-\frac {1}{4} (b n) \int \frac {\log (1-e x)}{x^3} \, dx+\frac {1}{8} (b e n) \int \left (\frac {1}{x^2}+\frac {e}{x}-\frac {e^2}{-1+e x}\right ) \, dx-\frac {1}{4} \left (b e^2 n\right ) \int \frac {\log (x)}{x} \, dx+\frac {1}{4} \left (b e^2 n\right ) \int \frac {\log (1-e x)}{x} \, dx\\ &=-\frac {3 b e n}{8 x}+\frac {1}{8} b e^2 n \log (x)-\frac {1}{8} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {1}{4} 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)}{4 x^2}-\frac {1}{4} 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)}{2 x^2}+\frac {1}{8} (b e n) \int \frac {1}{x^2 (1-e x)} \, dx\\ &=-\frac {3 b e n}{8 x}+\frac {1}{8} b e^2 n \log (x)-\frac {1}{8} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {1}{4} 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)}{4 x^2}-\frac {1}{4} 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)}{2 x^2}+\frac {1}{8} (b e n) \int \left (\frac {1}{x^2}+\frac {e}{x}-\frac {e^2}{-1+e x}\right ) \, dx\\ &=-\frac {b e n}{2 x}+\frac {1}{4} b e^2 n \log (x)-\frac {1}{8} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {1}{4} 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)}{4 x^2}-\frac {1}{4} 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)}{2 x^2}\\ \end {align*}
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Mathematica [A]
time = 0.15, size = 163, normalized size = 0.81 \begin {gather*} \frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (-e x+e^2 x^2 \log (x)+\log (1-e x)-e^2 x^2 \log (1-e x)-2 \text {Li}_2(e x)\right )}{4 x^2}+\frac {b n \left (-4 e x+e^2 x^2 \log ^2(x)+2 \log (1-e x)-2 e^2 x^2 \log (1-e x)-2 (-1+e x) \log (x) (-e x+(1+e x) \log (1-e x))-2 \left (1+e^2 x^2+2 \log (x)\right ) \text {Li}_2(e x)\right )}{8 x^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \polylog \left (2, e x \right )}{x^{3}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 182, normalized size = 0.90 \begin {gather*} \frac {b n x^{2} e^{2} \log \left (x\right )^{2} - 2 \, {\left (2 \, b n + a\right )} x e - 2 \, {\left (b n x^{2} e^{2} + b n + 2 \, a\right )} {\rm Li}_2\left (x e\right ) - 2 \, {\left ({\left (b n + a\right )} x^{2} e^{2} - b n - a\right )} \log \left (-x e + 1\right ) - 2 \, {\left (b x e + 2 \, b {\rm Li}_2\left (x e\right ) + {\left (b x^{2} e^{2} - b\right )} \log \left (-x e + 1\right )\right )} \log \left (c\right ) + 2 \, {\left (b x^{2} e^{2} \log \left (c\right ) + {\left (b n + a\right )} x^{2} e^{2} - b n x e - 2 \, b n {\rm Li}_2\left (x e\right ) - {\left (b n x^{2} e^{2} - b n\right )} \log \left (-x e + 1\right )\right )} \log \left (x\right )}{8 \, x^{2}} \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 \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \operatorname {Li}_{2}\left (e x\right )}{x^{3}}\, 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 \frac {\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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