Optimal. Leaf size=106 \[ x \text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )-\frac {(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )-b n x \text {Li}_2(e x)-\frac {b n \text {Li}_2(e x)}{e}+\frac {2 b n (1-e x) \log (1-e x)}{e}+3 b n x \]
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Rubi [A] time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2381, 2389, 2295, 2370, 2411, 43, 2351, 2315} \[ x \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-b n x \text {PolyLog}(2,e x)-\frac {b n \text {PolyLog}(2,e x)}{e}-\frac {(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n (1-e x) \log (1-e x)}{e}+3 b n x \]
Antiderivative was successfully verified.
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Rule 43
Rule 2295
Rule 2315
Rule 2351
Rule 2370
Rule 2381
Rule 2389
Rule 2411
Rubi steps
\begin {align*} \int \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \, dx &=-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-(b n) \int \log (1-e x) \, dx+\int \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx\\ &=-x \left (a+b \log \left (c x^n\right )\right )-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-(b n) \int \left (-1-\frac {(1-e x) \log (1-e x)}{e x}\right ) \, dx+\frac {(b n) \operatorname {Subst}(\int \log (x) \, dx,x,1-e x)}{e}\\ &=2 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)+\frac {(b n) \int \frac {(1-e x) \log (1-e x)}{x} \, dx}{e}\\ &=2 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {(b n) \operatorname {Subst}\left (\int \frac {x \log (x)}{\frac {1}{e}-\frac {x}{e}} \, dx,x,1-e x\right )}{e^2}\\ &=2 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {(b n) \operatorname {Subst}\left (\int \left (-e \log (x)-\frac {e \log (x)}{-1+x}\right ) \, dx,x,1-e x\right )}{e^2}\\ &=2 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)+\frac {(b n) \operatorname {Subst}(\int \log (x) \, dx,x,1-e x)}{e}+\frac {(b n) \operatorname {Subst}\left (\int \frac {\log (x)}{-1+x} \, dx,x,1-e x\right )}{e}\\ &=3 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-\frac {b n \text {Li}_2(e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)\\ \end {align*}
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Mathematica [A] time = 0.08, size = 113, normalized size = 1.07 \[ \left (x \text {Li}_2(e x)+\left (x-\frac {1}{e}\right ) \log (1-e x)-x\right ) \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )+\frac {b n (\text {Li}_2(e x) (-e x+e x \log (x)-1)+3 e x-2 e x \log (1-e x)+2 \log (1-e x)+\log (x) ((e x-1) \log (1-e x)-e x))}{e} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.04, size = 137, normalized size = 1.29 \[ \frac {{\left (3 \, b e n - a e\right )} x - {\left (b n + {\left (b e n - a e\right )} x\right )} {\rm Li}_2\left (e x\right ) + {\left (2 \, b n - {\left (2 \, b e n - a e\right )} x - a\right )} \log \left (-e x + 1\right ) + {\left (b e x {\rm Li}_2\left (e x\right ) - b e x + {\left (b e x - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + {\left (b e n x {\rm Li}_2\left (e x\right ) - b e n x + {\left (b e n x - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x)}{e} \]
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 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) \polylog \left (2, e x \right )\, dx \]
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 \[ b {\left (\frac {{\left (e x \log \left (x^{n}\right ) - {\left (e n - e \log \relax (c)\right )} x\right )} {\rm Li}_2\left (e x\right ) - {\left ({\left (2 \, e n - e \log \relax (c)\right )} x - n \log \relax (x)\right )} \log \left (-e x + 1\right ) - {\left (e x - {\left (e x - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{e} - \int -\frac {{\left (3 \, e n - e \log \relax (c)\right )} x - n \log \relax (x) - n}{e x - 1}\,{d x}\right )} + \frac {{\left (e x {\rm Li}_2\left (e x\right ) - e x + {\left (e x - 1\right )} \log \left (-e x + 1\right )\right )} a}{e} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 23.64, size = 172, normalized size = 1.62 \[ \begin {cases} - a x \operatorname {Li}_{1}\left (e x\right ) + a x \operatorname {Li}_{2}\left (e x\right ) - a x + \frac {a \operatorname {Li}_{1}\left (e x\right )}{e} - b n x \log {\relax (x )} \operatorname {Li}_{1}\left (e x\right ) + b n x \log {\relax (x )} \operatorname {Li}_{2}\left (e x\right ) - b n x \log {\relax (x )} + 2 b n x \operatorname {Li}_{1}\left (e x\right ) - b n x \operatorname {Li}_{2}\left (e x\right ) + 3 b n x - b x \log {\relax (c )} \operatorname {Li}_{1}\left (e x\right ) + b x \log {\relax (c )} \operatorname {Li}_{2}\left (e x\right ) - b x \log {\relax (c )} + \frac {b n \log {\relax (x )} \operatorname {Li}_{1}\left (e x\right )}{e} - \frac {2 b n \operatorname {Li}_{1}\left (e x\right )}{e} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{e} + \frac {b \log {\relax (c )} \operatorname {Li}_{1}\left (e x\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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