Optimal. Leaf size=202 \[ \frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {\text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac {1}{4} b e^2 n \text {Li}_2(e x)-\frac {1}{8} b e^2 n \log ^2(x)+\frac {1}{4} b e^2 n \log (x)-\frac {1}{4} b e^2 n \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}+\frac {b n \log (1-e x)}{4 x^2}-\frac {b e n}{2 x} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.16, 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 = {2385, 2395, 44, 2376, 2301, 2391} \[ -\frac {\text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {1}{4} b e^2 n \text {PolyLog}(2,e x)-\frac {b n \text {PolyLog}(2,e x)}{4 x^2}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac {1}{8} b e^2 n \log ^2(x)+\frac {1}{4} b e^2 n \log (x)-\frac {1}{4} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {b e n}{2 x} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 44
Rule 2301
Rule 2376
Rule 2385
Rule 2391
Rule 2395
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*}
________________________________________________________________________________________
Mathematica [A] time = 0.20, size = 163, normalized size = 0.81 \[ \frac {\left (e^2 x^2 \log (x)-e^2 x^2 \log (1-e x)-2 \text {Li}_2(e x)-e x+\log (1-e x)\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{4 x^2}+\frac {b n \left (-2 \text {Li}_2(e x) \left (e^2 x^2+2 \log (x)+1\right )+e^2 x^2 \log ^2(x)-2 e^2 x^2 \log (1-e x)-4 e x+2 \log (1-e x)-2 (e x-1) \log (x) ((e x+1) \log (1-e x)-e x)\right )}{8 x^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.94, size = 190, normalized size = 0.94 \[ \frac {b e^{2} n x^{2} \log \relax (x)^{2} - 2 \, {\left (2 \, b e n + a e\right )} x - 2 \, {\left (b e^{2} n x^{2} + b n + 2 \, a\right )} {\rm Li}_2\left (e x\right ) - 2 \, {\left ({\left (b e^{2} n + a e^{2}\right )} x^{2} - b n - a\right )} \log \left (-e x + 1\right ) - 2 \, {\left (b e x + 2 \, b {\rm Li}_2\left (e x\right ) + {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + 2 \, {\left (b e^{2} x^{2} \log \relax (c) - b e n x + {\left (b e^{2} n + a e^{2}\right )} x^{2} - 2 \, b n {\rm Li}_2\left (e x\right ) - {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x)}{8 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right )}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \polylog \left (2, e x \right )}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left (e^{2} \log \relax (x) - \frac {e x + {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right ) + 2 \, {\rm Li}_2\left (e x\right )}{x^{2}}\right )} a - \frac {1}{4} \, b {\left (\frac {{\left (n + 2 \, \log \relax (c) + 2 \, \log \left (x^{n}\right )\right )} {\rm Li}_2\left (e x\right ) - {\left (e^{2} n x^{2} \log \relax (x) + n + \log \relax (c)\right )} \log \left (-e x + 1\right ) - {\left (e^{2} x^{2} \log \relax (x) - e x - {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{x^{2}} + 4 \, \int -\frac {e^{2} n x - 2 \, e n - e \log \relax (c) - {\left (2 \, e^{3} n x^{2} - e^{2} n x\right )} \log \relax (x)}{4 \, {\left (e x^{3} - x^{2}\right )}}\,{d x}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \operatorname {Li}_{2}\left (e x\right )}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________