Optimal. Leaf size=185 \[ -\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}+\frac {1}{2} x^2 \text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {1}{4} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2(e x)}{4 e^2}+\frac {b n \log (1-e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} b n x^2 \log (1-e x)+\frac {b n x}{2 e}+\frac {3}{16} b n x^2 \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.13, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2385, 2395, 43, 2376, 2391} \[ \frac {1}{2} x^2 \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {PolyLog}(2,e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {PolyLog}(2,e x)-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}+\frac {1}{4} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{4 e^2}-\frac {1}{4} b n x^2 \log (1-e x)+\frac {b n x}{2 e}+\frac {3}{16} b n x^2 \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 43
Rule 2376
Rule 2385
Rule 2391
Rule 2395
Rubi steps
\begin {align*} \int x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \, dx &=-\frac {1}{4} b n x^2 \text {Li}_2(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)+\frac {1}{2} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx-\frac {1}{4} (b n) \int x \log (1-e x) \, dx\\ &=-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} 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}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{2} (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-\frac {1}{8} (b e n) \int \frac {x^2}{1-e x} \, dx\\ &=\frac {b n x}{4 e}+\frac {1}{16} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} 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}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} (b n) \int x \log (1-e x) \, dx+\frac {(b n) \int \frac {\log (1-e x)}{x} \, dx}{4 e^2}-\frac {1}{8} (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}{8 e}+\frac {1}{8} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{8 e^2}-\frac {1}{4} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {Li}_2(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{8} (b e n) \int \frac {x^2}{1-e x} \, dx\\ &=\frac {3 b n x}{8 e}+\frac {1}{8} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{8 e^2}-\frac {1}{4} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {Li}_2(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{8} (b e n) \int \left (-\frac {1}{e^2}-\frac {x}{e}-\frac {1}{e^2 (-1+e x)}\right ) \, dx\\ &=\frac {b n x}{2 e}+\frac {3}{16} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{4 e^2}-\frac {1}{4} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {Li}_2(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.34, size = 168, normalized size = 0.91 \[ \frac {\left (4 e^2 x^2 \text {Li}_2(e x)+2 \left (e^2 x^2-1\right ) \log (1-e x)-e x (e x+2)\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{8 e^2}+\frac {b n \left (\text {Li}_2(e x) \left (-4 e^2 x^2+8 e^2 x^2 \log (x)-4\right )+3 e^2 x^2-4 e^2 x^2 \log (1-e x)+\log (x) \left (4 \left (e^2 x^2-1\right ) \log (1-e x)-2 e x (e x+2)\right )+8 e x+4 \log (1-e x)\right )}{16 e^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.56, size = 207, normalized size = 1.12 \[ \frac {{\left (3 \, b e^{2} n - 2 \, a e^{2}\right )} x^{2} + 4 \, {\left (2 \, b e n - a e\right )} x - 4 \, {\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2} + b n\right )} {\rm Li}_2\left (e x\right ) - 4 \, {\left ({\left (b e^{2} n - a e^{2}\right )} x^{2} - b n + a\right )} \log \left (-e x + 1\right ) + 2 \, {\left (4 \, b e^{2} x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} x^{2} - 2 \, b e x + 2 \, {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + 2 \, {\left (4 \, b e^{2} n x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} n x^{2} - 2 \, b e n x + 2 \, {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x)}{16 \, e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} x {\rm Li}_2\left (e x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.25, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) x \polylog \left (2, e x \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, b {\left (\frac {2 \, {\left (2 \, e^{2} x^{2} \log \left (x^{n}\right ) - {\left (e^{2} n - 2 \, e^{2} \log \relax (c)\right )} x^{2}\right )} {\rm Li}_2\left (e x\right ) - 2 \, {\left ({\left (e^{2} n - e^{2} \log \relax (c)\right )} x^{2} - n \log \relax (x)\right )} \log \left (-e x + 1\right ) - {\left (e^{2} x^{2} + 2 \, e x - 2 \, {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{e^{2}} - 8 \, \int -\frac {e n x + {\left (3 \, e^{2} n - 2 \, e^{2} \log \relax (c)\right )} x^{2} - 2 \, n \log \relax (x) - 2 \, n}{8 \, {\left (e^{2} x - e\right )}}\,{d x}\right )} + \frac {{\left (4 \, e^{2} x^{2} {\rm Li}_2\left (e x\right ) - e^{2} x^{2} - 2 \, e x + 2 \, {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} a}{8 \, e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 83.20, size = 264, normalized size = 1.43 \[ \begin {cases} - \frac {a x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {a x^{2} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {a x^{2}}{8} - \frac {a x}{4 e} + \frac {a \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} - \frac {b n x^{2} \log {\relax (x )} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {b n x^{2} \log {\relax (x )} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {b n x^{2} \log {\relax (x )}}{8} + \frac {b n x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} - \frac {b n x^{2} \operatorname {Li}_{2}\left (e x\right )}{4} + \frac {3 b n x^{2}}{16} - \frac {b x^{2} \log {\relax (c )} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {b x^{2} \log {\relax (c )} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {b x^{2} \log {\relax (c )}}{8} - \frac {b n x \log {\relax (x )}}{4 e} + \frac {b n x}{2 e} - \frac {b x \log {\relax (c )}}{4 e} + \frac {b n \log {\relax (x )} \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} - \frac {b n \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{4 e^{2}} + \frac {b \log {\relax (c )} \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________