Optimal. Leaf size=146 \[ -\frac{b n \text{PolyLog}(2,-e x)}{2 e^2}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{1}{2} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (e x+1)}{4 e^2}-\frac{1}{4} b n x^2 \log (e x+1)-\frac{3 b n x}{4 e}+\frac{1}{4} b n x^2 \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.0756235, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2395, 43, 2376, 2391} \[ -\frac{b n \text{PolyLog}(2,-e x)}{2 e^2}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{1}{2} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (e x+1)}{4 e^2}-\frac{1}{4} b n x^2 \log (e x+1)-\frac{3 b n x}{4 e}+\frac{1}{4} b n x^2 \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2395
Rule 43
Rule 2376
Rule 2391
Rubi steps
\begin{align*} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx &=\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 e^2}+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-(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{b n x}{2 e}+\frac{1}{8} b n x^2+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 e^2}+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{1}{2} (b n) \int x \log (1+e x) \, dx+\frac{(b n) \int \frac{\log (1+e x)}{x} \, dx}{2 e^2}\\ &=-\frac{b n x}{2 e}+\frac{1}{8} b n x^2+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\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)}{2 e^2}+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{b n \text{Li}_2(-e x)}{2 e^2}+\frac{1}{4} (b e n) \int \frac{x^2}{1+e x} \, dx\\ &=-\frac{b n x}{2 e}+\frac{1}{8} b n x^2+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\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)}{2 e^2}+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{b n \text{Li}_2(-e x)}{2 e^2}+\frac{1}{4} (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}{4 e}+\frac{1}{4} b n x^2+\frac{x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{1}{4} 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)}{2 e^2}+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{b n \text{Li}_2(-e x)}{2 e^2}\\ \end{align*}
Mathematica [A] time = 0.0598828, size = 131, normalized size = 0.9 \[ \frac{-2 b n \text{PolyLog}(2,-e x)-a e^2 x^2+2 a e^2 x^2 \log (e x+1)+2 a e x-2 a \log (e x+1)+b \left (2 \left (e^2 x^2-1\right ) \log (e x+1)+e x (2-e x)\right ) \log \left (c x^n\right )+b e^2 n x^2-b e^2 n x^2 \log (e x+1)-3 b e n x+b n \log (e x+1)}{4 e^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [C] time = 0.072, size = 725, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.34027, size = 240, normalized size = 1.64 \begin{align*} -\frac{{\left (\log \left (e x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-e x\right )\right )} b n}{2 \, e^{2}} + \frac{{\left (b{\left (n - 2 \, \log \left (c\right )\right )} - 2 \, a\right )} \log \left (e x + 1\right )}{4 \, e^{2}} - \frac{{\left (a e^{2} -{\left (e^{2} n - e^{2} \log \left (c\right )\right )} b\right )} x^{2} +{\left ({\left (3 \, e n - 2 \, e \log \left (c\right )\right )} b - 2 \, a e\right )} x -{\left ({\left (2 \, a e^{2} -{\left (e^{2} n - 2 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} + 2 \, b n \log \left (x\right )\right )} \log \left (e x + 1\right ) +{\left (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 \left (x^{n}\right )}{4 \, e^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a x \log \left (e x + 1\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left (e x + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]