Integrand size = 10, antiderivative size = 59 \[ \int x \log \left (a+b e^x\right ) \, dx=\frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )-x \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right ) \]
[Out]
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2612, 2611, 2320, 6724} \[ \int x \log \left (a+b e^x\right ) \, dx=-x \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )+\frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (\frac {b e^x}{a}+1\right ) \]
[In]
[Out]
Rule 2320
Rule 2611
Rule 2612
Rule 6724
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )+\int x \log \left (1+\frac {b e^x}{a}\right ) \, dx \\ & = \frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )-x \text {Li}_2\left (-\frac {b e^x}{a}\right )+\int \text {Li}_2\left (-\frac {b e^x}{a}\right ) \, dx \\ & = \frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )-x \text {Li}_2\left (-\frac {b e^x}{a}\right )+\text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a}\right )}{x} \, dx,x,e^x\right ) \\ & = \frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )-x \text {Li}_2\left (-\frac {b e^x}{a}\right )+\text {Li}_3\left (-\frac {b e^x}{a}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x \log \left (a+b e^x\right ) \, dx=\frac {1}{2} x^2 \log \left (a+b e^x\right )-\frac {1}{2} x^2 \log \left (1+\frac {b e^x}{a}\right )-x \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right ) \]
[In]
[Out]
Time = 0.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {x^{2} \ln \left (a +b \,{\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{2}-x \,\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(52\) |
risch | \(\frac {x^{2} \ln \left (a +b \,{\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{2}-x \,\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(52\) |
parts | \(\frac {x^{2} \ln \left (a +b \,{\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{2}-x \,\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(52\) |
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int x \log \left (a+b e^x\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (b e^{x} + a\right ) - \frac {1}{2} \, x^{2} \log \left (\frac {b e^{x} + a}{a}\right ) - x {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) + {\rm polylog}\left (3, -\frac {b e^{x}}{a}\right ) \]
[In]
[Out]
\[ \int x \log \left (a+b e^x\right ) \, dx=- \frac {b \int \frac {x^{2} e^{x}}{a + b e^{x}}\, dx}{2} + \frac {x^{2} \log {\left (a + b e^{x} \right )}}{2} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int x \log \left (a+b e^x\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (b e^{x} + a\right ) - \frac {1}{2} \, x^{2} \log \left (\frac {b e^{x}}{a} + 1\right ) - x {\rm Li}_2\left (-\frac {b e^{x}}{a}\right ) + {\rm Li}_{3}(-\frac {b e^{x}}{a}) \]
[In]
[Out]
\[ \int x \log \left (a+b e^x\right ) \, dx=\int { x \log \left (b e^{x} + a\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int x \log \left (a+b e^x\right ) \, dx=\int x\,\ln \left (a+b\,{\mathrm {e}}^x\right ) \,d x \]
[In]
[Out]