Integrand size = 10, antiderivative size = 18 \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+\left (1+e^x\right ) \log \left (1+e^x\right ) \]
[Out]
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2225, 2634, 2280, 45} \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+e^x \log \left (e^x+1\right )+\log \left (e^x+1\right ) \]
[In]
[Out]
Rule 45
Rule 2225
Rule 2280
Rule 2634
Rubi steps \begin{align*} \text {integral}& = e^x \log \left (1+e^x\right )-\int \frac {e^{2 x}}{1+e^x} \, dx \\ & = e^x \log \left (1+e^x\right )-\text {Subst}\left (\int \frac {x}{1+x} \, dx,x,e^x\right ) \\ & = e^x \log \left (1+e^x\right )-\text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,e^x\right ) \\ & = -e^x+\log \left (1+e^x\right )+e^x \log \left (1+e^x\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+\left (1+e^x\right ) \log \left (1+e^x\right ) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\left (1+{\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{x}\right )-1-{\mathrm e}^{x}\) | \(17\) |
default | \(\left (1+{\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{x}\right )-1-{\mathrm e}^{x}\) | \(17\) |
norman | \({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )\) | \(19\) |
risch | \({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )\) | \(19\) |
parallelrisch | \({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )+1\) | \(20\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} \]
[In]
[Out]
Timed out. \[ \int e^x \log \left (1+e^x\right ) \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \]
[In]
[Out]
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (1+e^x\right ) \, dx=\ln \left ({\mathrm {e}}^x+1\right )-{\mathrm {e}}^x+{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+1\right ) \]
[In]
[Out]