Integrand size = 16, antiderivative size = 102 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \]
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Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2320, 272, 36, 29, 31, 2463, 2438, 266, 2441, 2440} \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 272
Rule 2320
Rule 2438
Rule 2440
Rule 2441
Rule 2463
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\log (1+x)}{x \left (1+x^2\right )} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (\frac {\log (1+x)}{x}-\frac {x \log (1+x)}{1+x^2}\right ) \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {x \log (1+x)}{1+x^2} \, dx,x,e^x\right ) \\ & = -\operatorname {PolyLog}\left (2,-e^x\right )-\text {Subst}\left (\int \left (-\frac {\log (1+x)}{2 (i-x)}+\frac {\log (1+x)}{2 (i+x)}\right ) \, dx,x,e^x\right ) \\ & = -\operatorname {PolyLog}\left (2,-e^x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\log (1+x)}{i-x} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {\log (1+x)}{i+x} \, dx,x,e^x\right ) \\ & = -\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (i-x)\right )}{1+x} \, dx,x,e^x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) (i+x)\right )}{1+x} \, dx,x,e^x\right ) \\ & = -\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\log \left (1-\left (\frac {1}{2}+\frac {i}{2}\right ) x\right )}{x} \, dx,x,1+e^x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\log \left (1-\left (\frac {1}{2}-\frac {i}{2}\right ) x\right )}{x} \, dx,x,1+e^x\right ) \\ & = -\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \]
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Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}\) | \(83\) |
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\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
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\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\log {\left (e^{x} + 1 \right )}}{e^{2 x} + 1}\, dx \]
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\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
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\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
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Timed out. \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\ln \left ({\mathrm {e}}^x+1\right )}{{\mathrm {e}}^{2\,x}+1} \,d x \]
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