Optimal. Leaf size=102 \[ -\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\text {Li}_2\left (-e^x\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \]
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Rubi [A]
time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2320, 272,
36, 29, 31, 2463, 2438, 266, 2441, 2440} \begin {gather*} -\text {Li}_2\left (-e^x\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \text {Li}_2\left (\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 ) \end {gather*}
Antiderivative was successfully verified.
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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*} \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx &=\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 )\\ &=-\text {Li}_2\left (-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 )\\ &=-\text {Li}_2\left (-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 )-\text {Li}_2\left (-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 )-\text {Li}_2\left (-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 )-\text {Li}_2\left (-e^x\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right )\\ \end {align*}
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Mathematica [A]
time = 0.01, size = 102, normalized size = 1.00 \begin {gather*} -\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 )-\text {Li}_2\left (-e^x\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \text {Li}_2\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.02, size = 83, normalized size = 0.81
method | result | size |
risch | \(-\dilog \left (1+{\mathrm e}^{x}\right )-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\dilog \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\dilog \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}\) | \(83\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left ({\mathrm {e}}^x+1\right )}{{\mathrm {e}}^{2\,x}+1} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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