Optimal. Leaf size=102 \[ -\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
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Rubi [A] time = 0.15, 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 = {2282, 266, 36, 29, 31, 2416, 2391, 260, 2394, 2393} \[ -\text {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \text {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \text {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 260
Rule 266
Rule 2282
Rule 2391
Rule 2393
Rule 2394
Rule 2416
Rubi steps
\begin {align*} \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx &=\operatorname {Subst}\left (\int \frac {\log (1+x)}{x \left (1+x^2\right )} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {\log (1+x)}{x}-\frac {x \log (1+x)}{1+x^2}\right ) \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^x\right )-\operatorname {Subst}\left (\int \frac {x \log (1+x)}{1+x^2} \, dx,x,e^x\right )\\ &=-\text {Li}_2\left (-e^x\right )-\operatorname {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} \operatorname {Subst}\left (\int \frac {\log (1+x)}{i-x} \, dx,x,e^x\right )-\frac {1}{2} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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 = 0.02, size = 102, normalized size = 1.00 \[ -\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (e^x+1\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 83, normalized size = 0.81 \[ -\frac {\ln \left ({\mathrm e}^{x}+1\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}+\frac {1}{2}-\frac {i \left ({\mathrm e}^{x}+1\right )}{2}\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}+1\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}+\frac {1}{2}+\frac {i \left ({\mathrm e}^{x}+1\right )}{2}\right )}{2}-\dilog \left ({\mathrm e}^{x}+1\right )-\frac {\dilog \left (-\frac {{\mathrm e}^{x}}{2}+\frac {1}{2}-\frac {i \left ({\mathrm e}^{x}+1\right )}{2}\right )}{2}-\frac {\dilog \left (-\frac {{\mathrm e}^{x}}{2}+\frac {1}{2}+\frac {i \left ({\mathrm e}^{x}+1\right )}{2}\right )}{2} \]
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 \[ \int \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \]
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 \[ \int \frac {\ln \left ({\mathrm {e}}^x+1\right )}{{\mathrm {e}}^{2\,x}+1} \,d x \]
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 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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