Optimal. Leaf size=102 \[ -\text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}+\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
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Rubi [A] time = 0.239488, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625 \[ -\text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}+\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]
Antiderivative was successfully verified.
[In] Int[Log[1 + E^x]/(1 + E^(2*x)),x]
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Rubi in Sympy [F] time = 0., size = 0, normalized size = 0. \[ - \frac{\log{\left (e^{x} + 1 \right )} \log{\left (e^{2 x} + 1 \right )}}{2} + \frac{\log{\left (e^{x} + 1 \right )} \log{\left (e^{2 x} \right )}}{2} + \frac{\int ^{e^{x}} \frac{- \log{\left (x^{2} \right )} + \log{\left (x^{2} + 1 \right )}}{x + 1}\, dx}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(ln(1+exp(x))/(1+exp(2*x)),x)
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Mathematica [A] time = 0.133675, size = 0, normalized size = 0. \[ \int \frac{\log \left (1+e^x\right )}{1+e^{2 x}} \, dx \]
Verification is Not applicable to the result.
[In] Integrate[Log[1 + E^x]/(1 + E^(2*x)),x]
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Maple [C] time = 0.073, size = 83, normalized size = 0.8 \[ -{\frac{\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}\ln \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}+{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}\ln \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}-{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}+{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}-{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\it dilog} \left ( 1+{{\rm e}^{x}} \right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(ln(1+exp(x))/(1+exp(2*x)),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(log(e^x + 1)/(e^(2*x) + 1),x, algorithm="maxima")
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Fricas [F] time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(log(e^x + 1)/(e^(2*x) + 1),x, algorithm="fricas")
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Sympy [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{\log{\left (e^{x} + 1 \right )}}{e^{2 x} + 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(ln(1+exp(x))/(1+exp(2*x)),x)
[Out]
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GIAC/XCAS [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(log(e^x + 1)/(e^(2*x) + 1),x, algorithm="giac")
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