Integrand size = 20, antiderivative size = 32 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx=-2 e^x+e^{-x} \log \left (1+e^{2 x}\right )+e^x \log \left (1+e^{2 x}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2320, 2526, 2498, 327, 209, 2505} \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx=-2 e^x+e^{-x} \log \left (e^{2 x}+1\right )+e^x \log \left (e^{2 x}+1\right ) \]
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Rule 209
Rule 327
Rule 2320
Rule 2498
Rule 2505
Rule 2526
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\left (-1+x^2\right ) \log \left (1+x^2\right )}{x^2} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (\log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{x^2}\right ) \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \log \left (1+x^2\right ) \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {\log \left (1+x^2\right )}{x^2} \, dx,x,e^x\right ) \\ & = e^{-x} \log \left (1+e^{2 x}\right )+e^x \log \left (1+e^{2 x}\right )-2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )-2 \text {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,e^x\right ) \\ & = -2 e^x-2 \arctan \left (e^x\right )+e^{-x} \log \left (1+e^{2 x}\right )+e^x \log \left (1+e^{2 x}\right )+2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right ) \\ & = -2 e^x+e^{-x} \log \left (1+e^{2 x}\right )+e^x \log \left (1+e^{2 x}\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx=-2 e^x+\left (e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\left (1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x} \ln \left (1+{\mathrm e}^{2 x}\right )-2 \,{\mathrm e}^{x}\) | \(24\) |
norman | \(\left ({\mathrm e}^{2 x} \ln \left (1+{\mathrm e}^{2 x}\right )-2 \,{\mathrm e}^{2 x}+\ln \left (1+{\mathrm e}^{2 x}\right )\right ) {\mathrm e}^{-x}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx={\left ({\left (e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{\left (2 \, x\right )} + 1\right ) - 2 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} \]
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Time = 81.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx=e^{x} \log {\left (e^{2 x} + 1 \right )} - 2 e^{x} + e^{- x} \log {\left (e^{2 x} + 1 \right )} \]
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Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx={\left (e^{\left (-x\right )} + e^{x}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right ) - 2 \, e^{x} \]
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Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx={\left (e^{\left (-x\right )} + e^{x}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right ) - 2 \, e^{x} \]
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Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \left (-e^{-x}+e^x\right ) \log \left (1+e^{2 x}\right ) \, dx=2\,\ln \left ({\mathrm {e}}^{2\,x}+1\right )\,\mathrm {cosh}\left (x\right )-\frac {{\mathrm {e}}^{2\,x}+1}{\mathrm {cosh}\left (x\right )} \]
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