Integrand size = 16, antiderivative size = 23 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {1}{3} \log \left (1-2 e^x\right )-\frac {1}{3} \log \left (1+e^x\right ) \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2320, 630, 31} \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {1}{3} \log \left (1-2 e^x\right )-\frac {1}{3} \log \left (e^x+1\right ) \]
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Rule 31
Rule 630
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{-1+x+2 x^2} \, dx,x,e^x\right ) \\ & = \frac {2}{3} \text {Subst}\left (\int \frac {1}{-1+2 x} \, dx,x,e^x\right )-\frac {2}{3} \text {Subst}\left (\int \frac {1}{2+2 x} \, dx,x,e^x\right ) \\ & = \frac {1}{3} \log \left (1-2 e^x\right )-\frac {1}{3} \log \left (1+e^x\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {2}{3} \text {arctanh}\left (\frac {1}{3}-\frac {2 e^{-x}}{3}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {\ln \left (-\frac {1}{2}+{\mathrm e}^{x}\right )}{3}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{3}\) | \(16\) |
parallelrisch | \(\frac {\ln \left (-\frac {1}{2}+{\mathrm e}^{x}\right )}{3}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{3}\) | \(16\) |
derivativedivides | \(\frac {\ln \left (2 \,{\mathrm e}^{x}-1\right )}{3}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{3}\) | \(18\) |
default | \(\frac {\ln \left (2 \,{\mathrm e}^{x}-1\right )}{3}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{3}\) | \(18\) |
norman | \(\frac {\ln \left (2 \,{\mathrm e}^{x}-1\right )}{3}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{3}\) | \(18\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {1}{3} \, \log \left (2 \, e^{x} - 1\right ) - \frac {1}{3} \, \log \left (e^{x} + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {\log {\left (-2 + e^{- x} \right )}}{3} - \frac {\log {\left (1 + e^{- x} \right )}}{3} \]
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Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=-\frac {1}{3} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {1}{3} \, \log \left (e^{\left (-x\right )} - 2\right ) \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=-\frac {1}{3} \, \log \left (e^{x} + 1\right ) + \frac {1}{3} \, \log \left ({\left | 2 \, e^{x} - 1 \right |}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1}{1-e^{-x}+2 e^x} \, dx=\frac {\ln \left (2\,{\mathrm {e}}^x-1\right )}{3}-\frac {\ln \left ({\mathrm {e}}^x+1\right )}{3} \]
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