Integrand size = 42, antiderivative size = 17 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (e+x)} \]
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Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2326} \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (x+e)} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{2-2 x}}{2 \log (e+x)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (e+x)} \]
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Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94
method | result | size |
norman | \(\frac {{\mathrm e}^{2} {\mathrm e}^{-2 x}}{2 \ln \left (x +{\mathrm e}\right )}\) | \(16\) |
risch | \(\frac {{\mathrm e}^{2-2 x}}{2 \ln \left (x +{\mathrm e}\right )}\) | \(16\) |
parallelrisch | \(\frac {{\mathrm e}^{2} {\mathrm e}^{-2 x}}{2 \ln \left (x +{\mathrm e}\right )}\) | \(16\) |
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none
Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2} e^{- 2 x}}{2 \log {\left (x + e \right )}} \]
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none
Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]
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none
Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]
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Time = 9.50 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2}{2\,\ln \left (x+\mathrm {e}\right )} \]
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