Integrand size = 60, antiderivative size = 17 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \left (e^7+x\right )}{\log \left (e^x-x\right )} \]
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\[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 (-1+x) \left (e^7+x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}-\frac {2 \left (e^7+x-\log \left (e^x-x\right )\right )}{\log ^2\left (e^x-x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {(-1+x) \left (e^7+x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {e^7+x-\log \left (e^x-x\right )}{\log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \left (-\frac {e^7}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}+\frac {\left (-1+e^7\right ) x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}+\frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}\right ) \, dx\right )-2 \int \left (\frac {e^7+x}{\log ^2\left (e^x-x\right )}-\frac {1}{\log \left (e^x-x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {e^7+x}{\log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \left (\frac {e^7}{\log ^2\left (e^x-x\right )}+\frac {x}{\log ^2\left (e^x-x\right )}\right ) \, dx\right )-2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \frac {x}{\log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx-\left (2 e^7\right ) \int \frac {1}{\log ^2\left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \left (e^7+x\right )}{\log \left (e^x-x\right )} \]
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Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\) | \(16\) |
norman | \(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\) | \(19\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\) | \(19\) |
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Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 x + 2 e^{7}}{\log {\left (- x + e^{x} \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]
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Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]
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Time = 11.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2\,\left (x+{\mathrm {e}}^7\right )}{\ln \left ({\mathrm {e}}^x-x\right )} \]
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