Integrand size = 109, antiderivative size = 21 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (4 \left (-4+\frac {2}{\log \left (e^x-x\right )}\right )\right ) \]
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\[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=\int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-\left (\left (-e^x+x\right ) \log \left (e^x-x\right )\right )-\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )-\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (e^x-x\right ) \left (1-2 \log \left (e^x-x\right )\right ) \log \left (e^x-x\right )} \, dx \\ & = \int \left (\frac {2 (-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx \\ & = 2 \int \frac {(-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+\int \frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx \\ & = 2 \int \left (-\frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx+\int \left (1+\frac {2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx \\ & = x+2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx-2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+2 \int \frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.97 (sec) , antiderivative size = 567, normalized size of antiderivative = 27.00
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=\log \left (-\frac {8 \, {\left (2 \, \log \left (-x + e^{x}\right ) - 1\right )}}{\log \left (-x + e^{x}\right )}\right )^{2} + x \]
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Time = 0.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x + \log {\left (\frac {8 - 16 \log {\left (- x + e^{x} \right )}}{\log {\left (- x + e^{x} \right )}} \right )}^{2} \]
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Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.52 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-2 \, {\left (-i \, \pi - 3 \, \log \left (2\right ) + \log \left (\log \left (-x + e^{x}\right )\right )\right )} \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) + \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} - 2 \, {\left (i \, \pi + 3 \, \log \left (2\right )\right )} \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]
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Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (18) = 36\).
Time = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.62 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-\log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} + 2 \, \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) - 2 \, \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]
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Time = 9.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx={\ln \left (\frac {8}{\ln \left ({\mathrm {e}}^x-x\right )}-16\right )}^2+x \]
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