Integrand size = 120, antiderivative size = 25 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=e^{5+e^x+x} x \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \]
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\[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=\int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int e^{5+e^x+x} \left (2 \left (2+x+e^x x\right )-\frac {2 x}{(-2+x) \log (-2+x)}-\left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx \\ & = \int \left (2 e^{5+e^x+x} \left (2+x+e^x x\right )-\frac {2 e^{5+e^x+x} x}{(-2+x) \log (-2+x)}-e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx \\ & = 2 \int e^{5+e^x+x} \left (2+x+e^x x\right ) \, dx-2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-\int e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right ) \, dx \\ & = -\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int \left (2 e^{5+e^x+x}+e^{5+e^x+x} x+e^{5+e^x+2 x} x\right ) \, dx-2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx+\int 2 e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx \\ & = -\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+4 \int e^{5+e^x+x} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx \\ & = -\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (-e^{5+e^x+x}+\frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx+4 \text {Subst}\left (\int e^{5+x} \, dx,x,e^x\right ) \\ & = 4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}-2 \int e^{5+e^x+x} \, dx+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx \\ & = 4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-2 \text {Subst}\left (\int e^{5+x} \, dx,x,e^x\right )-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx \\ & = 2 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx \\ \end{align*}
Time = 5.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=-e^{5+e^x+x} x \left (-2+\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 627, normalized size of antiderivative = 25.08
\[\text {Expression too large to display}\]
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=x e^{\left (x + e^{x} + \log \left (-\log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) + 2\right ) + 5\right )} \]
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Timed out. \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=\text {Timed out} \]
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Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=-2 \, x e^{\left (x + e^{x} + 5\right )} \log \left (\log \left (x - 2\right )\right ) + 2 \, {\left (x e^{5} \log \left (x\right ) + x e^{5}\right )} e^{\left (x + e^{x}\right )} \]
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\[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=\int { \frac {{\left ({\left (x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - x - 2\right )} \log \left (x - 2\right ) \log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) - 2 \, {\left (x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - 4\right )} \log \left (x - 2\right ) + 2 \, x\right )} e^{\left (x + e^{x} + \log \left (-\log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) + 2\right ) + 5\right )}}{{\left (x - 2\right )} \log \left (x - 2\right ) \log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) - 2 \, {\left (x - 2\right )} \log \left (x - 2\right )} \,d x } \]
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Timed out. \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=-\int \frac {{\mathrm {e}}^{x+\ln \left (2-\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\right )+{\mathrm {e}}^x+5}\,\left (2\,x+\ln \left (x-2\right )\,\left ({\mathrm {e}}^x\,\left (4\,x-2\,x^2\right )-2\,x^2+8\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x+{\mathrm {e}}^x\,\left (2\,x-x^2\right )-x^2+2\right )\right )}{\ln \left (x-2\right )\,\left (2\,x-4\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x-2\right )} \,d x \]
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