Integrand size = 98, antiderivative size = 26 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \]
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\[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx \\ & = \int \left (-e^{e^x+x} \log (16) \log \left (\log \left (e^{-x} x\right )\right )+\frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx \\ & = -\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx \\ & = -\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {-e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )-2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(5-x) x \log \left (e^{-x} x\right )} \, dx \\ & = -\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \left (\frac {e^{e^x} (-1+x) \log (16)}{x \log \left (e^{-x} x\right )}+\frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx \\ & = \log (16) \int \frac {e^{e^x} (-1+x)}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx \\ & = \log (16) \int \left (\frac {e^{e^x}}{\log \left (e^{-x} x\right )}-\frac {e^{e^x}}{x \log \left (e^{-x} x\right )}\right ) \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {\frac {\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}-2 \log \left (\log \left (e^{-x} x\right )\right )}{5-x} \, dx \\ & = \log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \left (-\frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}+\frac {2 \log \left (\log \left (e^{-x} x\right )\right )}{-5+x}\right ) \, dx \\ & = 2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx \\ & = 2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \left (\frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )}-\frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}\right ) \, dx \\ & = 2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )} \, dx+\int \frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 220, normalized size of antiderivative = 8.46
\[\left (-4 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (-5+x \right )\right ) \ln \left (\ln \left (x \right )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right ) \left (\operatorname {csgn}\left (i \left (-5+x \right )\right )^{2}-2 \,\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (-5+x \right )\right )+\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right )^{2}\right ) \ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-2 i \ln \left (x \right )\right )}{2}\right )}{2}\]
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=-{\left (4 \, e^{\left (x + e^{x}\right )} \log \left (2\right ) - e^{x} \log \left (x^{2} - 10 \, x + 25\right )\right )} e^{\left (-x\right )} \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) \]
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Timed out. \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\text {Timed out} \]
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Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=-2 \, {\left (2 \, e^{\left (e^{x}\right )} \log \left (2\right ) - \log \left (x - 5\right )\right )} \log \left (-x + \log \left (x\right )\right ) \]
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\[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\int { \frac {4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (e^{x}\right )} \log \left (2\right ) - 2 \, {\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \left (2\right ) - x\right )} \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) - {\left (x^{2} - 6 \, x + 5\right )} \log \left (x^{2} - 10 \, x + 25\right )}{{\left (x^{2} - 5 \, x\right )} \log \left (x e^{\left (-x\right )}\right )} \,d x } \]
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Time = 9.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\ln \left (x^2-10\,x+25\right )\,\ln \left (\ln \left (x\right )-x\right )-4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \left (2\right )\,\ln \left (\ln \left (x\right )-x\right ) \]
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