Integrand size = 144, antiderivative size = 24 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\frac {1}{\log \left (\left (2+\frac {x}{4}-\log \left (3+e^{e^x}\right )\right )^2\right )}} \]
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\[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \left (-3-e^{e^x}+4 e^{e^x+x}\right ) \log (20)}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx \\ & = \log (20) \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \left (-3-e^{e^x}+4 e^{e^x+x}\right )}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx \\ & = \log (20) \int \left (-\frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}}}{\left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}+\frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} e^{e^x+x}}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}\right ) \, dx \\ & = -\left (\log (20) \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}}}{\left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx\right )+\log (20) \int \frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} e^{e^x+x}}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 5.50
\[\left (\frac {1}{16}\right )^{\frac {1}{2 i \operatorname {csgn}\left (i {\left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}^{2}\right ) \pi -2 i \pi \,\operatorname {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )+8 \ln \left (2\right )-4 \ln \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}} \left (\frac {1}{25}\right )^{\frac {1}{2 i \operatorname {csgn}\left (i {\left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}^{2}\right ) \pi -2 i \pi \,\operatorname {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )+8 \ln \left (2\right )-4 \ln \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}}\]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )} \]
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Timed out. \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\int { \frac {2 \, {\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}} \,d x } \]
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\[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\int { \frac {2 \, {\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}} \,d x } \]
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Time = 11.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx={20}^{\frac {1}{\ln \left (\frac {x^2}{16}-\frac {x\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}{2}+x+{\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}^2-4\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )+4\right )}} \]
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