Integrand size = 131, antiderivative size = 26 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{e^{\frac {e^{4 e^x+x}}{\left (4-\log \left (x^2\right )\right )^2}}} \]
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\[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}\right ) \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \left (-4 \left (1+x+4 e^x x\right )+\left (1+4 e^x\right ) x \log \left (x^2\right )\right )}{x \left (4-\log \left (x^2\right )\right )^3} \, dx \\ & = \int \left (\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3}+\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3}-\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2}-\frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \log \left (x^2\right )}{\left (-4+\log \left (x^2\right )\right )^3}\right ) \, dx \\ & = 4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx+4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \log \left (x^2\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx \\ & = 4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3} \, dx+4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \left (\frac {4 \exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^3}+\frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2}\right ) \, dx \\ & = 4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{x \left (-4+\log \left (x^2\right )\right )^3} \, dx-4 \int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+2 x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx-\int \frac {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}+x+\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}\right )}{\left (-4+\log \left (x^2\right )\right )^2} \, dx \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73
\[-{\mathrm e}^{{\mathrm e}^{\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{x}+x}}{{\left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+4 \ln \left (x \right )-8\right )}^{2}}}}\]
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Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (20) = 40\).
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.62 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{\left (\frac {4 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )} - 32 \, e^{x} \log \left (x^{2}\right ) + e^{\left (x + 4 \, e^{x}\right )} + 64 \, e^{x}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - \frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - 4 \, e^{x}\right )} \]
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Timed out. \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=\text {Timed out} \]
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Time = 0.83 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{\left (e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{4 \, {\left (\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4\right )}}\right )}\right )} \]
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\[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=\int { \frac {{\left (16 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x + 1\right )} e^{x} - {\left (4 \, x e^{\left (2 \, x\right )} + x e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} + 4 \, e^{x} + e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )}\right )}}{x \log \left (x^{2}\right )^{3} - 12 \, x \log \left (x^{2}\right )^{2} + 48 \, x \log \left (x^{2}\right ) - 64 \, x} \,d x } \]
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Time = 15.59 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^x}{{\ln \left (x^2\right )}^2-8\,\ln \left (x^2\right )+16}}} \]
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