Integrand size = 171, antiderivative size = 28 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=e^{\frac {e^{-x} (5+x)}{x+\frac {4 \left (2+\log \left (x^2\right )\right )}{x^3}}}+x \]
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\[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=\int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (e^x \left (8+x^4\right )^2+8 e^x \left (8+x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )-e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6+4 \left (-15+x+x^2\right ) \log \left (x^2\right )\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx \\ & = \int \left (1-\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = x-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx \\ & = x-\int \left (\frac {4 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx \\ & = x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )} \, dx \\ & = x-4 \int \left (\frac {10 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {2 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {5 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx-\int \left (-\frac {15 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx \\ & = x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-8 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx+15 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )} \, dx-20 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-40 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )} \, dx \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}}+x \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.68
\[x +{\mathrm e}^{\frac {x^{3} \left (5+x \right ) {\mathrm e}^{-x}}{-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-2 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+x^{4}+8 \ln \left (x \right )+8}}\]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=x + e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\right )} \]
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Time = 0.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=x + e^{\frac {x^{4} + 5 x^{3}}{\left (x^{4} + 8\right ) e^{x} + 4 e^{x} \log {\left (x^{2} \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.14 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx={\left (x e^{\left (\frac {8 \, \log \left (x\right )}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \left (x\right )} + \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \left (x\right )}\right )} + e^{\left (\frac {5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \left (x\right )} + e^{\left (-x\right )}\right )}\right )} e^{\left (-\frac {8 \, \log \left (x\right )}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \left (x\right )} - \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \left (x\right )}\right )} \]
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\[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=\int { \frac {8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x} - {\left (x^{8} + 5 \, x^{7} + 5 \, x^{6} + 8 \, x^{4} + 16 \, x^{3} - 80 \, x^{2} + 4 \, {\left (x^{4} + x^{3} - 15 \, x^{2}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\right )}}{8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x}} \,d x } \]
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Time = 9.92 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx=x+{\mathrm {e}}^{\frac {x^4}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {5\,x^3}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}} \]
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