Integrand size = 265, antiderivative size = 44 \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\frac {e^{-x+\frac {(1-x) \left (-e^x+x\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}}}{x} \]
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Timed out. \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
\[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 3.84
\[\frac {{\mathrm e}^{-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 x \ln \left (3\right )-4 x \ln \left (x \right )+2 x \ln \left (-x^{2}+1\right )+2 \,{\mathrm e}^{x} x -2 x^{2}-2 \,{\mathrm e}^{x}+2 x}{i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 \ln \left (3\right )-4 \ln \left (x \right )+2 \ln \left (-x^{2}+1\right )}}}{x}\]
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Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.34 \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\frac {e^{\left (-\frac {x^{2} - {\left (x - 1\right )} e^{x} + x \log \left (\frac {1}{3} \, x^{2}\right ) - x \log \left (-x^{2} + 1\right ) - x}{\log \left (\frac {1}{3} \, x^{2}\right ) - \log \left (-x^{2} + 1\right )}\right )}}{x} \]
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Time = 8.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\frac {e^{\frac {- x^{2} - x \log {\left (\frac {x^{2}}{3} \right )} + x \log {\left (1 - x^{2} \right )} + x + \left (x - 1\right ) e^{x}}{\log {\left (\frac {x^{2}}{3} \right )} - \log {\left (1 - x^{2} \right )}}}}{x} \]
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\[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{3} \, x^{2}\right )^{2} + {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x^{2} + 1\right )^{2} + {\left (2 \, x^{3} + x^{2} - {\left (x^{3} + x^{2}\right )} e^{x} - 2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x^{2} + 1\right ) - x\right )} \log \left (\frac {1}{3} \, x^{2}\right ) - {\left (2 \, x^{3} + x^{2} - {\left (x^{3} + x^{2}\right )} e^{x} - x\right )} \log \left (-x^{2} + 1\right ) + 2 \, x - 2 \, e^{x}\right )} e^{\left (-\frac {x^{2} - {\left (x - 1\right )} e^{x} + x \log \left (\frac {1}{3} \, x^{2}\right ) - x \log \left (-x^{2} + 1\right ) - x}{\log \left (\frac {1}{3} \, x^{2}\right ) - \log \left (-x^{2} + 1\right )}\right )}}{{\left (x^{3} + x^{2}\right )} \log \left (\frac {1}{3} \, x^{2}\right )^{2} - 2 \, {\left (x^{3} + x^{2}\right )} \log \left (\frac {1}{3} \, x^{2}\right ) \log \left (-x^{2} + 1\right ) + {\left (x^{3} + x^{2}\right )} \log \left (-x^{2} + 1\right )^{2}} \,d x } \]
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Time = 4.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\frac {e^{\left (-x + \frac {x^{2} - x e^{x} - x + e^{x}}{\log \left (3\right ) - \log \left (x^{2}\right ) + \log \left (-x^{2} + 1\right )}\right )}}{x} \]
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Time = 9.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 3.57 \[ \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx=\frac {{\mathrm {e}}^{-\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {x^2}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}\,{\left (\frac {x^2}{3}\right )}^{\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}}{x\,{\left (1-x^2\right )}^{\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \left (3\right )}}} \]
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