Integrand size = 232, antiderivative size = 29 \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \]
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\[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=\int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {125+15 e^{2 x^2}+e^{3 x^2}-e^{x^2} \left (-75-4 e^8 (-1+x) x^2\right )-\left (5+e^{x^2}\right )^3 (-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{\left (5+e^{x^2}\right )^3 (1-x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx \\ & = \int \left (\frac {20 e^8 x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}-\frac {4 e^8 x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}\right ) \, dx \\ & = -\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx \\ & = -\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \left (-\frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )\right ) \, dx \\ & = -\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx-\int \frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \, dx \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 302, normalized size of antiderivative = 10.41
\[\ln \left (-2 \ln \left (2\right )+\ln \left (x \right )+\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )-\ln \left (-1+x \right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right )+\operatorname {csgn}\left (\frac {i}{-1+x}\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right ) \left (-\operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right )+\operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{-1+x}\right )\right )}{2}\right ) x\]
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=x \log \left (\log \left (\frac {x e^{\left (\frac {e^{8}}{e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25}\right )}}{4 \, {\left (x - 1\right )}}\right )\right ) \]
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Timed out. \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.62 \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=x \log \left (-{\left (2 \, \log \left (2\right ) + \log \left (x - 1\right )\right )} e^{\left (2 \, x^{2}\right )} - 10 \, {\left (2 \, \log \left (2\right ) + \log \left (x - 1\right )\right )} e^{\left (x^{2}\right )} + {\left (e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25\right )} \log \left (x\right ) + e^{8} - 50 \, \log \left (2\right ) - 25 \, \log \left (x - 1\right )\right ) - 2 \, x \log \left (e^{\left (x^{2}\right )} + 5\right ) \]
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Timed out. \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx=-\int \frac {15\,{\mathrm {e}}^{2\,x^2}+{\mathrm {e}}^{3\,x^2}-{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^8\,\left (4\,x^2-4\,x^3\right )-75\right )-\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\right )\,\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )+125}{\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )} \,d x \]
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