Integrand size = 175, antiderivative size = 32 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=2 \left (-e^{\frac {2}{3-e^x \left (\log (3)-\frac {x}{\log (4)}\right )^2}}+\log (2)\right ) \]
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\[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=\int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}\right ) \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) \log ^2(4) \left (-x^2+\log (3) \log (4) (2-\log (3) \log (4))-x (2-\log (4) \log (9))\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2} \, dx \\ & = \left (4 \log ^2(4)\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) \left (-x^2+\log (3) \log (4) (2-\log (3) \log (4))-x (2-\log (4) \log (9))\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2} \, dx \\ & = \left (4 \log ^2(4)\right ) \int \left (-\frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x^2}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2}-\frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) \log (3) \log (4) (-2+\log (3) \log (4))}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2}+\frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x (-2+\log (4) \log (9))}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2}\right ) \, dx \\ & = -\left (\left (4 \log ^2(4)\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x^2}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2} \, dx\right )+\left (4 \log (3) \log ^3(4) (2-\log (3) \log (4))\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right )}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2} \, dx-\left (4 \log ^2(4) (2-\log (4) \log (9))\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x}{\left (e^x x^2-2 e^x x \log (3) \log (4)-3 \log ^2(4)+e^x \log ^2(3) \log ^2(4)\right )^2} \, dx \\ & = -\left (\left (4 \log ^2(4)\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x^2}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2} \, dx\right )+\left (4 \log (3) \log ^3(4) (2-\log (3) \log (4))\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2} \, dx-\left (4 \log ^2(4) (2-\log (4) \log (9))\right ) \int \frac {\exp \left (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x-\log (3) \log (4))^2}\right ) x}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2} \, dx \\ \end{align*}
\[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=\int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx \]
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Time = 18.94 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44
method | result | size |
risch | \(-2 \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{4 \,{\mathrm e}^{x} \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 \,{\mathrm e}^{x} \ln \left (3\right ) \ln \left (2\right ) x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}}\) | \(46\) |
parallelrisch | \(-2 \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{4 \,{\mathrm e}^{x} \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 \,{\mathrm e}^{x} \ln \left (3\right ) \ln \left (2\right ) x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}}\) | \(46\) |
norman | \(\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}-2 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}-8 \,{\mathrm e}^{x} \ln \left (3\right )^{2} \ln \left (2\right )^{2} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}+8 \,{\mathrm e}^{x} \ln \left (3\right ) \ln \left (2\right ) x \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}}{4 \,{\mathrm e}^{x} \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 \,{\mathrm e}^{x} \ln \left (3\right ) \ln \left (2\right ) x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}\) | \(233\) |
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Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.56 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-x - \frac {4 \, {\left (3 \, x + 2\right )} \log \left (2\right )^{2} - {\left (4 \, x \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x^{2} \log \left (3\right ) \log \left (2\right ) + x^{3}\right )} e^{x}}{{\left (4 \, \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x \log \left (3\right ) \log \left (2\right ) + x^{2}\right )} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \]
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Time = 0.41 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=- 2 e^{- \frac {8 \log {\left (2 \right )}^{2}}{\left (x^{2} - 4 x \log {\left (2 \right )} \log {\left (3 \right )} + 4 \log {\left (2 \right )}^{2} \log {\left (3 \right )}^{2}\right ) e^{x} - 12 \log {\left (2 \right )}^{2}}} \]
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Time = 0.54 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-\frac {8 \, \log \left (2\right )^{2}}{{\left (4 \, \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x \log \left (3\right ) \log \left (2\right ) + x^{2}\right )} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \]
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Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-\frac {8 \, \log \left (2\right )^{2}}{4 \, e^{x} \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x e^{x} \log \left (3\right ) \log \left (2\right ) + x^{2} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \]
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Time = 9.45 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2\,{\mathrm {e}}^{-\frac {8\,{\ln \left (2\right )}^2}{x^2\,{\mathrm {e}}^x-12\,{\ln \left (2\right )}^2+4\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2\,{\ln \left (3\right )}^2-4\,x\,{\mathrm {e}}^x\,\ln \left (2\right )\,\ln \left (3\right )}} \]
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