Integrand size = 121, antiderivative size = 32 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{x-\frac {x^2 (1-2 x-\log (5))}{3 \left (1-e^{20}-x\right )}} \]
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\[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=\int \frac {\exp \left (\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}\right ) \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{3 \left (1-e^{20}\right )^2-6 \left (1-e^{20}\right ) x+3 x^2} \, dx \\ & = \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{3 \left (-1+e^{20}+x\right )^2} \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{\left (-1+e^{20}+x\right )^2} \, dx \\ & = \frac {1}{3} \int \left (-4 \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) x-\frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (-1+e^{20}\right )^2 \left (-1+2 e^{20}-\log (5)\right )}{\left (-1+e^{20}+x\right )^2}+2 \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (1+e^{20}-\frac {\log (5)}{2}\right )\right ) \, dx \\ & = -\left (\frac {4}{3} \int \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) x \, dx\right )+\frac {1}{3} \left (2+2 e^{20}-\log (5)\right ) \int \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \, dx+\frac {1}{3} \left (\left (1-e^{20}\right )^2 \left (1-2 e^{20}+\log (5)\right )\right ) \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right )}{\left (-1+e^{20}+x\right )^2} \, dx \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=5^{-\frac {x^2}{3 \left (-1+e^{20}+x\right )}} e^{-\frac {x \left (3-3 e^{20}-4 x+2 x^2\right )}{3 \left (-1+e^{20}+x\right )}} \]
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Time = 7.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
method | result | size |
gosper | \({\mathrm e}^{-\frac {x \left (x \ln \left (5\right )+2 x^{2}-3 \,{\mathrm e}^{20}-4 x +3\right )}{3 \left ({\mathrm e}^{20}+x -1\right )}}\) | \(30\) |
risch | \({\mathrm e}^{\frac {\left (-x \ln \left (5\right )-2 x^{2}+3 \,{\mathrm e}^{20}+4 x -3\right ) x}{3 \,{\mathrm e}^{20}+3 x -3}}\) | \(31\) |
parallelrisch | \({\mathrm e}^{\frac {\left (-x \ln \left (5\right )-2 x^{2}+3 \,{\mathrm e}^{20}+4 x -3\right ) x}{3 \,{\mathrm e}^{20}+3 x -3}}\) | \(31\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-x^{2} \ln \left (5\right )+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}+\left ({\mathrm e}^{20}-1\right ) {\mathrm e}^{\frac {-x^{2} \ln \left (5\right )+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}}{{\mathrm e}^{20}+x -1}\) | \(95\) |
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Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\left (-\frac {2 \, x^{3} + x^{2} \log \left (5\right ) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \]
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Time = 0.39 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\frac {- 2 x^{3} - x^{2} \log {\left (5 \right )} + 4 x^{2} - 3 x + 3 x e^{20}}{3 x - 3 + 3 e^{20}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (22) = 44\).
Time = 0.74 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.59 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=5^{\frac {1}{3} \, e^{20} - \frac {1}{3}} e^{\left (-\frac {2}{3} \, x^{2} + \frac {2}{3} \, x e^{20} - \frac {1}{3} \, x \log \left (5\right ) + \frac {2}{3} \, x - \frac {e^{40} \log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{20} \log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{60}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {5 \, e^{40}}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {4 \, e^{20}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {\log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {1}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {2}{3} \, e^{40} + e^{20} - \frac {1}{3}\right )} \]
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Time = 0.59 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\left (-\frac {2 \, x^{3} + x^{2} \log \left (5\right ) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \]
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Time = 14.73 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.69 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=\frac {{\mathrm {e}}^{-\frac {3\,x}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{-\frac {2\,x^3}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {4\,x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {3\,x\,{\mathrm {e}}^{20}}{3\,x+3\,{\mathrm {e}}^{20}-3}}}{5^{\frac {x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}} \]
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