Integrand size = 80, antiderivative size = 26 \[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=e^{4 \left (4 e^{-(-5+x)^2} x^2+\frac {2+x}{5}\right )} \]
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\[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=\int \frac {1}{5} \exp \left (-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )\right ) \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \exp \left (-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )\right ) \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx \\ & = \frac {1}{5} \int 4 \exp \left (-\frac {117}{5}+\frac {4 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \left (e^{25+x^2}-40 e^{10 x} x \left (-1-5 x+x^2\right )\right ) \, dx \\ & = \frac {4}{5} \int \exp \left (-\frac {117}{5}+\frac {4 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \left (e^{25+x^2}-40 e^{10 x} x \left (-1-5 x+x^2\right )\right ) \, dx \\ & = \frac {4}{5} \int \left (\exp \left (\frac {8}{5}+\frac {4 x}{5}+x^2+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right )-40 \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \left (-1-5 x+x^2\right )\right ) \, dx \\ & = \frac {4}{5} \int \exp \left (\frac {8}{5}+\frac {4 x}{5}+x^2+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \, dx-32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \left (-1-5 x+x^2\right ) \, dx \\ & = \frac {4}{5} \int e^{\frac {4}{5} \left (2+x+20 e^{-(-5+x)^2} x^2\right )} \, dx-32 \int \left (-\exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x-5 \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^2+\exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^3\right ) \, dx \\ & = \frac {4}{5} \int e^{\frac {4}{5} \left (2+x+20 e^{-(-5+x)^2} x^2\right )} \, dx+32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \, dx-32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^3 \, dx+160 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^2 \, dx \\ \end{align*}
Time = 2.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=e^{\frac {8}{5}+\frac {4 x}{5}+16 e^{-(-5+x)^2} x^2} \]
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Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \({\mathrm e}^{\frac {4 \left ({\mathrm e}^{\left (-5+x \right )^{2}} x +20 x^{2}+2 \,{\mathrm e}^{\left (-5+x \right )^{2}}\right ) {\mathrm e}^{-\left (-5+x \right )^{2}}}{5}}\) | \(34\) |
default | \({\mathrm e}^{\frac {\left (\left (4 x +8\right ) {\mathrm e}^{x^{2}-10 x +25}+80 x^{2}\right ) {\mathrm e}^{-x^{2}+10 x -25}}{5}}\) | \(36\) |
norman | \({\mathrm e}^{\frac {\left (\left (4 x +8\right ) {\mathrm e}^{x^{2}-10 x +25}+80 x^{2}\right ) {\mathrm e}^{-x^{2}+10 x -25}}{5}}\) | \(36\) |
parallelrisch | \({\mathrm e}^{\frac {\left (\left (4 x +8\right ) {\mathrm e}^{x^{2}-10 x +25}+80 x^{2}\right ) {\mathrm e}^{-x^{2}+10 x -25}}{5}}\) | \(36\) |
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=e^{\left (x^{2} + \frac {1}{5} \, {\left (80 \, x^{2} - {\left (5 \, x^{2} - 54 \, x + 117\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} - 10 \, x + 25\right )} \]
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Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=e^{\left (16 x^{2} + \frac {\left (4 x + 8\right ) e^{x^{2} - 10 x + 25}}{5}\right ) e^{- x^{2} + 10 x - 25}} \]
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\[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=\int { -\frac {4}{5} \, {\left (40 \, x^{3} - 200 \, x^{2} - 40 \, x - e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + \frac {4}{5} \, {\left (20 \, x^{2} + {\left (x + 2\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} + 10 \, x - 25\right )} \,d x } \]
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\[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx=\int { -\frac {4}{5} \, {\left (40 \, x^{3} - 200 \, x^{2} - 40 \, x - e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + \frac {4}{5} \, {\left (20 \, x^{2} + {\left (x + 2\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} + 10 \, x - 25\right )} \,d x } \]
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Time = 12.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx={\mathrm {e}}^{\frac {4\,x}{5}}\,{\mathrm {e}}^{8/5}\,{\mathrm {e}}^{16\,x^2\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-25}\,{\mathrm {e}}^{-x^2}} \]
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