Integrand size = 104, antiderivative size = 28 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x-\frac {5 e^{-\frac {x^2}{(1-x)^2}} x}{(4+x) \log (5)} \]
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\[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{-16+40 x-25 x^2-5 x^3+5 x^4+x^5} \, dx}{\log (5)} \\ & = \frac {\int \frac {e^{-\frac {x^2}{(-1+x)^2}} \left (-20+60 x-20 x^2+30 x^3-e^{\frac {x^2}{(-1+x)^2}} (-1+x)^3 (4+x)^2 \log (5)\right )}{(1-x)^3 (4+x)^2} \, dx}{\log (5)} \\ & = \frac {\int \left (\frac {20 e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3 (4+x)^2}-\frac {60 e^{-\frac {x^2}{(-1+x)^2}} x}{(-1+x)^3 (4+x)^2}+\frac {20 e^{-\frac {x^2}{(-1+x)^2}} x^2}{(-1+x)^3 (4+x)^2}-\frac {30 e^{-\frac {x^2}{(-1+x)^2}} x^3}{(-1+x)^3 (4+x)^2}+\log (5)\right ) \, dx}{\log (5)} \\ & = x+\frac {20 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}+\frac {20 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x^2}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}-\frac {30 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x^3}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}-\frac {60 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)} \\ & = x+\frac {20 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}-\frac {16 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}+\frac {20 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}-\frac {2 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}-\frac {e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}-\frac {30 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {13 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {48 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}+\frac {64 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {48 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}-\frac {60 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}-\frac {7 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}+\frac {4 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}+\frac {7 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)} \\ & = x+\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {4 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}+\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {8 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}+\frac {84 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {84 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}+2 \frac {4 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}-\frac {6 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}+\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {36 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {48 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}-\frac {288 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}+\frac {288 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}-\frac {64 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}-\frac {78 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {384 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)} \\ \end{align*}
Time = 10.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {-\frac {5 e^{-\frac {x^2}{(-1+x)^2}} x}{4+x}+(-1+x) \log (5)}{\log (5)} \]
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Time = 1.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
risch | \(x -\frac {5 x \,{\mathrm e}^{-\frac {x^{2}}{\left (-1+x \right )^{2}}}}{\ln \left (5\right ) \left (4+x \right )}\) | \(26\) |
parts | \(x +\frac {\left (\frac {10 x^{2}}{\ln \left (5\right )}-\frac {5 x}{\ln \left (5\right )}-\frac {5 x^{3}}{\ln \left (5\right )}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (4+x \right ) \left (-1+x \right )^{2}}\) | \(57\) |
norman | \(\frac {\left (x^{4} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-8 \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+18 x \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-11 x^{2} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-\frac {5 x}{\ln \left (5\right )}+\frac {10 x^{2}}{\ln \left (5\right )}-\frac {5 x^{3}}{\ln \left (5\right )}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (4+x \right ) \left (-1+x \right )^{2}}\) | \(129\) |
parallelrisch | \(\frac {\left (\ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{4}+4 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{3}-3 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{2}-10 \ln \left (5\right ) x \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-5 x^{3}+8 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+10 x^{2}-5 x \right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\ln \left (5\right ) \left (4+x \right ) \left (-1+x \right )^{2}}\) | \(151\) |
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {{\left ({\left (x^{2} + 4 \, x\right )} e^{\left (\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} \log \left (5\right ) - 5 \, x\right )} e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )}}{{\left (x + 4\right )} \log \left (5\right )} \]
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Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x - \frac {5 x e^{- \frac {x^{2}}{x^{2} - 2 x + 1}}}{x \log {\left (5 \right )} + 4 \log {\left (5 \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (25) = 50\).
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=-\frac {{\left (5 \, x e^{\left (-\frac {1}{x^{2} - 2 \, x + 1}\right )} - {\left (x^{2} e \log \left (5\right ) + 4 \, x e \log \left (5\right )\right )} e^{\left (\frac {2}{x - 1}\right )}\right )} e^{\left (-\frac {2}{x - 1}\right )}}{{\left (x e + 4 \, e\right )} \log \left (5\right )} \]
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Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {x^{2} \log \left (5\right ) - 5 \, x e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} + 4 \, x \log \left (5\right )}{{\left (x + 4\right )} \log \left (5\right )} \]
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Time = 16.70 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x-\frac {5\,x\,{\mathrm {e}}^{-\frac {x^2}{x^2-2\,x+1}}}{\ln \left (5\right )\,\left (x+4\right )} \]
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