Integrand size = 96, antiderivative size = 27 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 e^{e^{\frac {x^2}{5+x}}-x} \left (1+\frac {10}{e^4}+x\right ) \]
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\[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{(5+x)^2} \, dx \\ & = \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2-2 e^4 x (5+x)^2+2 e^{\frac {x^2}{5+x}} x (10+x) \left (10+e^4 (1+x)\right )\right )}{(5+x)^2} \, dx \\ & = \int \left (-2 e^{-4+e^{\frac {x^2}{5+x}}-x} \left (10+e^4 x\right )+\frac {2 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x (10+x) \left (10+e^4+e^4 x\right )}{(5+x)^2}\right ) \, dx \\ & = -\left (2 \int e^{-4+e^{\frac {x^2}{5+x}}-x} \left (10+e^4 x\right ) \, dx\right )+2 \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x (10+x) \left (10+e^4+e^4 x\right )}{(5+x)^2} \, dx \\ & = -\left (2 \int \left (10 e^{-4+e^{\frac {x^2}{5+x}}-x}+e^{e^{\frac {x^2}{5+x}}-x} x\right ) \, dx\right )+2 \int \left (10 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \left (1+\frac {e^4}{10}\right )+e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x+\frac {50 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \left (-5+2 e^4\right )}{(5+x)^2}-\frac {25 e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{5+x}\right ) \, dx \\ & = -\left (2 \int e^{e^{\frac {x^2}{5+x}}-x} x \, dx\right )+2 \int e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x \, dx-20 \int e^{-4+e^{\frac {x^2}{5+x}}-x} \, dx-50 \int \frac {e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{5+x} \, dx-\left (100 \left (5-2 e^4\right )\right ) \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{(5+x)^2} \, dx+\left (2 \left (10+e^4\right )\right ) \int e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \, dx \\ \end{align*}
Time = 5.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=e^{-4+e^{\frac {x^2}{5+x}}-x} \left (20+2 e^4 (1+x)\right ) \]
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Time = 3.62 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\left (2 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}+20\right ) {\mathrm e}^{-4+{\mathrm e}^{\frac {x^{2}}{5+x}}-x}\) | \(29\) |
parallelrisch | \({\mathrm e}^{-4} \left (2 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}+20\right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{5+x}}-x}\) | \(34\) |
norman | \(\frac {\left (2 x^{2}+10 \left (10+{\mathrm e}^{4}\right ) {\mathrm e}^{-4}+4 \left (5+3 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4} x \right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{5+x}}-x}}{5+x}\) | \(53\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 \, {\left ({\left (x + 1\right )} e^{4} + 10\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )} \]
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Time = 7.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\frac {\left (2 x e^{4} + 20 + 2 e^{4}\right ) e^{- x + e^{\frac {x^{2}}{x + 5}}}}{e^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 \, {\left (x e^{4} + e^{4} + 10\right )} e^{\left (-x + e^{\left (x + \frac {25}{x + 5} - 5\right )} - 4\right )} \]
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\[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\int { -\frac {2 \, {\left (10 \, x^{2} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{4} - {\left (10 \, x^{2} + {\left (x^{3} + 11 \, x^{2} + 10 \, x\right )} e^{4} + 100 \, x\right )} e^{\left (\frac {x^{2}}{x + 5}\right )} + 100 \, x + 250\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )}}{x^{2} + 10 \, x + 25} \,d x } \]
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Time = 9.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {x^2}{x+5}}-x}\,\left (2\,x+{\mathrm {e}}^{-4}\,\left (2\,{\mathrm {e}}^4+20\right )\right ) \]
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