Integrand size = 115, antiderivative size = 32 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=\frac {x}{e^{\frac {5}{2} \left (e^{\frac {1}{x^2}} x-x^2-x^3\right )}-x} \]
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Time = 2.54 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 6843, 32} \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {1}{1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (x+1)}}{x}} \]
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Rule 12
Rule 32
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {5}{2} x \left (e^{\frac {1}{x^2}}+x+x^2\right )} \left (-5 e^{\frac {1}{x^2}} \left (-2+x^2\right )+x \left (2+10 x^2+15 x^3\right )\right )}{2 x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (1+x)} x\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{\frac {5}{2} x \left (e^{\frac {1}{x^2}}+x+x^2\right )} \left (-5 e^{\frac {1}{x^2}} \left (-2+x^2\right )+x \left (2+10 x^2+15 x^3\right )\right )}{x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (1+x)} x\right )^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (1+x)}}{x}\right ) \\ & = -\frac {1}{1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (1+x)}}{x}} \\ \end{align*}
Time = 3.59 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {e^{\frac {5}{2} x^2 (1+x)} x}{-e^{\frac {5}{2} e^{\frac {1}{x^2}} x}+e^{\frac {5}{2} x^2 (1+x)} x} \]
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Time = 0.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {x}{x -{\mathrm e}^{-\frac {5 x \left (x^{2}-{\mathrm e}^{\frac {1}{x^{2}}}+x \right )}{2}}}\) | \(25\) |
parallelrisch | \(-\frac {{\mathrm e}^{\frac {5 x \left (-x^{2}+{\mathrm e}^{\frac {1}{x^{2}}}-x \right )}{2}}}{x -{\mathrm e}^{\frac {5 x \left (-x^{2}+{\mathrm e}^{\frac {1}{x^{2}}}-x \right )}{2}}}\) | \(43\) |
norman | \(-\frac {{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}{x -{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}\) | \(47\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=\frac {x}{- x + e^{- \frac {5 x^{3}}{2} - \frac {5 x^{2}}{2} + \frac {5 x e^{\frac {1}{x^{2}}}}{2}}} \]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )}}{x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )} - e^{\left (\frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \]
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Time = 0.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \]
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Time = 11.89 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.97 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x^4\,\left (2\,x+10\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^2\,{\mathrm {e}}^{\frac {1}{x^2}}+10\,x^3+15\,x^4\right )}{\left (x-{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}-\frac {5\,x^2}{2}-\frac {5\,x^3}{2}}\right )\,\left (10\,x^3\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^5\,{\mathrm {e}}^{\frac {1}{x^2}}+2\,x^4+10\,x^6+15\,x^7\right )} \]
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