Integrand size = 39, antiderivative size = 28 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=\frac {3}{5} e^{\frac {4}{x}-x} (-2-e-2 x)-\log (2) \]
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\[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=\int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \frac {e^{\frac {4}{x}-x} \left (12 (2+e)+24 x+3 e x^2+6 x^3\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (3 e^{1+\frac {4}{x}-x}+\frac {12 e^{\frac {4}{x}-x} (2+e)}{x^2}+\frac {24 e^{\frac {4}{x}-x}}{x}+6 e^{\frac {4}{x}-x} x\right ) \, dx \\ & = \frac {3}{5} \int e^{1+\frac {4}{x}-x} \, dx+\frac {6}{5} \int e^{\frac {4}{x}-x} x \, dx+\frac {24}{5} \int \frac {e^{\frac {4}{x}-x}}{x} \, dx+\frac {1}{5} (12 (2+e)) \int \frac {e^{\frac {4}{x}-x}}{x^2} \, dx \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=\frac {3}{5} e^{\frac {4}{x}-x} (-2-e-2 x) \]
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Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
gosper | \(-\frac {3 \,{\mathrm e}^{-\frac {x^{2}-4}{x}} \left (2 x +{\mathrm e}+2\right )}{5}\) | \(21\) |
risch | \(\frac {\left (-3 \,{\mathrm e}-6-6 x \right ) {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}}}{5}\) | \(24\) |
norman | \(\frac {\left (-\frac {6}{5}-\frac {3 \,{\mathrm e}}{5}\right ) x \,{\mathrm e}^{\frac {-x^{2}+4}{x}}-\frac {6 x^{2} {\mathrm e}^{\frac {-x^{2}+4}{x}}}{5}}{x}\) | \(43\) |
parallelrisch | \(-\frac {3 \,{\mathrm e}^{-\frac {x^{2}-4}{x}} {\mathrm e}}{5}-\frac {6 x \,{\mathrm e}^{-\frac {x^{2}-4}{x}}}{5}-\frac {6 \,{\mathrm e}^{-\frac {x^{2}-4}{x}}}{5}\) | \(44\) |
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=-\frac {3}{5} \, {\left (2 \, x + e + 2\right )} e^{\left (-\frac {x^{2} - 4}{x}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=\frac {\left (- 6 x - 3 e - 6\right ) e^{\frac {4 - x^{2}}{x}}}{5} \]
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=-\frac {3}{5} \, {\left (2 \, x + e + 2\right )} e^{\left (-x + \frac {4}{x}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=-\frac {6}{5} \, x e^{\left (-\frac {x^{2} - 4}{x}\right )} - \frac {3}{5} \, e^{\left (-\frac {x^{2} - x - 4}{x}\right )} - \frac {6}{5} \, e^{\left (-\frac {x^{2} - 4}{x}\right )} \]
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Time = 7.95 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\frac {4-x^2}{x}} \left (24+24 x+6 x^3+e \left (12+3 x^2\right )\right )}{5 x^2} \, dx=-{\mathrm {e}}^{\frac {4}{x}-x}\,\left (\frac {6\,x}{5}+\frac {3\,\mathrm {e}}{5}+\frac {6}{5}\right ) \]
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