Integrand size = 52, antiderivative size = 33 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=\frac {(-3+x) \left (-1+5 e^{-\frac {e^2}{x}}+x^2+\frac {-2+2 x}{x}\right )}{x} \]
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Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.97, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6874, 2243, 2240, 14} \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=x^2+\frac {6}{x^2}-3 x-15 e^{-\frac {e^2}{x}-2}+5 \left (3+e^2\right ) e^{-\frac {e^2}{x}-2}-\frac {15 e^{-\frac {e^2}{x}}}{x}-\frac {5}{x} \]
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Rule 14
Rule 2240
Rule 2243
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5 e^{-\frac {e^2}{x}} \left (-3 e^2+\left (3+e^2\right ) x\right )}{x^3}+\frac {-12+5 x-3 x^3+2 x^4}{x^3}\right ) \, dx \\ & = 5 \int \frac {e^{-\frac {e^2}{x}} \left (-3 e^2+\left (3+e^2\right ) x\right )}{x^3} \, dx+\int \frac {-12+5 x-3 x^3+2 x^4}{x^3} \, dx \\ & = 5 \int \left (-\frac {3 e^{2-\frac {e^2}{x}}}{x^3}+\frac {e^{-\frac {e^2}{x}} \left (3+e^2\right )}{x^2}\right ) \, dx+\int \left (-3-\frac {12}{x^3}+\frac {5}{x^2}+2 x\right ) \, dx \\ & = \frac {6}{x^2}-\frac {5}{x}-3 x+x^2-15 \int \frac {e^{2-\frac {e^2}{x}}}{x^3} \, dx+\left (5 \left (3+e^2\right )\right ) \int \frac {e^{-\frac {e^2}{x}}}{x^2} \, dx \\ & = 5 e^{-2-\frac {e^2}{x}} \left (3+e^2\right )+\frac {6}{x^2}-\frac {5}{x}-\frac {15 e^{-\frac {e^2}{x}}}{x}-3 x+x^2-\frac {15 \int \frac {e^{2-\frac {e^2}{x}}}{x^2} \, dx}{e^2} \\ & = -15 e^{-2-\frac {e^2}{x}}+5 e^{-2-\frac {e^2}{x}} \left (3+e^2\right )+\frac {6}{x^2}-\frac {5}{x}-\frac {15 e^{-\frac {e^2}{x}}}{x}-3 x+x^2 \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=\frac {6-5 x+5 e^{-\frac {e^2}{x}} (-3+x) x-3 x^3+x^4}{x^2} \]
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x^{2}-3 x +\frac {-5 x +6}{x^{2}}+\frac {5 \left (-3+x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x}\) | \(33\) |
norman | \(\frac {\left (x^{4} {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}-15 x +5 x^{2}-5 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x -3 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x^{3}+6 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x^{2}}\) | \(65\) |
parallelrisch | \(\frac {\left (x^{4} {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}-15 x +5 x^{2}-5 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x -3 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x^{3}+6 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x^{2}}\) | \(65\) |
parts | \(x^{2}-3 x -\frac {5}{x}+\frac {6}{x^{2}}-5 \,{\mathrm e}^{-4} \left (-{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{4}-3 \,{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{2}-3 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{2} {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x}-{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}\right )\right )\) | \(85\) |
derivativedivides | \(-{\mathrm e}^{-4} \left (-\frac {6 \,{\mathrm e}^{4}}{x^{2}}-x^{2} {\mathrm e}^{4}+3 x \,{\mathrm e}^{4}-15 \,{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{2}-5 \,{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{4}-15 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{2} {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x}-{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}\right )+\frac {5 \,{\mathrm e}^{4}}{x}\right )\) | \(102\) |
default | \(-{\mathrm e}^{-4} \left (-\frac {6 \,{\mathrm e}^{4}}{x^{2}}-x^{2} {\mathrm e}^{4}+3 x \,{\mathrm e}^{4}-15 \,{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{2}-5 \,{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{4}-15 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{2} {\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}}{x}-{\mathrm e}^{-\frac {{\mathrm e}^{2}}{x}}\right )+\frac {5 \,{\mathrm e}^{4}}{x}\right )\) | \(102\) |
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Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=\frac {{\left (5 \, x^{2} + {\left (x^{4} - 3 \, x^{3} - 5 \, x + 6\right )} e^{\left (\frac {e^{2}}{x}\right )} - 15 \, x\right )} e^{\left (-\frac {e^{2}}{x}\right )}}{x^{2}} \]
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Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=x^{2} - 3 x + \frac {\left (5 x - 15\right ) e^{- \frac {e^{2}}{x}}}{x} + \frac {6 - 5 x}{x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=x^{2} - 15 \, e^{\left (-2\right )} \Gamma \left (2, \frac {e^{2}}{x}\right ) - 3 \, x - \frac {5}{x} + \frac {6}{x^{2}} + 15 \, e^{\left (-\frac {e^{2}}{x} - 2\right )} + 5 \, e^{\left (-\frac {e^{2}}{x}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (30) = 60\).
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.91 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=-x^{2} {\left (\frac {3 \, e^{8}}{x} - \frac {5 \, e^{\left (-\frac {e^{2}}{x} + 8\right )}}{x^{2}} + \frac {5 \, e^{8}}{x^{3}} + \frac {15 \, e^{\left (-\frac {e^{2}}{x} + 8\right )}}{x^{3}} - \frac {6 \, e^{8}}{x^{4}} - e^{8}\right )} e^{\left (-8\right )} \]
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Time = 11.39 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-\frac {e^2}{x}} \left (15 x+e^2 (-15+5 x)+e^{\frac {e^2}{x}} \left (-12+5 x-3 x^3+2 x^4\right )\right )}{x^3} \, dx=5\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^2}{x}}-3\,x-\frac {x\,\left (15\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^2}{x}}+5\right )-6}{x^2}+x^2 \]
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