Integrand size = 57, antiderivative size = 22 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=e^{-x} \left (-2+\frac {5 e^2}{2}+x\right ) (-x+\log (x)) \]
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Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(22)=44\).
Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.73, number of steps used = 16, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6874, 2230, 2225, 2209, 2207, 2634} \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=-e^{-x} x^2-2 e^{-x} x+\frac {1}{2} \left (8-5 e^2\right ) e^{-x} x-e^{-x}+\frac {1}{2} \left (8-5 e^2\right ) e^{-x}-\frac {1}{2} \left (6-5 e^2\right ) e^{-x}+e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (-2 x-5 e^2+6\right ) \log (x) \]
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Rule 2207
Rule 2209
Rule 2225
Rule 2230
Rule 2634
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{-x} \left (-4+5 e^2+\left (6-5 e^2\right ) x-\left (8-5 e^2\right ) x^2+2 x^3\right )}{2 x}-\frac {1}{2} e^{-x} \left (-6+5 e^2+2 x\right ) \log (x)\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^{-x} \left (-4+5 e^2+\left (6-5 e^2\right ) x-\left (8-5 e^2\right ) x^2+2 x^3\right )}{x} \, dx-\frac {1}{2} \int e^{-x} \left (-6+5 e^2+2 x\right ) \log (x) \, dx \\ & = e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (6-5 e^2-2 x\right ) \log (x)+\frac {1}{2} \int \frac {e^{-x} \left (4-5 e^2-2 x\right )}{x} \, dx+\frac {1}{2} \int \left (e^{-x} \left (6-5 e^2\right )+\frac {e^{-x} \left (-4+5 e^2\right )}{x}-e^{-x} \left (8-5 e^2\right ) x+2 e^{-x} x^2\right ) \, dx \\ & = e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (6-5 e^2-2 x\right ) \log (x)+\frac {1}{2} \int \left (-2 e^{-x}+\frac {e^{-x} \left (4-5 e^2\right )}{x}\right ) \, dx+\frac {1}{2} \left (6-5 e^2\right ) \int e^{-x} \, dx+\frac {1}{2} \left (-8+5 e^2\right ) \int e^{-x} x \, dx+\frac {1}{2} \left (-4+5 e^2\right ) \int \frac {e^{-x}}{x} \, dx+\int e^{-x} x^2 \, dx \\ & = -\frac {1}{2} e^{-x} \left (6-5 e^2\right )+\frac {1}{2} e^{-x} \left (8-5 e^2\right ) x-e^{-x} x^2-\frac {1}{2} \left (4-5 e^2\right ) \text {Ei}(-x)+e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (6-5 e^2-2 x\right ) \log (x)+2 \int e^{-x} x \, dx+\frac {1}{2} \left (4-5 e^2\right ) \int \frac {e^{-x}}{x} \, dx+\frac {1}{2} \left (-8+5 e^2\right ) \int e^{-x} \, dx-\int e^{-x} \, dx \\ & = e^{-x}-\frac {1}{2} e^{-x} \left (6-5 e^2\right )+\frac {1}{2} e^{-x} \left (8-5 e^2\right )-2 e^{-x} x+\frac {1}{2} e^{-x} \left (8-5 e^2\right ) x-e^{-x} x^2+e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (6-5 e^2-2 x\right ) \log (x)+2 \int e^{-x} \, dx \\ & = -e^{-x}-\frac {1}{2} e^{-x} \left (6-5 e^2\right )+\frac {1}{2} e^{-x} \left (8-5 e^2\right )-2 e^{-x} x+\frac {1}{2} e^{-x} \left (8-5 e^2\right ) x-e^{-x} x^2+e^{-x} \log (x)-\frac {1}{2} e^{-x} \left (6-5 e^2-2 x\right ) \log (x) \\ \end{align*}
Time = 1.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=-\frac {1}{2} e^{-x} \left (-4+5 e^2+2 x\right ) (x-\log (x)) \]
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Time = 1.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50
method | result | size |
norman | \(\left (x \ln \left (x \right )+\left (-\frac {5 \,{\mathrm e}^{2}}{2}+2\right ) x +\left (\frac {5 \,{\mathrm e}^{2}}{2}-2\right ) \ln \left (x \right )-x^{2}\right ) {\mathrm e}^{-x}\) | \(33\) |
risch | \(\frac {\left (5 \,{\mathrm e}^{2}-4+2 x \right ) {\mathrm e}^{-x} \ln \left (x \right )}{2}-\frac {x \left (5 \,{\mathrm e}^{2}-4+2 x \right ) {\mathrm e}^{-x}}{2}\) | \(35\) |
parallelrisch | \(\left (-x \,{\mathrm e}^{\ln \left (\frac {5}{2}\right )+2}+\ln \left (x \right ) {\mathrm e}^{\ln \left (\frac {5}{2}\right )+2}-x^{2}+x \ln \left (x \right )+2 x -2 \ln \left (x \right )\right ) {\mathrm e}^{-x}\) | \(39\) |
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).
Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=-{\left (x^{2} - 2 \, x\right )} e^{\left (-x\right )} - x e^{\left (-x + \log \left (\frac {5}{2}\right ) + 2\right )} + {\left ({\left (x - 2\right )} e^{\left (-x\right )} + e^{\left (-x + \log \left (\frac {5}{2}\right ) + 2\right )}\right )} \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).
Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=\frac {\left (- 2 x^{2} + 2 x \log {\left (x \right )} - 5 x e^{2} + 4 x - 4 \log {\left (x \right )} + 5 e^{2} \log {\left (x \right )}\right ) e^{- x}}{2} \]
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\[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=\int { \frac {{\left (x^{3} - 4 \, x^{2} + {\left (x^{2} - x + 1\right )} e^{\left (\log \left (\frac {5}{2}\right ) + 2\right )} - {\left (x^{2} + x e^{\left (\log \left (\frac {5}{2}\right ) + 2\right )} - 3 \, x\right )} \log \left (x\right ) + 3 \, x - 2\right )} e^{\left (-x\right )}}{x} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.36 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=-x^{2} e^{\left (-x\right )} + x e^{\left (-x\right )} \log \left (x\right ) + 2 \, x e^{\left (-x\right )} - \frac {5}{2} \, x e^{\left (-x + 2\right )} - 2 \, e^{\left (-x\right )} \log \left (x\right ) + \frac {5}{2} \, e^{\left (-x + 2\right )} \log \left (x\right ) \]
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Time = 9.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-x} \left (-2+3 x-4 x^2+x^3+\frac {5}{2} e^2 \left (1-x+x^2\right )+\left (3 x-\frac {5 e^2 x}{2}-x^2\right ) \log (x)\right )}{x} \, dx=-\frac {{\mathrm {e}}^{-x}\,\left (x-\ln \left (x\right )\right )\,\left (2\,x+5\,{\mathrm {e}}^2-4\right )}{2} \]
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