Integrand size = 44, antiderivative size = 28 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=2+e^{-x}+\frac {1}{5} \left (e^4-e^{4/x}-x\right ) x \]
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Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 6820, 2225, 2326} \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=-\frac {x^2}{5}-\frac {1}{5} e^{4/x} x+\frac {e^4 x}{5}+e^{-x} \]
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Rule 12
Rule 2225
Rule 2326
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{x} \, dx \\ & = \frac {1}{5} \int \left (e^4-5 e^{-x}+e^{4/x} \left (-1+\frac {4}{x}\right )-2 x\right ) \, dx \\ & = \frac {e^4 x}{5}-\frac {x^2}{5}+\frac {1}{5} \int e^{4/x} \left (-1+\frac {4}{x}\right ) \, dx-\int e^{-x} \, dx \\ & = e^{-x}+\frac {e^4 x}{5}-\frac {1}{5} e^{4/x} x-\frac {x^2}{5} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=\frac {1}{5} \left (5 e^{-x}+e^4 x-e^{4/x} x-x^2\right ) \]
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Time = 0.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {x \,{\mathrm e}^{4}}{5}-\frac {x^{2}}{5}+{\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{\frac {4}{x}}}{5}\) | \(25\) |
risch | \(\frac {x \,{\mathrm e}^{4}}{5}-\frac {x^{2}}{5}+{\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{\frac {4}{x}}}{5}\) | \(25\) |
parts | \(\frac {x \,{\mathrm e}^{4}}{5}-\frac {x^{2}}{5}+{\mathrm e}^{-x}-\frac {x \,{\mathrm e}^{\frac {4}{x}}}{5}\) | \(25\) |
norman | \(\left (1-\frac {{\mathrm e}^{x} x^{2}}{5}+\frac {x \,{\mathrm e}^{4} {\mathrm e}^{x}}{5}-\frac {{\mathrm e}^{x} {\mathrm e}^{\frac {4}{x}} x}{5}\right ) {\mathrm e}^{-x}\) | \(33\) |
parallelrisch | \(\frac {\left (x \,{\mathrm e}^{4} {\mathrm e}^{x}+5-{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} {\mathrm e}^{\frac {4}{x}} x \right ) {\mathrm e}^{-x}}{5}\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=-\frac {1}{5} \, {\left ({\left (x^{2} - x e^{4} + x e^{\frac {4}{x}}\right )} e^{x} - 5\right )} e^{\left (-x\right )} \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=- \frac {x^{2}}{5} - \frac {x e^{\frac {4}{x}}}{5} + \frac {x e^{4}}{5} + e^{- x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=-\frac {1}{5} \, x^{2} + \frac {1}{5} \, x e^{4} - \frac {4}{5} \, {\rm Ei}\left (\frac {4}{x}\right ) + e^{\left (-x\right )} + \frac {4}{5} \, \Gamma \left (-1, -\frac {4}{x}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx=-\frac {1}{5} \, x^{2} + \frac {1}{5} \, x e^{4} - \frac {1}{5} \, x e^{\frac {4}{x}} + e^{\left (-x\right )} \]
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Time = 17.93 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x} \left (-5 x+e^x \left (e^{4/x} (4-x)+e^4 x-2 x^2\right )\right )}{5 x} \, dx={\mathrm {e}}^{-x}+\frac {x\,{\mathrm {e}}^4}{5}-\frac {x\,{\mathrm {e}}^{4/x}}{5}-\frac {x^2}{5} \]
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