Integrand size = 54, antiderivative size = 26 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=-e^{e^{1+e^{\frac {e x^2}{5}}}-x}+2 x \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12, 6838} \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2 x-e^{e^{e^{\frac {e x^2}{5}}+1}-x} \]
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Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx \\ & = 2 x+\frac {1}{5} \int e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right ) \, dx \\ & = -e^{e^{1+e^{\frac {e x^2}{5}}}-x}+2 x \\ \end{align*}
Time = 0.44 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=-e^{e^{1+e^{\frac {e x^2}{5}}}-x}+2 x \]
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Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
default | \(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\) | \(23\) |
norman | \(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\) | \(23\) |
risch | \(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\) | \(23\) |
parallelrisch | \(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\) | \(23\) |
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2 \, x - e^{\left (-{\left (x e^{\left (\frac {1}{5} \, x^{2} e + 1\right )} - e^{\left (\frac {1}{5} \, x^{2} e + e^{\left (\frac {1}{5} \, x^{2} e\right )} + 2\right )}\right )} e^{\left (-\frac {1}{5} \, x^{2} e - 1\right )}\right )} \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2 x - e^{- x + e^{e^{\frac {e x^{2}}{5}} + 1}} \]
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none
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2 \, x - e^{\left (-x + e^{\left (e^{\left (\frac {1}{5} \, x^{2} e\right )} + 1\right )}\right )} \]
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none
Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2 \, x - e^{\left (-x + e^{\left (e^{\left (\frac {1}{5} \, x^{2} e\right )} + 1\right )}\right )} \]
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Time = 8.87 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1}{5} \left (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} \left (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x\right )\right ) \, dx=2\,x-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{{\left ({\mathrm {e}}^{x^2\,\mathrm {e}}\right )}^{1/5}}} \]
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