Integrand size = 35, antiderivative size = 21 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{2+e^{-e^4}+x} \left (5+x^2\right ) \]
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Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{x+e^{-e^4}+2} x^2+2 e^{x+e^{-e^4}+2} \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx \\ & = \frac {1}{5} \int \left (10 e^{2+e^{-e^4}+x}+4 e^{2+e^{-e^4}+x} x+2 e^{2+e^{-e^4}+x} x^2\right ) \, dx \\ & = \frac {2}{5} \int e^{2+e^{-e^4}+x} x^2 \, dx+\frac {4}{5} \int e^{2+e^{-e^4}+x} x \, dx+2 \int e^{2+e^{-e^4}+x} \, dx \\ & = 2 e^{2+e^{-e^4}+x}+\frac {4}{5} e^{2+e^{-e^4}+x} x+\frac {2}{5} e^{2+e^{-e^4}+x} x^2-\frac {4}{5} \int e^{2+e^{-e^4}+x} \, dx-\frac {4}{5} \int e^{2+e^{-e^4}+x} x \, dx \\ & = \frac {6}{5} e^{2+e^{-e^4}+x}+\frac {2}{5} e^{2+e^{-e^4}+x} x^2+\frac {4}{5} \int e^{2+e^{-e^4}+x} \, dx \\ & = 2 e^{2+e^{-e^4}+x}+\frac {2}{5} e^{2+e^{-e^4}+x} x^2 \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{2+e^{-e^4}+x} \left (5+x^2\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {\left (2 x^{2}+10\right ) {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}+2+x}}{5}\) | \(19\) |
gosper | \(\frac {2 \,{\mathrm e}^{x +\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (x^{2}+5\right )}{5}\) | \(24\) |
default | \(\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left ({\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x}\right )}{5}\) | \(28\) |
norman | \(2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}} {\mathrm e}^{x}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}} x^{2} {\mathrm e}^{x}}{5}\) | \(29\) |
parallelrisch | \(\frac {{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (2 \,{\mathrm e}^{x} x^{2}+10 \,{\mathrm e}^{x}\right )}{5}\) | \(29\) |
parts | \(\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} {\mathrm e}^{x} x^{2}}{5}+2 \,{\mathrm e}^{x} {\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}}\) | \(41\) |
meijerg | \(-2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (1-{\mathrm e}^{x}\right )-\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )}{5}+\frac {4 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (1-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}\right )}{5}\) | \(83\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} + 5\right )} e^{\left ({\left ({\left (x + 2\right )} e^{\left (e^{4}\right )} + 1\right )} e^{\left (-e^{4}\right )}\right )} \]
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Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {\left (2 x^{2} e^{2} e^{e^{- e^{4}}} + 10 e^{2} e^{e^{- e^{4}}}\right ) e^{x}}{5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (16) = 32\).
Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.57 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} e^{\left (e^{\left (-e^{4}\right )} + 2\right )} - 2 \, x e^{\left (e^{\left (-e^{4}\right )} + 2\right )} + 2 \, e^{\left (e^{\left (-e^{4}\right )} + 2\right )}\right )} e^{x} + \frac {4}{5} \, {\left (x e^{\left (e^{\left (-e^{4}\right )} + 2\right )} - e^{\left (e^{\left (-e^{4}\right )} + 2\right )}\right )} e^{x} + 2 \, e^{\left (x + e^{\left (-e^{4}\right )} + 2\right )} \]
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Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} + 5\right )} e^{\left (x + e^{\left (-e^{4}\right )} + 2\right )} \]
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Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2\,{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^4}}\,{\mathrm {e}}^x\,\left (x^2+5\right )}{5} \]
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