Integrand size = 53, antiderivative size = 22 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 e^{-2+e^x+\frac {x^4}{25 e^2}} x^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(22)=44\).
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=\frac {9 e^{-\frac {-x^4-25 e^{x+2}+50 e^2}{25 e^2}} \left (4 x^5+25 e^{x+2} x^2\right )}{4 x^3+25 e^{x+2}} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx \\ & = \frac {9 e^{-\frac {50 e^2-25 e^{2+x}-x^4}{25 e^2}} \left (25 e^{2+x} x^2+4 x^5\right )}{25 e^{2+x}+4 x^3} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 e^{-2+e^x+\frac {x^4}{25 e^2}} x^2 \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23
method | result | size |
risch | \(9 x^{2} {\mathrm e}^{-\frac {\left (-25 \,{\mathrm e}^{2+x}+50 \,{\mathrm e}^{2}-x^{4}\right ) {\mathrm e}^{-2}}{25}}\) | \(27\) |
norman | \(9 x^{2} {\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{2} {\mathrm e}^{x}-50 \,{\mathrm e}^{2}+x^{4}\right ) {\mathrm e}^{-2}}{25}}\) | \(31\) |
parallelrisch | \(9 x^{2} {\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{2} {\mathrm e}^{x}-50 \,{\mathrm e}^{2}+x^{4}\right ) {\mathrm e}^{-2}}{25}}\) | \(31\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 \, x^{2} e^{\left (\frac {1}{25} \, {\left (x^{4} - 100 \, e^{2} + 25 \, e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} + 2\right )} \]
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Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 x^{2} e^{\frac {\frac {x^{4}}{25} + e^{2} e^{x} - 2 e^{2}}{e^{2}}} \]
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Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9 \, x^{2} e^{\left (\frac {1}{25} \, x^{4} e^{\left (-2\right )} + e^{x} - 2\right )} \]
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\[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=\int { \frac {9}{25} \, {\left (4 \, x^{5} + 25 \, x^{2} e^{\left (x + 2\right )} + 50 \, x e^{2}\right )} e^{\left (\frac {1}{25} \, {\left (x^{4} - 50 \, e^{2} + 25 \, e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} - 2\right )} \,d x } \]
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Time = 7.83 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} e^{-2+\frac {-50 e^2+25 e^{2+x}+x^4}{25 e^2}} \left (450 e^2 x+225 e^{2+x} x^2+36 x^5\right ) \, dx=9\,x^2\,{\mathrm {e}}^{\frac {x^4\,{\mathrm {e}}^{-2}}{25}}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-2} \]
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