Integrand size = 36, antiderivative size = 18 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=e^{-3+e-x} \left (4-\frac {2 x^2}{3}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 21, 2227, 2225, 2207} \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=4 e^{-x+e-3}-\frac {2}{3} e^{-x+e-3} x^2 \]
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Rule 12
Rule 21
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{-6+x^2} \, dx \\ & = -\left (\frac {2}{3} \int e^{-3+e-x} \left (6+2 x-x^2\right ) \, dx\right ) \\ & = -\left (\frac {2}{3} \int \left (6 e^{-3+e-x}+2 e^{-3+e-x} x-e^{-3+e-x} x^2\right ) \, dx\right ) \\ & = \frac {2}{3} \int e^{-3+e-x} x^2 \, dx-\frac {4}{3} \int e^{-3+e-x} x \, dx-4 \int e^{-3+e-x} \, dx \\ & = 4 e^{-3+e-x}+\frac {4}{3} e^{-3+e-x} x-\frac {2}{3} e^{-3+e-x} x^2-\frac {4}{3} \int e^{-3+e-x} \, dx+\frac {4}{3} \int e^{-3+e-x} x \, dx \\ & = \frac {16}{3} e^{-3+e-x}-\frac {2}{3} e^{-3+e-x} x^2+\frac {4}{3} \int e^{-3+e-x} \, dx \\ & = 4 e^{-3+e-x}-\frac {2}{3} e^{-3+e-x} x^2 \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=\frac {2}{3} e^{-3+e-x} \left (6-x^2\right ) \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94
method | result | size |
gosper | \({\mathrm e}^{\ln \left (-\frac {2 x^{2}}{3}+4\right )+{\mathrm e}-3-x}\) | \(17\) |
norman | \({\mathrm e}^{\ln \left (-\frac {2 x^{2}}{3}+4\right )+{\mathrm e}-3-x}\) | \(17\) |
risch | \(\left (-\frac {2 x^{2}}{3}+4\right ) {\mathrm e}^{{\mathrm e}-3-x}\) | \(17\) |
parallelrisch | \({\mathrm e}^{\ln \left (-\frac {2 x^{2}}{3}+4\right )+{\mathrm e}-3-x}\) | \(17\) |
default | \(\text {Expression too large to display}\) | \(5846\) |
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Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=e^{\left (-x + e + \log \left (-\frac {2}{3} \, x^{2} + 4\right ) - 3\right )} \]
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Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=\frac {\left (12 - 2 x^{2}\right ) e^{- x - 3 + e}}{3} \]
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Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=-\frac {2}{3} \, {\left (x^{2} e^{e} - 6 \, e^{e}\right )} e^{\left (-x - 3\right )} \]
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=-\frac {2}{3} \, x^{2} e^{\left (-x + e - 3\right )} + 4 \, e^{\left (-x + e - 3\right )} \]
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Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{3 \left (-6+x^2\right )} \, dx=-\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{\mathrm {e}}\,\left (x^2-6\right )}{3} \]
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