Integrand size = 36, antiderivative size = 24 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=e^{-2 e} \left (4-x+\frac {1}{3} \left (-4+\frac {3}{(2+x)^2}\right )\right ) \]
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Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2099} \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {e^{-2 e}}{(x+2)^2}-e^{-2 e} x \]
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Rule 12
Rule 2099
Rubi steps \begin{align*} \text {integral}& = e^{-2 e} \int \frac {-10-12 x-6 x^2-x^3}{8+12 x+6 x^2+x^3} \, dx \\ & = e^{-2 e} \int \left (-1-\frac {2}{(2+x)^3}\right ) \, dx \\ & = -e^{-2 e} x+\frac {e^{-2 e}}{(2+x)^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-e^{-2 e} \left (x-\frac {1}{(2+x)^2}\right ) \]
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Time = 0.54 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67
method | result | size |
default | \({\mathrm e}^{-2 \,{\mathrm e}} \left (-x +\frac {1}{\left (2+x \right )^{2}}\right )\) | \(16\) |
gosper | \(-\frac {\left (x^{3}-12 x -17\right ) {\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\) | \(26\) |
risch | \(-{\mathrm e}^{-2 \,{\mathrm e}} x +\frac {{\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\) | \(26\) |
parallelrisch | \(-\frac {\left (x^{3}-12 x -17\right ) {\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\) | \(26\) |
norman | \(\frac {\left (12 \,{\mathrm e}^{-{\mathrm e}} x -{\mathrm e}^{-{\mathrm e}} x^{3}+17 \,{\mathrm e}^{-{\mathrm e}}\right ) {\mathrm e}^{-{\mathrm e}}}{\left (2+x \right )^{2}}\) | \(38\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-\frac {{\left (x^{3} + 4 \, x^{2} + 4 \, x - 1\right )} e^{\left (-2 \, e\right )}}{x^{2} + 4 \, x + 4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=- \frac {x}{e^{2 e}} + \frac {1}{x^{2} e^{2 e} + 4 x e^{2 e} + 4 e^{2 e}} \]
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Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-{\left (x - \frac {1}{x^{2} + 4 \, x + 4}\right )} e^{\left (-2 \, e\right )} \]
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Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-{\left (x - \frac {1}{{\left (x + 2\right )}^{2}}\right )} e^{\left (-2 \, e\right )} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {{\mathrm {e}}^{-2\,\mathrm {e}}}{{\left (x+2\right )}^2}-x\,{\mathrm {e}}^{-2\,\mathrm {e}} \]
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