Integrand size = 61, antiderivative size = 22 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=-e^{\frac {x^3}{e^4 \left (4+x^2\right )}}+2 x \]
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 28, 6820, 6838} \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=2 x-e^{\frac {x^3}{e^4 \left (x^2+4\right )}} \]
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Rule 12
Rule 28
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{16+8 x^2+x^4} \, dx}{e^4} \\ & = \frac {\int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{\left (4+x^2\right )^2} \, dx}{e^4} \\ & = \frac {\int \left (2 e^4-\frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} x^2 \left (12+x^2\right )}{\left (4+x^2\right )^2}\right ) \, dx}{e^4} \\ & = 2 x-\frac {\int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} x^2 \left (12+x^2\right )}{\left (4+x^2\right )^2} \, dx}{e^4} \\ & = -e^{\frac {x^3}{e^4 \left (4+x^2\right )}}+2 x \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=-e^{\frac {x}{e^4}-\frac {4 x}{e^4 \left (4+x^2\right )}}+2 x \]
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Time = 0.86 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(2 x -{\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}}\) | \(21\) |
parallelrisch | \({\mathrm e}^{-4} \left (2 x \,{\mathrm e}^{4}-{\mathrm e}^{4} {\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}}\right )\) | \(32\) |
parts | \(2 x +\frac {-{\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}} x^{2}-4 \,{\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}}}{x^{2}+4}\) | \(53\) |
norman | \(\frac {8 x +2 x^{3}-{\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}} x^{2}-4 \,{\mathrm e}^{\frac {x^{3} {\mathrm e}^{-4}}{x^{2}+4}}}{x^{2}+4}\) | \(57\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=2 \, x - e^{\left (\frac {x^{3} e^{\left (-4\right )}}{x^{2} + 4}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=2 x - e^{\frac {x^{3}}{\left (x^{2} + 4\right ) e^{4}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (20) = 40\).
Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 4.14 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx={\left (2 \, {\left (x + \frac {2 \, x}{x^{2} + 4} - 3 \, \arctan \left (\frac {1}{2} \, x\right )\right )} e^{4} + 2 \, {\left (\frac {2 \, x}{x^{2} + 4} + \arctan \left (\frac {1}{2} \, x\right )\right )} e^{4} - 4 \, {\left (\frac {2 \, x}{x^{2} + 4} - \arctan \left (\frac {1}{2} \, x\right )\right )} e^{4} - e^{\left (x e^{\left (-4\right )} - \frac {4 \, x}{x^{2} e^{4} + 4 \, e^{4}} + 4\right )}\right )} e^{\left (-4\right )} \]
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Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx={\left (2 \, x e^{4} - e^{\left (\frac {x^{3}}{x^{2} e^{4} + 4 \, e^{4}} + 4\right )}\right )} e^{\left (-4\right )} \]
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Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x^3}{e^4 \left (4+x^2\right )}} \left (-12 x^2-x^4\right )+e^4 \left (32+16 x^2+2 x^4\right )}{e^4 \left (16+8 x^2+x^4\right )} \, dx=2\,x-{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{-4}}{x^2+4}} \]
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