Integrand size = 53, antiderivative size = 25 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=e^3+\frac {x^2}{e^6 \left (-4+e^5\right )^2 (-8-x)} \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1607, 2008, 27, 12, 75} \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {x^2}{e^6 \left (4-e^5\right )^2 (x+8)} \]
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Rule 12
Rule 27
Rule 75
Rule 1607
Rule 2008
Rubi steps \begin{align*} \text {integral}& = \int \frac {(-16-x) x}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx \\ & = \int \frac {(-16-x) x}{64 e^6 \left (4-e^5\right )^2+16 e^6 \left (4-e^5\right )^2 x+e^6 \left (4-e^5\right )^2 x^2} \, dx \\ & = \int \frac {(-16-x) x}{e^6 \left (-4+e^5\right )^2 (8+x)^2} \, dx \\ & = \frac {\int \frac {(-16-x) x}{(8+x)^2} \, dx}{e^6 \left (4-e^5\right )^2} \\ & = -\frac {x^2}{e^6 \left (4-e^5\right )^2 (8+x)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {x+\frac {64}{8+x}}{e^6 \left (-4+e^5\right )^2} \]
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Time = 0.78 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
norman | \(-\frac {{\mathrm e}^{-6} x^{2}}{\left ({\mathrm e}^{5}-4\right )^{2} \left (x +8\right )}\) | \(21\) |
parallelrisch | \(-\frac {x^{2} {\mathrm e}^{-6}}{\left (-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}+16\right ) \left (x +8\right )}\) | \(27\) |
default | \(\frac {{\mathrm e}^{-6} \left (-x -\frac {64}{x +8}\right )}{-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}+16}\) | \(29\) |
gosper | \(-\frac {x^{2} {\mathrm e}^{-6}}{x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128}\) | \(38\) |
meijerg | \(-\frac {8 \left (\frac {x \left (\frac {3 x}{8}+6\right )}{24+3 x}-2 \ln \left (1+\frac {x}{8}\right )\right )}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}-\frac {16 \left (-\frac {x}{8 \left (1+\frac {x}{8}\right )}+\ln \left (1+\frac {x}{8}\right )\right )}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}\) | \(73\) |
risch | \(-\frac {x}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}-\frac {64 \,{\mathrm e}^{-6} {\mathrm e}^{16}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}+\frac {512 \,{\mathrm e}^{-6} {\mathrm e}^{11}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}-\frac {1024 \,{\mathrm e}^{-6} {\mathrm e}^{6}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}\) | \(147\) |
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Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {x^{2} + 8 \, x + 64}{{\left (x + 8\right )} e^{16} - 8 \, {\left (x + 8\right )} e^{11} + 16 \, {\left (x + 8\right )} e^{6}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=- \frac {x}{- 8 e^{11} + 16 e^{6} + e^{16}} - \frac {64}{x \left (- 8 e^{11} + 16 e^{6} + e^{16}\right ) - 64 e^{11} + 128 e^{6} + 8 e^{16}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {x}{e^{16} - 8 \, e^{11} + 16 \, e^{6}} - \frac {64}{x {\left (e^{16} - 8 \, e^{11} + 16 \, e^{6}\right )} + 8 \, e^{16} - 64 \, e^{11} + 128 \, e^{6}} \]
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {x}{e^{16} - 8 \, e^{11} + 16 \, e^{6}} - \frac {64}{{\left (x + 8\right )} {\left (e^{16} - 8 \, e^{11} + 16 \, e^{6}\right )}} \]
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Time = 9.91 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-16 x-x^2}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx=-\frac {{\mathrm {e}}^{-6}\,\left (x^2+8\,x+64\right )}{{\left ({\mathrm {e}}^5-4\right )}^2\,\left (x+8\right )} \]
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