Integrand size = 35, antiderivative size = 24 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=\left (-1+\frac {5}{e^2 (-4+x)}-\frac {27}{4 x}-2 x\right ) x \]
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Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 27, 1864} \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=-2 x^2-x-\frac {20}{e^2 (4-x)} \]
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Rule 12
Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{16-8 x+x^2} \, dx}{e^2} \\ & = \frac {\int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{(-4+x)^2} \, dx}{e^2} \\ & = \frac {\int \left (-e^2-\frac {20}{(-4+x)^2}-4 e^2 x\right ) \, dx}{e^2} \\ & = -\frac {20}{e^2 (4-x)}-x-2 x^2 \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=\frac {20}{e^2 (-4+x)}-x-2 x^2 \]
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Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-2 x^{2}-x +\frac {20 \,{\mathrm e}^{-2}}{x -4}\) | \(19\) |
default | \({\mathrm e}^{-2} \left (-2 x^{2} {\mathrm e}^{2}-{\mathrm e}^{2} x +\frac {20}{x -4}\right )\) | \(26\) |
norman | \(\frac {7 x^{2}-2 x^{3}+4 \left (4 \,{\mathrm e}^{2}+5\right ) {\mathrm e}^{-2}}{x -4}\) | \(30\) |
gosper | \(-\frac {\left (2 x^{3} {\mathrm e}^{2}-7 x^{2} {\mathrm e}^{2}-20-16 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-2}}{x -4}\) | \(32\) |
parallelrisch | \(-\frac {\left (2 x^{3} {\mathrm e}^{2}-7 x^{2} {\mathrm e}^{2}-20-16 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-2}}{x -4}\) | \(32\) |
meijerg | \(-\frac {5 \,{\mathrm e}^{-2} x}{4 \left (-\frac {x}{4}+1\right )}-\frac {4 x \left (-\frac {1}{8} x^{2}-\frac {3}{2} x +12\right )}{-\frac {x}{4}+1}+\frac {31 x \left (-\frac {3 x}{4}+6\right )}{3 \left (-\frac {x}{4}+1\right )}-\frac {15 x}{-\frac {x}{4}+1}\) | \(59\) |
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Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=-\frac {{\left ({\left (2 \, x^{3} - 7 \, x^{2} - 4 \, x\right )} e^{2} - 20\right )} e^{\left (-2\right )}}{x - 4} \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=- 2 x^{2} - x + \frac {20}{x e^{2} - 4 e^{2}} \]
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Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=-{\left (2 \, x^{2} e^{2} + x e^{2} - \frac {20}{x - 4}\right )} e^{\left (-2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=-{\left (2 \, x^{2} e^{2} + x e^{2} - \frac {20}{x - 4}\right )} e^{\left (-2\right )} \]
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Time = 8.73 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{e^2 \left (16-8 x+x^2\right )} \, dx=-x-\frac {20}{4\,{\mathrm {e}}^2-x\,{\mathrm {e}}^2}-2\,x^2 \]
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