Integrand size = 40, antiderivative size = 25 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=4+\frac {9 x^2}{(-4+e) \left (\frac {4}{5 x}-2 x\right )} \]
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Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1607, 2017, 28, 460} \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=-\frac {45 x^3}{2 (4-e) \left (2-5 x^2\right )} \]
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Rule 28
Rule 460
Rule 1607
Rule 2017
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (270-225 x^2\right )}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx \\ & = \int \frac {x^2 \left (270-225 x^2\right )}{-8 (4-e)+40 (4-e) x^2-50 (4-e) x^4} \, dx \\ & = -\left ((50 (4-e)) \int \frac {x^2 \left (270-225 x^2\right )}{\left (20 (4-e)-50 (4-e) x^2\right )^2} \, dx\right ) \\ & = -\frac {45 x^3}{2 (4-e) \left (2-5 x^2\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=-\frac {45 x^3}{2 (-4+e) \left (-2+5 x^2\right )} \]
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
norman | \(-\frac {45 x^{3}}{2 \left ({\mathrm e}-4\right ) \left (5 x^{2}-2\right )}\) | \(21\) |
parallelrisch | \(-\frac {45 x^{3}}{2 \left ({\mathrm e}-4\right ) \left (5 x^{2}-2\right )}\) | \(21\) |
default | \(\frac {-9 x -\frac {18 x}{5 \left (x^{2}-\frac {2}{5}\right )}}{2 \,{\mathrm e}-8}\) | \(25\) |
gosper | \(-\frac {45 x^{3}}{2 \left (5 x^{2} {\mathrm e}-20 x^{2}-2 \,{\mathrm e}+8\right )}\) | \(26\) |
risch | \(-\frac {9 x}{2 \,{\mathrm e}-8}+\frac {\left (-\frac {18 \,{\mathrm e}}{5}+\frac {72}{5}\right ) x}{\left (2 \,{\mathrm e}-8\right ) \left (x^{2} {\mathrm e}-4 x^{2}-\frac {2 \,{\mathrm e}}{5}+\frac {8}{5}\right )}\) | \(48\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=\frac {45 \, x^{3}}{2 \, {\left (20 \, x^{2} - {\left (5 \, x^{2} - 2\right )} e - 8\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=- \frac {9 x}{-8 + 2 e} - \frac {9 x}{x^{2} \left (-20 + 5 e\right ) - 2 e + 8} \]
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Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=-\frac {9 \, x}{5 \, x^{2} {\left (e - 4\right )} - 2 \, e + 8} - \frac {9 \, x}{2 \, {\left (e - 4\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=-\frac {9 \, x}{2 \, {\left (e - 4\right )}} - \frac {9 \, x}{{\left (5 \, x^{2} - 2\right )} {\left (e - 4\right )}} \]
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Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx=-\frac {45\,x^3}{2\,\left (5\,x^2-2\right )\,\left (\mathrm {e}-4\right )} \]
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