Integrand size = 28, antiderivative size = 19 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=e^2-\frac {x^2}{8 e^4 (-1+x)} \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 27, 697} \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=\frac {1}{8 e^4 (1-x)}-\frac {x}{8 e^4} \]
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Rule 12
Rule 27
Rule 697
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {2 x-x^2}{4-8 x+4 x^2} \, dx}{2 e^4} \\ & = \frac {\int \frac {2 x-x^2}{4 (-1+x)^2} \, dx}{2 e^4} \\ & = \frac {\int \frac {2 x-x^2}{(-1+x)^2} \, dx}{8 e^4} \\ & = \frac {\int \left (-1+\frac {1}{(-1+x)^2}\right ) \, dx}{8 e^4} \\ & = \frac {1}{8 e^4 (1-x)}-\frac {x}{8 e^4} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=-\frac {\frac {1}{-1+x}+x}{8 e^4} \]
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Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79
method | result | size |
norman | \(-\frac {x^{2} {\mathrm e}^{-4}}{8 \left (-1+x \right )}\) | \(15\) |
risch | \(-\frac {x \,{\mathrm e}^{-4}}{8}-\frac {{\mathrm e}^{-4}}{8 \left (-1+x \right )}\) | \(16\) |
gosper | \(-\frac {x^{2} {\mathrm e}^{-4}}{8 \left (-1+x \right )}\) | \(18\) |
parallelrisch | \(-\frac {x^{2} {\mathrm e}^{-4}}{8 \left (-1+x \right )}\) | \(18\) |
default | \(\frac {{\mathrm e}^{-4} \left (-x -\frac {1}{-1+x}\right )}{8}\) | \(21\) |
meijerg | \(\frac {{\mathrm e}^{-4-\ln \left (2\right )} \left (-\frac {x \left (-3 x +6\right )}{3 \left (1-x \right )}-2 \ln \left (1-x \right )\right )}{4}+\frac {{\mathrm e}^{-4-\ln \left (2\right )} \left (\frac {x}{1-x}+\ln \left (1-x \right )\right )}{2}\) | \(60\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=-\frac {{\left (x^{2} - x + 1\right )} e^{\left (-\log \left (2\right ) - 4\right )}}{4 \, {\left (x - 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=- \frac {x}{8 e^{4}} - \frac {1}{8 x e^{4} - 8 e^{4}} \]
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Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=-\frac {1}{8} \, {\left (x + \frac {1}{x - 1}\right )} e^{\left (-4\right )} \]
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Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=-\frac {1}{4} \, {\left (x + \frac {1}{x - 1}\right )} e^{\left (-\log \left (2\right ) - 4\right )} \]
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Time = 10.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {2 x-x^2}{2 e^4 \left (4-8 x+4 x^2\right )} \, dx=-\frac {x\,{\mathrm {e}}^{-4}}{8}-\frac {{\mathrm {e}}^{-4}}{8\,\left (x-1\right )} \]
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