Integrand size = 29, antiderivative size = 22 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {1}{2} x \left (e^{e^4}+\frac {x}{-2 x+x^2}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {27, 12, 1864} \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {e^{e^4} x}{2}-\frac {1}{2-x} \]
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Rule 12
Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{2 (-2+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{(-2+x)^2} \, dx \\ & = \frac {1}{2} \int \left (e^{e^4}-\frac {2}{(-2+x)^2}\right ) \, dx \\ & = -\frac {1}{2-x}+\frac {e^{e^4} x}{2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {1}{2} \left (\frac {2}{-2+x}+e^{e^4} (-2+x)\right ) \]
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Time = 0.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {x \,{\mathrm e}^{{\mathrm e}^{4}}}{2}+\frac {1}{-2+x}\) | \(13\) |
risch | \(\frac {x \,{\mathrm e}^{{\mathrm e}^{4}}}{2}+\frac {1}{-2+x}\) | \(13\) |
gosper | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{4}}+2-4 \,{\mathrm e}^{{\mathrm e}^{4}}}{2 x -4}\) | \(22\) |
norman | \(\frac {\frac {x^{2} {\mathrm e}^{{\mathrm e}^{4}}}{2}+1-2 \,{\mathrm e}^{{\mathrm e}^{4}}}{-2+x}\) | \(22\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{4}}+2-4 \,{\mathrm e}^{{\mathrm e}^{4}}}{2 x -4}\) | \(22\) |
meijerg | \(-\frac {x}{4 \left (1-\frac {x}{2}\right )}-{\mathrm e}^{{\mathrm e}^{4}} \left (-\frac {x \left (-\frac {3 x}{2}+6\right )}{6 \left (1-\frac {x}{2}\right )}-2 \ln \left (1-\frac {x}{2}\right )\right )-2 \,{\mathrm e}^{{\mathrm e}^{4}} \left (\frac {x}{2-x}+\ln \left (1-\frac {x}{2}\right )\right )+\frac {x \,{\mathrm e}^{{\mathrm e}^{4}}}{2-x}\) | \(76\) |
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {{\left (x^{2} - 2 \, x\right )} e^{\left (e^{4}\right )} + 2}{2 \, {\left (x - 2\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {x e^{e^{4}}}{2} + \frac {1}{x - 2} \]
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Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {1}{2} \, x e^{\left (e^{4}\right )} + \frac {1}{x - 2} \]
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Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {1}{2} \, x e^{\left (e^{4}\right )} + \frac {1}{x - 2} \]
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Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55 \[ \int \frac {-2+e^{e^4} \left (4-4 x+x^2\right )}{8-8 x+2 x^2} \, dx=\frac {1}{x-2}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^4}}{2} \]
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