Integrand size = 51, antiderivative size = 22 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=\log \left (e^{-x} x \left (-x+\frac {4+3 e+x}{e^5}\right )\right ) \]
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Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6, 1607, 1834} \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=-x+\log (x)+\log \left (\left (1-e^5\right ) x+3 e+4\right ) \]
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Rule 6
Rule 1607
Rule 1834
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{(-4-3 e) x-x^2+e^5 x^2} \, dx \\ & = \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{(-4-3 e) x+\left (-1+e^5\right ) x^2} \, dx \\ & = \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{x \left (-4-3 e+\left (-1+e^5\right ) x\right )} \, dx \\ & = \int \left (-1+\frac {1}{x}+\frac {1-e^5}{4+3 e+\left (1-e^5\right ) x}\right ) \, dx \\ & = -x+\log (x)+\log \left (4+3 e+\left (1-e^5\right ) x\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=-x+\log (x)+\log \left (4+3 e+x-e^5 x\right ) \]
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Time = 0.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
default | \(-x +\ln \left (x \,{\mathrm e}^{5}-3 \,{\mathrm e}-x -4\right )+\ln \left (x \right )\) | \(21\) |
norman | \(-x +\ln \left (x \,{\mathrm e}^{5}-3 \,{\mathrm e}-x -4\right )+\ln \left (x \right )\) | \(21\) |
risch | \(-x +\ln \left (\left ({\mathrm e}^{5}-1\right ) x^{2}+\left (-3 \,{\mathrm e}-4\right ) x \right )\) | \(23\) |
parallelrisch | \(-x +\ln \left (x \right )+\ln \left (\frac {x \,{\mathrm e}^{5}-3 \,{\mathrm e}-x -4}{{\mathrm e}^{5}-1}\right )\) | \(28\) |
meijerg | \(\frac {\left (3 \,{\mathrm e}+4\right )^{2} \left (1-{\mathrm e}^{5}\right ) \left (\frac {x \left ({\mathrm e}^{5}-1\right )}{-3 \,{\mathrm e}-4}-\ln \left (1+\frac {x \left ({\mathrm e}^{5}-1\right )}{-3 \,{\mathrm e}-4}\right )\right )}{\left ({\mathrm e}^{5}-1\right )^{2} \left (-3 \,{\mathrm e}-4\right )}-\frac {\left (2 \,{\mathrm e}^{5}+3 \,{\mathrm e}+2\right ) \left (3 \,{\mathrm e}+4\right ) \ln \left (1+\frac {x \left ({\mathrm e}^{5}-1\right )}{-3 \,{\mathrm e}-4}\right )}{\left (-3 \,{\mathrm e}-4\right ) \left ({\mathrm e}^{5}-1\right )}+\frac {3 \,{\mathrm e} \left (3 \,{\mathrm e}+4\right ) \left (\ln \left (x \right )+\ln \left ({\mathrm e}^{5}-1\right )+\ln \left (-\frac {1}{-3 \,{\mathrm e}-4}\right )+i \pi -\ln \left (1+\frac {x \left ({\mathrm e}^{5}-1\right )}{-3 \,{\mathrm e}-4}\right )\right )}{\left (-3 \,{\mathrm e}-4\right )^{2}}+\frac {4 \left (3 \,{\mathrm e}+4\right ) \left (\ln \left (x \right )+\ln \left ({\mathrm e}^{5}-1\right )+\ln \left (-\frac {1}{-3 \,{\mathrm e}-4}\right )+i \pi -\ln \left (1+\frac {x \left ({\mathrm e}^{5}-1\right )}{-3 \,{\mathrm e}-4}\right )\right )}{\left (-3 \,{\mathrm e}-4\right )^{2}}\) | \(232\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=-x + \log \left (x^{2} e^{5} - x^{2} - 3 \, x e - 4 \, x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=- x + \log {\left (x^{2} \left (-1 + e^{5}\right ) + x \left (- 3 e - 4\right ) \right )} \]
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Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=-x + \log \left (x {\left (e^{5} - 1\right )} - 3 \, e - 4\right ) + \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=-\frac {x e^{5} - x}{e^{5} - 1} + \log \left ({\left | x e^{5} - x - 3 \, e - 4 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-4+2 x+x^2+e (-3+3 x)+e^5 \left (2 x-x^2\right )}{-4 x-3 e x-x^2+e^5 x^2} \, dx=\ln \left (x^2\,\left ({\mathrm {e}}^5-1\right )-x\,\left (3\,\mathrm {e}+4\right )\right )-x \]
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