Integrand size = 49, antiderivative size = 29 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-3-2 x+\log \left (\frac {x^2}{1-4 x-\frac {2 x}{3-e^5}}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2009, 1607, 907} \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-2 x+2 \log (x)-\log \left (-2 \left (7-2 e^5\right ) x-e^5+3\right ) \]
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Rule 907
Rule 1607
Rule 2009
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (3-e^5\right )-2 \left (10-3 e^5\right ) x+4 \left (7-2 e^5\right ) x^2}{\left (3-e^5\right ) x-2 \left (7-2 e^5\right ) x^2} \, dx \\ & = \int \frac {2 \left (3-e^5\right )-2 \left (10-3 e^5\right ) x+4 \left (7-2 e^5\right ) x^2}{x \left (3-e^5-2 \left (7-2 e^5\right ) x\right )} \, dx \\ & = \int \left (-2+\frac {2}{x}+\frac {2 \left (7-2 e^5\right )}{3-e^5-2 \left (7-2 e^5\right ) x}\right ) \, dx \\ & = -2 x+2 \log (x)-\log \left (3-e^5-2 \left (7-2 e^5\right ) x\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-2 x+2 \log (x)-\log \left (3-e^5-14 x+4 e^5 x\right ) \]
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Time = 0.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
norman | \(-2 x +2 \ln \left (x \right )-\ln \left (4 x \,{\mathrm e}^{5}-{\mathrm e}^{5}-14 x +3\right )\) | \(26\) |
risch | \(-2 x +2 \ln \left (-x \right )-\ln \left (x \left (4 \,{\mathrm e}^{5}-14\right )+3-{\mathrm e}^{5}\right )\) | \(28\) |
parallelrisch | \(-2 x +2 \ln \left (x \right )-\ln \left (\frac {4 x \,{\mathrm e}^{5}-{\mathrm e}^{5}-14 x +3}{4 \,{\mathrm e}^{5}-14}\right )\) | \(36\) |
default | \(-2 x +\frac {2 \left (-2 \,{\mathrm e}^{5}+7\right ) \ln \left (4 x \,{\mathrm e}^{5}-{\mathrm e}^{5}-14 x +3\right )}{4 \,{\mathrm e}^{5}-14}+2 \ln \left (x \right )\) | \(40\) |
meijerg | \(\frac {\left ({\mathrm e}^{5}-3\right )^{2} \left (-8 \,{\mathrm e}^{5}+28\right ) \left (\frac {2 x \left (2 \,{\mathrm e}^{5}-7\right )}{3-{\mathrm e}^{5}}-\ln \left (1+\frac {2 x \left (2 \,{\mathrm e}^{5}-7\right )}{3-{\mathrm e}^{5}}\right )\right )}{4 \left (2 \,{\mathrm e}^{5}-7\right )^{2} \left (3-{\mathrm e}^{5}\right )}-\frac {\left (6 \,{\mathrm e}^{5}-20\right ) \left ({\mathrm e}^{5}-3\right ) \ln \left (1+\frac {2 x \left (2 \,{\mathrm e}^{5}-7\right )}{3-{\mathrm e}^{5}}\right )}{2 \left (3-{\mathrm e}^{5}\right ) \left (2 \,{\mathrm e}^{5}-7\right )}+\frac {2 \,{\mathrm e}^{5} \left ({\mathrm e}^{5}-3\right ) \left (\ln \left (x \right )+\ln \left (2\right )+\ln \left (2 \,{\mathrm e}^{5}-7\right )+\ln \left (-\frac {1}{3-{\mathrm e}^{5}}\right )+i \pi -\ln \left (1+\frac {2 x \left (2 \,{\mathrm e}^{5}-7\right )}{3-{\mathrm e}^{5}}\right )\right )}{\left (3-{\mathrm e}^{5}\right )^{2}}-\frac {6 \left ({\mathrm e}^{5}-3\right ) \left (\ln \left (x \right )+\ln \left (2\right )+\ln \left (2 \,{\mathrm e}^{5}-7\right )+\ln \left (-\frac {1}{3-{\mathrm e}^{5}}\right )+i \pi -\ln \left (1+\frac {2 x \left (2 \,{\mathrm e}^{5}-7\right )}{3-{\mathrm e}^{5}}\right )\right )}{\left (3-{\mathrm e}^{5}\right )^{2}}\) | \(248\) |
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Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-2 \, x - \log \left ({\left (4 \, x - 1\right )} e^{5} - 14 \, x + 3\right ) + 2 \, \log \left (x\right ) \]
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Time = 0.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=- 2 x + 2 \log {\left (x \right )} - \log {\left (x + \frac {9 - 3 e^{5}}{-42 + 12 e^{5}} \right )} \]
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Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-2 \, x - \log \left (2 \, x {\left (2 \, e^{5} - 7\right )} - e^{5} + 3\right ) + 2 \, \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=-\frac {2 \, {\left (2 \, x e^{5} - 7 \, x\right )}}{2 \, e^{5} - 7} - \log \left ({\left | 4 \, x e^{5} - 14 \, x - e^{5} + 3 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \]
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Time = 12.68 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {6-20 x+28 x^2+e^5 \left (-2+6 x-8 x^2\right )}{3 x-14 x^2+e^5 \left (-x+4 x^2\right )} \, dx=2\,\ln \left (x\right )-\ln \left (12\,x\,{\mathrm {e}}^5-3\,{\mathrm {e}}^5-42\,x+9\right )+\frac {28\,x}{4\,{\mathrm {e}}^5-14}-\frac {8\,x\,{\mathrm {e}}^5}{4\,{\mathrm {e}}^5-14} \]
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