Integrand size = 37, antiderivative size = 17 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=\log \left (\frac {x \log (x)}{1-e^2+3 x}\right ) \]
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Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {6, 1607, 6874, 36, 31, 29, 2339} \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=\log (x)-\log \left (3 x-e^2+1\right )+\log (\log (x)) \]
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Rule 6
Rule 29
Rule 31
Rule 36
Rule 1607
Rule 2339
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (\left (-1+e^2\right ) x-3 x^2\right ) \log (x)} \, dx \\ & = \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-1+e^2-3 x\right ) x \log (x)} \, dx \\ & = \int \left (\frac {-1+e^2}{\left (-1+e^2-3 x\right ) x}+\frac {1}{x \log (x)}\right ) \, dx \\ & = \left (-1+e^2\right ) \int \frac {1}{\left (-1+e^2-3 x\right ) x} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = 3 \int \frac {1}{-1+e^2-3 x} \, dx+\int \frac {1}{x} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \log (x)-\log \left (1-e^2+3 x\right )+\log (\log (x)) \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=\log (x)-\log \left (1-e^2+3 x\right )+\log (\log (x)) \]
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Time = 0.45 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (x \right )-\ln \left ({\mathrm e}^{2}-1-3 x \right )+\ln \left (\ln \left (x \right )\right )\) | \(17\) |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )-\ln \left (-\frac {{\mathrm e}^{2}}{3}+\frac {1}{3}+x \right )+\ln \left (x \right )\) | \(17\) |
risch | \(-\ln \left (1+3 x -{\mathrm e}^{2}\right )+\ln \left (x \right )+\ln \left (\ln \left (x \right )\right )\) | \(19\) |
default | \(\ln \left (\ln \left (x \right )\right )+\left ({\mathrm e}^{2}-1\right ) \left (-\frac {\ln \left (1+3 x -{\mathrm e}^{2}\right )}{{\mathrm e}^{2}-1}+\frac {\ln \left (x \right )}{{\mathrm e}^{2}-1}\right )\) | \(38\) |
parts | \(\ln \left (\ln \left (x \right )\right )+\left ({\mathrm e}^{2}-1\right ) \left (-\frac {\ln \left (1+3 x -{\mathrm e}^{2}\right )}{{\mathrm e}^{2}-1}+\frac {\ln \left (x \right )}{{\mathrm e}^{2}-1}\right )\) | \(38\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=-\log \left (3 \, x - e^{2} + 1\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (14) = 28\).
Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 8.29 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=\left (1 - e^{2}\right ) \left (\frac {\log {\left (x - \frac {e^{4}}{6 \left (-1 + e\right ) \left (1 + e\right )} - \frac {e^{2}}{6} - \frac {1}{6 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6} + \frac {e^{2}}{3 \left (-1 + e\right ) \left (1 + e\right )} \right )}}{\left (-1 + e\right ) \left (1 + e\right )} - \frac {\log {\left (x - \frac {e^{2}}{6} - \frac {e^{2}}{3 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6} + \frac {e^{4}}{6 \left (-1 + e\right ) \left (1 + e\right )} \right )}}{\left (-1 + e\right ) \left (1 + e\right )}\right ) + \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=-\log \left (3 \, x - e^{2} + 1\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=-\log \left (3 \, x - e^{2} + 1\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 11.50 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-1+e^2-3 x+\left (-1+e^2\right ) \log (x)}{\left (-x+e^2 x-3 x^2\right ) \log (x)} \, dx=\ln \left (\ln \left (x\right )\right )-\ln \left (3\,x-{\mathrm {e}}^2+1\right )+\ln \left (x\right ) \]
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