Integrand size = 46, antiderivative size = 27 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=\frac {\left (3-e^x+6 x\right ) \log (\log (4))}{2 x (-4+2 x)} \]
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Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(27)=54\).
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1608, 27, 6874, 2208, 2209, 907} \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=\frac {e^x \log (\log (4))}{8 x}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {15 \log (\log (4))}{8 (2-x)} \]
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Rule 12
Rule 27
Rule 907
Rule 1608
Rule 2208
Rule 2209
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{16 x^2-16 x^3+4 x^4} \, dx \\ & = \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{x^2 \left (16-16 x+4 x^2\right )} \, dx \\ & = \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{4 (-2+x)^2 x^2} \, dx \\ & = \frac {1}{4} \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{(-2+x)^2 x^2} \, dx \\ & = \frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2}-\frac {6 \left (-1+x+x^2\right )}{(-2+x)^2 x^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \log (\log (4)) \int \frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2} \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \frac {-1+x+x^2}{(-2+x)^2 x^2} \, dx \\ & = -\left (\frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x}{2 (-2+x)^2}+\frac {e^x}{2 (-2+x)}+\frac {e^x}{2 x^2}-\frac {e^x}{2 x}\right ) \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \left (\frac {5}{4 (-2+x)^2}-\frac {1}{4 x^2}\right ) \, dx \\ & = -\frac {15 \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{(-2+x)^2} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x^2} \, dx+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx \\ & = -\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x}-\frac {1}{8} e^2 \text {Ei}(-2+x) \log (\log (4))+\frac {1}{8} \text {Ei}(x) \log (\log (4))+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx \\ & = -\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=-\frac {\left (-3+e^x-6 x\right ) \log (\log (4))}{4 (-2+x) x} \]
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Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(\frac {\ln \left (2 \ln \left (2\right )\right ) \left (3+6 x -{\mathrm e}^{x}\right )}{4 x \left (-2+x \right )}\) | \(25\) |
default | \(\frac {\ln \left (2 \ln \left (2\right )\right ) \left (-\frac {3}{2 x}+\frac {15}{2 \left (-2+x \right )}+\frac {{\mathrm e}^{x}}{2 x}-\frac {{\mathrm e}^{x}}{2 \left (-2+x \right )}\right )}{4}\) | \(37\) |
risch | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) \left (\frac {3 x}{2}+\frac {3}{4}\right )}{\left (-2+x \right ) x}-\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{x}}{4 \left (-2+x \right ) x}\) | \(40\) |
norman | \(\frac {\left (-\frac {\ln \left (2\right )}{4}-\frac {\ln \left (\ln \left (2\right )\right )}{4}\right ) {\mathrm e}^{x}+\left (\frac {3 \ln \left (2\right )}{2}+\frac {3 \ln \left (\ln \left (2\right )\right )}{2}\right ) x +\frac {3 \ln \left (2\right )}{4}+\frac {3 \ln \left (\ln \left (2\right )\right )}{4}}{\left (-2+x \right ) x}\) | \(45\) |
parts | \(-\frac {3 \ln \left (2 \ln \left (2\right )\right ) \left (\frac {1}{4 x}-\frac {5}{4 \left (-2+x \right )}\right )}{2}-\frac {\ln \left (2 \ln \left (2\right )\right ) \left (\frac {{\mathrm e}^{x}}{2 x -4}-\frac {{\mathrm e}^{x}}{2 x}\right )}{4}\) | \(46\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=\frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \left (2\right )\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 0.34 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=- \frac {x \left (- 6 \log {\left (2 \right )} - 6 \log {\left (\log {\left (2 \right )} \right )}\right ) - 3 \log {\left (2 \right )} - 3 \log {\left (\log {\left (2 \right )} \right )}}{4 x^{2} - 8 x} + \frac {\left (- \log {\left (2 \right )} - \log {\left (\log {\left (2 \right )} \right )}\right ) e^{x}}{4 x^{2} - 8 x} \]
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Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=-\frac {1}{4} \, {\left (\frac {3 \, {\left (x - 1\right )}}{x^{2} - 2 \, x} + \frac {e^{x}}{x^{2} - 2 \, x} - \frac {9}{x - 2}\right )} \log \left (2 \, \log \left (2\right )\right ) \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=\frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \left (2\right )\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \]
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Time = 11.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )\right ) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx=\frac {3\,\ln \left (\ln \left (4\right )\right )+x^2\,\ln \left ({\ln \left (4\right )}^3\right )-\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^x}{4\,x\,\left (x-2\right )} \]
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