Integrand size = 53, antiderivative size = 23 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\log \left (\frac {1}{5}-\frac {4}{3 x}+\log ((3-2 x) x \log (\log (2)))\right ) \]
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\[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-60-5 x+60 x^2}{(3-2 x) x (20-3 x-15 x \log ((3-2 x) x \log (\log (2))))} \, dx \\ & = \int \left (\frac {30}{-20+3 x+15 x \log ((3-2 x) x \log (\log (2)))}+\frac {20}{x (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))}+\frac {45}{(-3+2 x) (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))}\right ) \, dx \\ & = 20 \int \frac {1}{x (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))} \, dx+30 \int \frac {1}{-20+3 x+15 x \log ((3-2 x) x \log (\log (2)))} \, dx+45 \int \frac {1}{(-3+2 x) (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))} \, dx \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=-\log (x)+\log (20-3 x-15 x \log ((3-2 x) x \log (\log (2)))) \]
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Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17
method | result | size |
risch | \(\ln \left (\ln \left (\left (-2 x^{2}+3 x \right ) \ln \left (\ln \left (2\right )\right )\right )+\frac {3 x -20}{15 x}\right )\) | \(27\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (x \ln \left (\left (-2 x^{2}+3 x \right ) \ln \left (\ln \left (2\right )\right )\right )+\frac {x}{5}-\frac {4}{3}\right )\) | \(28\) |
norman | \(-\ln \left (x \right )+\ln \left (15 x \ln \left (\left (-2 x^{2}+3 x \right ) \ln \left (\ln \left (2\right )\right )\right )+3 x -20\right )\) | \(29\) |
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\log \left (\frac {15 \, x \log \left (-{\left (2 \, x^{2} - 3 \, x\right )} \log \left (\log \left (2\right )\right )\right ) + 3 \, x - 20}{x}\right ) \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\log {\left (\log {\left (\left (- 2 x^{2} + 3 x\right ) \log {\left (\log {\left (2 \right )} \right )} \right )} + \frac {3 x - 20}{15 x} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\log \left (\frac {3 \, x {\left (5 \, \log \left (\log \left (\log \left (2\right )\right )\right ) + 1\right )} + 15 \, x \log \left (x\right ) + 15 \, x \log \left (-2 \, x + 3\right ) - 20}{15 \, x}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\log \left (15 \, x \log \left (-2 \, x^{2} \log \left (\log \left (2\right )\right ) + 3 \, x \log \left (\log \left (2\right )\right )\right ) + 3 \, x - 20\right ) - \log \left (x\right ) \]
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Time = 10.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx=\ln \left (3\,x+15\,x\,\left (\ln \left (x\,\left (2\,x-3\right )\right )+\ln \left (-\ln \left (\ln \left (2\right )\right )\right )\right )-20\right )-\ln \left (x\right ) \]
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