Integrand size = 96, antiderivative size = 29 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=x+(2-x) \left (5+\frac {2}{\left (3-\frac {x}{5}\right ) \log (\log (\log (2 x)))}\right ) \]
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\[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=\int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{x \left (225-30 x+x^2\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx \\ & = \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{(-15+x)^2 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx \\ & = \int \left (-4-\frac {10 (-2+x)}{(-15+x) x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}-\frac {130}{(-15+x)^2 \log (\log (\log (2 x)))}\right ) \, dx \\ & = -4 x-10 \int \frac {-2+x}{(-15+x) x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x-10 \int \left (\frac {13}{15 (-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}+\frac {2}{15 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}\right ) \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x-\frac {4}{3} \int \frac {1}{x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-\frac {26}{3} \int \frac {1}{(-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x-\frac {4}{3} \text {Subst}\left (\int \frac {1}{x \log (x) \log ^2(\log (x))} \, dx,x,\log (2 x)\right )-\frac {26}{3} \int \frac {1}{(-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x-\frac {4}{3} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\log (\log (2 x))\right )-\frac {26}{3} \int \frac {1}{(-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x-\frac {4}{3} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (\log (\log (2 x)))\right )-\frac {26}{3} \int \frac {1}{(-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ & = -4 x+\frac {4}{3 \log (\log (\log (2 x)))}-\frac {26}{3} \int \frac {1}{(-15+x) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx-130 \int \frac {1}{(-15+x)^2 \log (\log (\log (2 x)))} \, dx \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-2 \left (2 x-\frac {5 (-2+x)}{(-15+x) \log (\log (\log (2 x)))}\right ) \]
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Time = 4.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-4 x +\frac {10 x -20}{\left (x -15\right ) \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}\) | \(23\) |
parallelrisch | \(\frac {-300-60 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) x^{2}+150 x +13500 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}{15 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) \left (x -15\right )}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \]
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Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=- 4 x + \frac {10 x - 20}{\left (x - 15\right ) \log {\left (\log {\left (\log {\left (2 x \right )} \right )} \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right )} \]
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Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-4 \, x + \frac {10 \, {\left (x - 2\right )}}{x \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 15 \, \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \]
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Time = 11.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=\frac {2\,\left (5\,x-2\,x^2\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )+30\,x\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )-10\right )}{\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )\,\left (x-15\right )} \]
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