Integrand size = 52, antiderivative size = 24 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {1}{5} x \left (5+x+x^2+\frac {x+\log (\log (x))}{5-2 x}\right ) \]
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\[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{5 (-5+2 x)^2 \log (x)} \, dx \\ & = \frac {1}{5} \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{(-5+2 x)^2 \log (x)} \, dx \\ & = \frac {1}{5} \int \left (\frac {5-2 x+125 \log (x)-40 x \log (x)+53 x^2 \log (x)-52 x^3 \log (x)+12 x^4 \log (x)}{(-5+2 x)^2 \log (x)}+\frac {5 \log (\log (x))}{(-5+2 x)^2}\right ) \, dx \\ & = \frac {1}{5} \int \frac {5-2 x+125 \log (x)-40 x \log (x)+53 x^2 \log (x)-52 x^3 \log (x)+12 x^4 \log (x)}{(-5+2 x)^2 \log (x)} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {125}{(-5+2 x)^2}-\frac {40 x}{(-5+2 x)^2}+\frac {53 x^2}{(-5+2 x)^2}-\frac {52 x^3}{(-5+2 x)^2}+\frac {12 x^4}{(-5+2 x)^2}-\frac {1}{(-5+2 x) \log (x)}\right ) \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {25}{2 (5-2 x)}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\frac {12}{5} \int \frac {x^4}{(-5+2 x)^2} \, dx-8 \int \frac {x}{(-5+2 x)^2} \, dx-\frac {52}{5} \int \frac {x^3}{(-5+2 x)^2} \, dx+\frac {53}{5} \int \frac {x^2}{(-5+2 x)^2} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {25}{2 (5-2 x)}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\frac {12}{5} \int \left (\frac {75}{16}+\frac {5 x}{4}+\frac {x^2}{4}+\frac {625}{16 (-5+2 x)^2}+\frac {125}{4 (-5+2 x)}\right ) \, dx-8 \int \left (\frac {5}{2 (-5+2 x)^2}+\frac {1}{2 (-5+2 x)}\right ) \, dx-\frac {52}{5} \int \left (\frac {5}{4}+\frac {x}{4}+\frac {125}{8 (-5+2 x)^2}+\frac {75}{8 (-5+2 x)}\right ) \, dx+\frac {53}{5} \int \left (\frac {1}{4}+\frac {25}{4 (-5+2 x)^2}+\frac {5}{2 (-5+2 x)}\right ) \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {5}{4 (5-2 x)}+\frac {9 x}{10}+\frac {x^2}{5}+\frac {x^3}{5}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {25+90 x-16 x^2+12 x^3-8 x^4+4 x \log (\log (x))}{5 (20-8 x)} \]
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Time = 1.51 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38
method | result | size |
parallelrisch | \(\frac {20 x^{4}-30 x^{3}-10 x \ln \left (\ln \left (x \right )\right )+40 x^{2}-625}{-250+100 x}\) | \(33\) |
risch | \(-\frac {\ln \left (\ln \left (x \right )\right )}{2 \left (-5+2 x \right )}-\frac {-8 x^{4}+12 x^{3}+4 x \ln \left (\ln \left (x \right )\right )-16 x^{2}-10 \ln \left (\ln \left (x \right )\right )+90 x +25}{20 \left (-5+2 x \right )}\) | \(54\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {8 \, x^{4} - 12 \, x^{3} + 16 \, x^{2} - 4 \, x \log \left (\log \left (x\right )\right ) - 90 \, x - 25}{20 \, {\left (2 \, x - 5\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {x^{3}}{5} + \frac {x^{2}}{5} + \frac {9 x}{10} - \frac {\log {\left (\log {\left (x \right )} \right )}}{10} - \frac {5}{8 x - 20} - \frac {\log {\left (\log {\left (x \right )} \right )}}{4 x - 10} \]
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Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {8 \, x^{4} - 12 \, x^{3} + 16 \, x^{2} - 4 \, x \log \left (\log \left (x\right )\right ) - 90 \, x - 25}{20 \, {\left (2 \, x - 5\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} + \frac {9}{10} \, x - \frac {\log \left (\log \left (x\right )\right )}{2 \, {\left (2 \, x - 5\right )}} - \frac {5}{4 \, {\left (2 \, x - 5\right )}} - \frac {1}{10} \, \log \left (\log \left (x\right )\right ) \]
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Time = 12.76 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {9\,x}{10}-\frac {\ln \left (\ln \left (x\right )\right )}{10}-\frac {5}{8\,\left (x-\frac {5}{2}\right )}+\frac {x^2}{5}+\frac {x^3}{5}-\frac {\ln \left (\ln \left (x\right )\right )}{4\,\left (x-\frac {5}{2}\right )} \]
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