Integrand size = 57, antiderivative size = 32 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=4+x-\frac {x}{5-x}+(14+2 x) \left (-x+\frac {\log (x)}{4 x}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.123, Rules used = {1608, 27, 12, 6874, 46, 45, 2341} \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=-2 x^2-13 x-\frac {5}{5-x}+\frac {\log (x)}{2}+\frac {7 \log (x)}{2 x} \]
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Rule 12
Rule 27
Rule 45
Rule 46
Rule 1608
Rule 2341
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{x^2 \left (50-20 x+2 x^2\right )} \, dx \\ & = \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{2 (-5+x)^2 x^2} \, dx \\ & = \frac {1}{2} \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{(-5+x)^2 x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {663}{(-5+x)^2}+\frac {175}{(-5+x)^2 x^2}-\frac {45}{(-5+x)^2 x}+\frac {61 x}{(-5+x)^2}+\frac {54 x^2}{(-5+x)^2}-\frac {8 x^3}{(-5+x)^2}-\frac {7 \log (x)}{x^2}\right ) \, dx \\ & = -\frac {663}{2 (5-x)}-\frac {7}{2} \int \frac {\log (x)}{x^2} \, dx-4 \int \frac {x^3}{(-5+x)^2} \, dx-\frac {45}{2} \int \frac {1}{(-5+x)^2 x} \, dx+27 \int \frac {x^2}{(-5+x)^2} \, dx+\frac {61}{2} \int \frac {x}{(-5+x)^2} \, dx+\frac {175}{2} \int \frac {1}{(-5+x)^2 x^2} \, dx \\ & = -\frac {663}{2 (5-x)}+\frac {7}{2 x}+\frac {7 \log (x)}{2 x}-4 \int \left (10+\frac {125}{(-5+x)^2}+\frac {75}{-5+x}+x\right ) \, dx-\frac {45}{2} \int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx+27 \int \left (1+\frac {25}{(-5+x)^2}+\frac {10}{-5+x}\right ) \, dx+\frac {61}{2} \int \left (\frac {5}{(-5+x)^2}+\frac {1}{-5+x}\right ) \, dx+\frac {175}{2} \int \left (\frac {1}{25 (-5+x)^2}-\frac {2}{125 (-5+x)}+\frac {1}{25 x^2}+\frac {2}{125 x}\right ) \, dx \\ & = -\frac {5}{5-x}-13 x-2 x^2+\frac {\log (x)}{2}+\frac {7 \log (x)}{2 x} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=\frac {1}{2} \left (-\frac {10}{5-x}-26 x-4 x^2+\log (x)+\frac {7 \log (x)}{x}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {7 \ln \left (x \right )}{2 x}-2 x^{2}-13 x +\frac {\ln \left (x \right )}{2}+\frac {5}{-5+x}\) | \(28\) |
parts | \(\frac {7 \ln \left (x \right )}{2 x}-2 x^{2}-13 x +\frac {\ln \left (x \right )}{2}+\frac {5}{-5+x}\) | \(28\) |
norman | \(\frac {330 x +x \ln \left (x \right )+\frac {x^{2} \ln \left (x \right )}{2}-3 x^{3}-2 x^{4}-\frac {35 \ln \left (x \right )}{2}}{\left (-5+x \right ) x}\) | \(39\) |
risch | \(\frac {7 \ln \left (x \right )}{2 x}+\frac {-4 x^{3}+x \ln \left (x \right )-6 x^{2}-5 \ln \left (x \right )+130 x +10}{2 x -10}\) | \(39\) |
parallelrisch | \(\frac {-4 x^{4}-6 x^{3}+x^{2} \ln \left (x \right )+2 x \ln \left (x \right )+660 x -35 \ln \left (x \right )}{2 x \left (-5+x \right )}\) | \(40\) |
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Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=-\frac {4 \, x^{4} + 6 \, x^{3} - 130 \, x^{2} - {\left (x^{2} + 2 \, x - 35\right )} \log \left (x\right ) - 10 \, x}{2 \, {\left (x^{2} - 5 \, x\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=- 2 x^{2} - 13 x + \frac {\log {\left (x \right )}}{2} + \frac {5}{x - 5} + \frac {7 \log {\left (x \right )}}{2 x} \]
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Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=-2 \, x^{2} - 13 \, x - \frac {7 \, {\left (2 \, x - 5\right )}}{2 \, {\left (x^{2} - 5 \, x\right )}} + \frac {7 \, {\left (\log \left (x\right ) + 1\right )}}{2 \, x} + \frac {17}{2 \, {\left (x - 5\right )}} + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=-2 \, x^{2} - 13 \, x + \frac {7 \, \log \left (x\right )}{2 \, x} + \frac {5}{x - 5} + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 14.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx=\frac {\ln \left (x\right )}{2}-13\,x-2\,x^2-\frac {\frac {35\,\ln \left (x\right )}{2}-x\,\left (\frac {7\,\ln \left (x\right )}{2}+5\right )}{x\,\left (x-5\right )} \]
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