Integrand size = 30, antiderivative size = 20 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=\frac {-4+x+\log (x)-4 x \log (x)}{\frac {1}{5}-5 x} \]
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Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1608, 27, 6873, 12, 6874, 907, 2351, 31} \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=-\frac {99}{5 (1-25 x)}+\frac {105 x \log (x)}{1-25 x}+5 \log (x) \]
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Rule 12
Rule 27
Rule 31
Rule 907
Rule 1608
Rule 2351
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {5-640 x+500 x^2+105 x \log (x)}{x \left (1-50 x+625 x^2\right )} \, dx \\ & = \int \frac {5-640 x+500 x^2+105 x \log (x)}{x (-1+25 x)^2} \, dx \\ & = \int \frac {5 \left (1-128 x+100 x^2+21 x \log (x)\right )}{(1-25 x)^2 x} \, dx \\ & = 5 \int \frac {1-128 x+100 x^2+21 x \log (x)}{(1-25 x)^2 x} \, dx \\ & = 5 \int \left (\frac {1-128 x+100 x^2}{x (-1+25 x)^2}+\frac {21 \log (x)}{(-1+25 x)^2}\right ) \, dx \\ & = 5 \int \frac {1-128 x+100 x^2}{x (-1+25 x)^2} \, dx+105 \int \frac {\log (x)}{(-1+25 x)^2} \, dx \\ & = \frac {105 x \log (x)}{1-25 x}+5 \int \left (\frac {1}{x}-\frac {99}{(-1+25 x)^2}-\frac {21}{-1+25 x}\right ) \, dx+105 \int \frac {1}{-1+25 x} \, dx \\ & = -\frac {99}{5 (1-25 x)}+5 \log (x)+\frac {105 x \log (x)}{1-25 x} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=\frac {5 (99+25 (-1+4 x) \log (x))}{-25+625 x} \]
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Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {-5 \ln \left (x \right )+20 x \ln \left (x \right )+\frac {99}{5}}{-1+25 x}\) | \(20\) |
parallelrisch | \(\frac {500 x \ln \left (x \right )+495-125 \ln \left (x \right )}{-25+625 x}\) | \(21\) |
default | \(5 \ln \left (x \right )+\frac {99}{5 \left (-1+25 x \right )}-\frac {105 \ln \left (x \right ) x}{-1+25 x}\) | \(27\) |
parts | \(5 \ln \left (x \right )+\frac {99}{5 \left (-1+25 x \right )}-\frac {105 \ln \left (x \right ) x}{-1+25 x}\) | \(27\) |
risch | \(-\frac {21 \ln \left (x \right )}{5 \left (-1+25 x \right )}+\frac {100 x \ln \left (x \right )-4 \ln \left (x \right )+99}{-5+125 x}\) | \(33\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=\frac {25 \, {\left (4 \, x - 1\right )} \log \left (x\right ) + 99}{5 \, {\left (25 \, x - 1\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=\frac {4 \log {\left (x \right )}}{5} - \frac {21 \log {\left (x \right )}}{125 x - 5} + \frac {99}{125 x - 5} \]
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Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=-\frac {21 \, \log \left (x\right )}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=-\frac {21 \, \log \left (x\right )}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \left (x\right ) \]
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Time = 8.77 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx=\frac {4\,\ln \left (x\right )}{5}-\frac {\frac {21\,\ln \left (x\right )}{5}-\frac {99}{5}}{25\,x-1} \]
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