Integrand size = 32, antiderivative size = 19 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x-\log (4) \log \left (\frac {5}{x}-\log (x)\right ) \]
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\[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=\int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{x^2 \left (-\frac {5}{x}+\log (x)\right )} \, dx \\ & = \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{x (-5+x \log (x))} \, dx \\ & = \int \left (6-\frac {(5+x) \log (4)}{x (-5+x \log (x))}\right ) \, dx \\ & = 6 x-\log (4) \int \frac {5+x}{x (-5+x \log (x))} \, dx \\ & = 6 x-\log (4) \int \left (\frac {1}{-5+x \log (x)}+\frac {5}{x (-5+x \log (x))}\right ) \, dx \\ & = 6 x-\log (4) \int \frac {1}{-5+x \log (x)} \, dx-(5 \log (4)) \int \frac {1}{x (-5+x \log (x))} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x+\log (4) \log (x)-\log (4) \log (5-x \log (x)) \]
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
risch | \(6 x -2 \ln \left (2\right ) \ln \left (\ln \left (x \right )-\frac {5}{x}\right )\) | \(18\) |
default | \(6 x -2 \ln \left (2\right ) \left (-\ln \left (x \right )+\ln \left (x \ln \left (x \right )-5\right )\right )\) | \(21\) |
norman | \(2 \ln \left (2\right ) \ln \left (x \right )+6 x -2 \ln \left (2\right ) \ln \left (x \ln \left (x \right )-5\right )\) | \(22\) |
parallelrisch | \(2 \ln \left (2\right ) \ln \left (x \right )+6 x -2 \ln \left (2\right ) \ln \left (x \ln \left (x \right )-5\right )\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (\frac {x \log \left (x\right ) - 5}{x}\right ) + 6 \, x \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x - 2 \log {\left (2 \right )} \log {\left (\log {\left (x \right )} - \frac {5}{x} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (\frac {x \log \left (x\right ) - 5}{x}\right ) + 6 \, x \]
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Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (x \log \left (x\right ) - 5\right ) + 2 \, \log \left (2\right ) \log \left (x\right ) + 6 \, x \]
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Time = 12.83 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6\,x-2\,\ln \left (2\right )\,\ln \left (x\,\ln \left (x\right )-5\right )+2\,\ln \left (2\right )\,\ln \left (x\right ) \]
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