Integrand size = 25, antiderivative size = 21 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (\frac {\left (-x+\frac {x^2}{16}\right ) \log (x)}{24 e^4}\right ) \]
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Time = 0.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1607, 6874, 78, 2339, 29} \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log (16-x)+\log (x)+\log (\log (x)) \]
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Rule 29
Rule 78
Rule 1607
Rule 2339
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-16+x+(-16+2 x) \log (x)}{(-16+x) x \log (x)} \, dx \\ & = \int \left (\frac {2 (-8+x)}{(-16+x) x}+\frac {1}{x \log (x)}\right ) \, dx \\ & = 2 \int \frac {-8+x}{(-16+x) x} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = 2 \int \left (\frac {1}{2 (-16+x)}+\frac {1}{2 x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \log (16-x)+\log (x)+\log (\log (x)) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log (16-x)+\log (x)+\log (\log (x)) \]
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Time = 1.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52
method | result | size |
default | \(\ln \left (\ln \left (x \right )\right )+\ln \left (x \left (x -16\right )\right )\) | \(11\) |
norman | \(\ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -16\right )\) | \(11\) |
parallelrisch | \(\ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -16\right )\) | \(11\) |
parts | \(\ln \left (\ln \left (x \right )\right )+\ln \left (x \left (x -16\right )\right )\) | \(11\) |
risch | \(\ln \left (x^{2}-16 x \right )+\ln \left (\ln \left (x \right )\right )\) | \(13\) |
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Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x^{2} - 16 \, x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log {\left (x^{2} - 16 x \right )} + \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x - 16\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x - 16\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 10.59 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\ln \left (x-16\right )+\ln \left (\ln \left (x\right )\right )+\ln \left (x\right ) \]
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