Integrand size = 13, antiderivative size = 11 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (2)+\log (2-x \log (x)) \]
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Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6816} \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (2-x \log (x)) \]
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Rule 6816
Rubi steps \begin{align*} \text {integral}& = \log (2-x \log (x)) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (-2+x \log (x)) \]
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Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\ln \left (x \ln \left (x \right )-2\right )\) | \(8\) |
default | \(\ln \left (x \ln \left (x \right )-2\right )\) | \(8\) |
norman | \(\ln \left (x \ln \left (x \right )-2\right )\) | \(8\) |
parallelrisch | \(\ln \left (x \ln \left (x \right )-2\right )\) | \(8\) |
risch | \(\ln \left (x \right )+\ln \left (\ln \left (x \right )-\frac {2}{x}\right )\) | \(13\) |
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none
Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x\right ) + \log \left (\frac {x \log \left (x\right ) - 2}{x}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \right )} - \frac {2}{x} \right )} \]
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none
Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x \log \left (x\right ) - 2\right ) \]
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none
Time = 0.29 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x \log \left (x\right ) - 2\right ) \]
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Time = 15.45 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\ln \left (x\,\ln \left (x\right )-2\right ) \]
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