Integrand size = 28, antiderivative size = 15 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{32} \left (-5 x+\frac {16}{\log (\log (x))}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6820, 2339, 30} \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2 \log (\log (x))}-\frac {5 x}{32} \]
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Rule 12
Rule 30
Rule 2339
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \frac {1}{32} \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{x \log (x) \log ^2(\log (x))} \, dx \\ & = \frac {1}{32} \int \left (-5-\frac {16}{x \log (x) \log ^2(\log (x))}\right ) \, dx \\ & = -\frac {5 x}{32}-\frac {1}{2} \int \frac {1}{x \log (x) \log ^2(\log (x))} \, dx \\ & = -\frac {5 x}{32}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\log (x)\right ) \\ & = -\frac {5 x}{32}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (\log (x))\right ) \\ & = -\frac {5 x}{32}+\frac {1}{2 \log (\log (x))} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5 x}{32}+\frac {1}{2 \log (\log (x))} \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\) | \(12\) |
risch | \(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\) | \(12\) |
parts | \(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\) | \(12\) |
norman | \(\frac {\frac {1}{2}-\frac {5 x \ln \left (\ln \left (x \right )\right )}{32}}{\ln \left (\ln \left (x \right )\right )}\) | \(15\) |
parallelrisch | \(-\frac {5 x \ln \left (\ln \left (x \right )\right )-16}{32 \ln \left (\ln \left (x \right )\right )}\) | \(16\) |
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Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5 \, x \log \left (\log \left (x\right )\right ) - 16}{32 \, \log \left (\log \left (x\right )\right )} \]
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Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=- \frac {5 x}{32} + \frac {1}{2 \log {\left (\log {\left (x \right )} \right )}} \]
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Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \left (x\right )\right )} \]
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Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \left (x\right )\right )} \]
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Time = 9.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2\,\ln \left (\ln \left (x\right )\right )}-\frac {5\,x}{32} \]
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