Integrand size = 109, antiderivative size = 24 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=x \log \left (\frac {4}{3 x+\frac {\log (x)}{2}+\log (4+\log (2 x))}\right ) \]
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\[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=\int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx \\ & = \int \left (\frac {-6-24 x-\log (2 x)-6 x \log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}+\log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )\right ) \, dx \\ & = \int \frac {-6-24 x-\log (2 x)-6 x \log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx+\int \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right ) \, dx \\ & = x \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )+\int \frac {x \left (6+\frac {1+\frac {2}{4+\log (2 x)}}{x}\right )}{6 x+\log (x)+2 \log (4+\log (2 x))} \, dx+\int \left (-\frac {6}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}-\frac {24 x}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}-\frac {\log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}-\frac {6 x \log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}\right ) \, dx \\ & = x \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )-6 \int \frac {1}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx-6 \int \frac {x \log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx-24 \int \frac {x}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx-\int \frac {\log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx+\int \left (\frac {1}{6 x+\log (x)+2 \log (4+\log (2 x))}+\frac {6 x}{6 x+\log (x)+2 \log (4+\log (2 x))}+\frac {2}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))}\right ) \, dx \\ & = x \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )+2 \int \frac {1}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx+6 \int \frac {x}{6 x+\log (x)+2 \log (4+\log (2 x))} \, dx-6 \int \frac {1}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx-6 \int \frac {x \log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx-24 \int \frac {x}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx+\int \frac {1}{6 x+\log (x)+2 \log (4+\log (2 x))} \, dx-\int \frac {\log (2 x)}{(4+\log (2 x)) (6 x+\log (x)+2 \log (4+\log (2 x)))} \, dx \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=x \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right ) \]
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Time = 14.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\ln \left (\frac {8}{2 \ln \left (\ln \left (2 x \right )+4\right )+\ln \left (x \right )+6 x}\right ) x\) | \(23\) |
risch | \(-x \ln \left (\frac {\ln \left (\ln \left (2\right )+\ln \left (x \right )+4\right )}{3}+\frac {\ln \left (x \right )}{6}+x \right )-x \ln \left (3\right )+2 x \ln \left (2\right )\) | \(31\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=x \log \left (\frac {8}{6 \, x + \log \left (x\right ) + 2 \, \log \left (\log \left (2\right ) + \log \left (x\right ) + 4\right )}\right ) \]
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Exception generated. \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=3 \, x \log \left (2\right ) - x \log \left (6 \, x + \log \left (x\right ) + 2 \, \log \left (\log \left (2\right ) + \log \left (x\right ) + 4\right )\right ) \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=3 \, x \log \left (2\right ) - x \log \left (6 \, x + \log \left (x\right ) + 2 \, \log \left (\log \left (2\right ) + \log \left (x\right ) + 4\right )\right ) \]
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Time = 9.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-6-24 x+(-1-6 x) \log (2 x)+(24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))) \log \left (\frac {8}{6 x+\log (x)+2 \log (4+\log (2 x))}\right )}{24 x+4 \log (x)+(6 x+\log (x)) \log (2 x)+(8+2 \log (2 x)) \log (4+\log (2 x))} \, dx=x\,\left (\ln \left (\frac {1}{6\,x+2\,\ln \left (\ln \left (2\,x\right )+4\right )+\ln \left (x\right )}\right )+3\,\ln \left (2\right )\right ) \]
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