Integrand size = 34, antiderivative size = 12 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\log (x)+x (-2+\log (1+\log (x))) \]
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\[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x (1+\log (x))} \, dx \\ & = \int \left (\frac {1-x+\log (x)-2 x \log (x)}{x (1+\log (x))}+\log (1+\log (x))\right ) \, dx \\ & = \int \frac {1-x+\log (x)-2 x \log (x)}{x (1+\log (x))} \, dx+\int \log (1+\log (x)) \, dx \\ & = \int \left (\frac {1-2 x}{x}+\frac {1}{1+\log (x)}\right ) \, dx+\int \log (1+\log (x)) \, dx \\ & = \int \frac {1-2 x}{x} \, dx+\int \frac {1}{1+\log (x)} \, dx+\int \log (1+\log (x)) \, dx \\ & = \int \left (-2+\frac {1}{x}\right ) \, dx+\int \log (1+\log (x)) \, dx+\text {Subst}\left (\int \frac {e^x}{1+x} \, dx,x,\log (x)\right ) \\ & = -2 x+\frac {\operatorname {ExpIntegralEi}(1+\log (x))}{e}+\log (x)+\int \log (1+\log (x)) \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=-2 x+\log (x)+x \log (1+\log (x)) \]
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Time = 0.82 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17
method | result | size |
default | \(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\) | \(14\) |
norman | \(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\) | \(14\) |
risch | \(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\) | \(14\) |
parallelrisch | \(\ln \left (\ln \left (x \right )+1\right ) x -2 x +\ln \left (x \right )-\frac {1}{2}\) | \(15\) |
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Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log {\left (\log {\left (x \right )} + 1 \right )} - 2 x + \log {\left (x \right )} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx={\left (x - 1\right )} \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) + \log \left (\log \left (x\right ) + 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) \]
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Time = 15.41 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\ln \left (x\right )-2\,x+x\,\ln \left (\ln \left (x\right )+1\right ) \]
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