Integrand size = 85, antiderivative size = 25 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-7+\frac {\log (2 x)}{x \log \left (\frac {4+x}{2 (1+x)}\right )} \]
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\[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{x^2 \left (4+5 x+x^2\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = \int \frac {\log (2 x) \left (\frac {3 x}{4+5 x+x^2}-\log \left (\frac {4+x}{2+2 x}\right )\right )+\log \left (\frac {4+x}{2+2 x}\right )}{x^2 \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = \int \left (\frac {3 \log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx \\ & = 3 \int \frac {\log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = 3 \int \left (\frac {\log (2 x)}{4 x \log ^2\left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{3 (1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {\log (2 x)}{12 (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}\right ) \, dx+\int \left (\frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {\log (2 x)}{(4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\frac {3}{4} \int \frac {\log (2 x)}{x \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{(1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log (2 x)}{x \log \left (\frac {4+x}{2+2 x}\right )} \]
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Time = 11.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {\ln \left (2 x \right )}{\ln \left (\frac {4+x}{2+2 x}\right ) x}\) | \(22\) |
risch | \(\frac {2 i \ln \left (2 x \right )}{x \left (\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )-\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{3}-2 i \ln \left (2\right )+2 i \ln \left (4+x \right )-2 i \ln \left (1+x \right )\right )}\) | \(129\) |
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log \left (2 \, x\right )}{x \log \left (\frac {x + 4}{2 \, {\left (x + 1\right )}}\right )} \]
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Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log {\left (2 x \right )}}{x \log {\left (\frac {x + 4}{2 x + 2} \right )}} \]
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Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \]
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Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \]
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Time = 15.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\ln \left (2\,x\right )}{x\,\ln \left (\frac {x+4}{2\,x+2}\right )} \]
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