Integrand size = 120, antiderivative size = 27 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^4-2 \left (x+\frac {\log (4-x (25+x))}{x \log \left (\frac {1}{x}\right )}\right ) \]
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\[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{x^2 \left (-4+25 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \\ & = \int \left (\frac {2 \left (-25-2 x+4 x \log \left (\frac {1}{x}\right )-25 x^2 \log \left (\frac {1}{x}\right )-x^3 \log \left (\frac {1}{x}\right )-8 x^4 \log \left (\frac {1}{x}\right )+50 x^5 \log \left (\frac {1}{x}\right )+2 x^6 \log \left (\frac {1}{x}\right )\right )}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )}+\frac {2 \left (-1+\log \left (\frac {1}{x}\right )\right ) \log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )}\right ) \, dx \\ & = 2 \int \frac {-25-2 x+4 x \log \left (\frac {1}{x}\right )-25 x^2 \log \left (\frac {1}{x}\right )-x^3 \log \left (\frac {1}{x}\right )-8 x^4 \log \left (\frac {1}{x}\right )+50 x^5 \log \left (\frac {1}{x}\right )+2 x^6 \log \left (\frac {1}{x}\right )}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\left (-1+\log \left (\frac {1}{x}\right )\right ) \log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx \\ & = 2 \int \frac {25+2 x-x \left (4-25 x-x^2-8 x^3+50 x^4+2 x^5\right ) \log \left (\frac {1}{x}\right )}{x \left (4-25 x-x^2\right ) \log \left (\frac {1}{x}\right )} \, dx+2 \int \left (-\frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )}+\frac {\log \left (4-25 x-x^2\right )}{x^2 \log \left (\frac {1}{x}\right )}\right ) \, dx \\ & = 2 \int \left (-1+2 x^3+\frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )}\right ) \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log \left (\frac {1}{x}\right )} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \frac {(-25-2 x) \text {li}\left (\frac {1}{x}\right )}{4-25 x-x^2} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \left (\frac {25 \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2}+\frac {2 x \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2}\right ) \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+4 \int \frac {x \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2} \, dx+50 \int \frac {\text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+4 \int \left (\frac {\left (1-\frac {25}{\sqrt {641}}\right ) \text {li}\left (\frac {1}{x}\right )}{25-\sqrt {641}+2 x}+\frac {\left (1+\frac {25}{\sqrt {641}}\right ) \text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x}\right ) \, dx+50 \int \left (-\frac {2 \text {li}\left (\frac {1}{x}\right )}{\sqrt {641} \left (-25+\sqrt {641}-2 x\right )}-\frac {2 \text {li}\left (\frac {1}{x}\right )}{\sqrt {641} \left (25+\sqrt {641}+2 x\right )}\right ) \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx-\frac {100 \int \frac {\text {li}\left (\frac {1}{x}\right )}{-25+\sqrt {641}-2 x} \, dx}{\sqrt {641}}-\frac {100 \int \frac {\text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x} \, dx}{\sqrt {641}}+\frac {1}{641} \left (4 \left (641-25 \sqrt {641}\right )\right ) \int \frac {\text {li}\left (\frac {1}{x}\right )}{25-\sqrt {641}+2 x} \, dx+\frac {1}{641} \left (4 \left (641+25 \sqrt {641}\right )\right ) \int \frac {\text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x} \, dx \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=2 \left (-x+\frac {x^4}{2}-\frac {\log \left (4-25 x-x^2\right )}{x \log \left (\frac {1}{x}\right )}\right ) \]
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Time = 23.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {2 \ln \left (-x^{2}-25 x +4\right )}{x \ln \left (x \right )}+x^{4}-2 x\) | \(28\) |
parallelrisch | \(\frac {50 x^{5} \ln \left (\frac {1}{x}\right )-100 x^{2} \ln \left (\frac {1}{x}\right )+1234 x \ln \left (\frac {1}{x}\right )-100 \ln \left (-x^{2}-25 x +4\right )}{50 x \ln \left (\frac {1}{x}\right )}\) | \(51\) |
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Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {{\left (x^{5} - 2 \, x^{2}\right )} \log \left (\frac {1}{x}\right ) - 2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (\frac {1}{x}\right )} \]
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Exception generated. \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {{\left (x^{5} - 2 \, x^{2}\right )} \log \left (x\right ) + 2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (x\right )} \]
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Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^{4} - 2 \, x + \frac {2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (x\right )} \]
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Time = 11.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^4-2\,x-\frac {2\,\ln \left (-x^2-25\,x+4\right )}{x\,\ln \left (\frac {1}{x}\right )} \]
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