Integrand size = 90, antiderivative size = 27 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=5-x^2 \left (4+\frac {4}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]
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\[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=\int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {68 x^2-8 x^3+x (-160-4 \log (7))+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx \\ & = \int \frac {4 x \left (17 x-2 x^2-40 \left (1+\frac {\log (7)}{40}\right )-2 (9-2 x) \log (7) \log \left (\frac {4}{x}\right )-2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )\right )}{\left (4-x+\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ & = 4 \int \frac {x \left (17 x-2 x^2-40 \left (1+\frac {\log (7)}{40}\right )-2 (9-2 x) \log (7) \log \left (\frac {4}{x}\right )-2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )\right )}{\left (4-x+\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ & = 4 \int \left (-2 x-\frac {x (x+\log (7))}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}+\frac {2 x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )}\right ) \, dx \\ & = -4 x^2-4 \int \frac {x (x+\log (7))}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx \\ & = -4 x^2-4 \int \left (\frac {x^2}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}+\frac {x \log (7)}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}\right ) \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx \\ & = -4 x^2-4 \int \frac {x^2}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx-(4 \log (7)) \int \frac {x}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \left (x^2+\frac {x^2}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]
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Time = 1.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-4 x^{2}-\frac {4 x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(28\) |
norman | \(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(42\) |
parallelrisch | \(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(42\) |
derivativedivides | \(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\) | \(48\) |
default | \(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\) | \(48\) |
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Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (\frac {4}{x}\right ) - x^{3} + 5 \, x^{2}\right )}}{\log \left (7\right ) \log \left (\frac {4}{x}\right ) - x + 4} \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=- 4 x^{2} - \frac {4 x^{2}}{- x + \log {\left (7 \right )} \log {\left (\frac {4}{x} \right )} + 4} \]
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Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (x\right ) - {\left (2 \, \log \left (7\right ) \log \left (2\right ) + 5\right )} x^{2} + x^{3}\right )}}{2 \, \log \left (7\right ) \log \left (2\right ) - \log \left (7\right ) \log \left (x\right ) - x + 4} \]
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Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \, x^{2} - \frac {4}{\frac {\log \left (7\right ) \log \left (\frac {4}{x}\right )}{x^{2}} - \frac {1}{x} + \frac {4}{x^{2}}} \]
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Time = 11.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.15 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=8\,x-\frac {\frac {8\,x^2\,\ln \left (\frac {4}{x}\right )}{x+\ln \left (7\right )}+\frac {4\,x\,\left (8\,x+x\,\ln \left (7\right )-x^2\right )}{\ln \left (7\right )\,\left (x+\ln \left (7\right )\right )}}{\ln \left (\frac {4}{x}\right )-\frac {x-4}{\ln \left (7\right )}}+\frac {8\,{\ln \left (7\right )}^2}{x+\ln \left (7\right )}-4\,x^2 \]
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