Integrand size = 52, antiderivative size = 35 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=\log \left (\frac {4}{\frac {x-x \left (-x^2+\left (x+x^4\right )^2\right )}{\log (3)}+\frac {5}{\log (x)}}\right ) \]
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\[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=\int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \log (3)-\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{x \log (x) \left (5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)\right )} \, dx \\ & = \int \left (\frac {1-12 x^5-9 x^8}{x \left (-1+2 x^5+x^8\right )}+\frac {1}{x \log (x)}+\frac {1-2 x^5-x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)}-\frac {5 \left (-1+12 x^5+9 x^8\right ) \log (3)}{x \left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )}\right ) \, dx \\ & = -\left ((5 \log (3)) \int \frac {-1+12 x^5+9 x^8}{x \left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx\right )+\int \frac {1-12 x^5-9 x^8}{x \left (-1+2 x^5+x^8\right )} \, dx+\int \frac {1}{x \log (x)} \, dx+\int \frac {1-2 x^5-x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx \\ & = -\log \left (-x \left (1-2 x^5-x^8\right )\right )-(5 \log (3)) \int \left (\frac {1}{x \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )}+\frac {2 x^4 \left (5+4 x^3\right )}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )}\right ) \, dx+\int \left (-\frac {1}{5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)}-\frac {2 x^5}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)}-\frac {x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -\log \left (-x \left (1-2 x^5-x^8\right )\right )+\log (\log (x))-2 \int \frac {x^5}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx-(5 \log (3)) \int \frac {1}{x \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-(10 \log (3)) \int \frac {x^4 \left (5+4 x^3\right )}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-\int \frac {1}{5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)} \, dx-\int \frac {x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx \\ & = -\log \left (-x \left (1-2 x^5-x^8\right )\right )+\log (\log (x))-2 \int \frac {x^5}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx-(5 \log (3)) \int \frac {1}{x \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-(10 \log (3)) \int \left (\frac {5 x^4}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )}+\frac {4 x^7}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )}\right ) \, dx-\int \frac {1}{5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)} \, dx-\int \frac {x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx \\ & = -\log \left (-x \left (1-2 x^5-x^8\right )\right )+\log (\log (x))-2 \int \frac {x^5}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx-(5 \log (3)) \int \frac {1}{x \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-(40 \log (3)) \int \frac {x^7}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-(50 \log (3)) \int \frac {x^4}{\left (-1+2 x^5+x^8\right ) \left (-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)\right )} \, dx-\int \frac {1}{5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)} \, dx-\int \frac {x^8}{-5 \log (3)-x \log (x)+2 x^6 \log (x)+x^9 \log (x)} \, dx \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.83 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=-\log (x)-\log \left (1-2 x^5-x^8\right )+\log \left (x \left (1-2 x^5-x^8\right )\right )+\log (\log (x))-\log \left (5 \log (3)+x \log (x)-2 x^6 \log (x)-x^9 \log (x)\right ) \]
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Time = 1.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.69
method | result | size |
default | \(-\ln \left (x^{9}+2 x^{6}-x -\frac {5 \ln \left (3\right )}{\ln \left (x \right )}\right )\) | \(24\) |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )-\ln \left (x^{9} \ln \left (x \right )+2 x^{6} \ln \left (x \right )-x \ln \left (x \right )-5 \ln \left (3\right )\right )\) | \(31\) |
risch | \(-\ln \left (x^{9}+2 x^{6}-x \right )+\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )-\frac {5 \ln \left (3\right )}{x \left (x^{8}+2 x^{5}-1\right )}\right )\) | \(45\) |
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Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.63 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=-\log \left (x^{9} + 2 \, x^{6} - x\right ) - \log \left (\frac {{\left (x^{9} + 2 \, x^{6} - x\right )} \log \left (x\right ) - 5 \, \log \left (3\right )}{x^{9} + 2 \, x^{6} - x}\right ) + \log \left (\log \left (x\right )\right ) \]
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Exception generated. \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \]
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Time = 0.36 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=-\log \left (x^{8} + 2 \, x^{5} - 1\right ) - \log \left (x\right ) - \log \left (\frac {{\left (x^{9} + 2 \, x^{6} - x\right )} \log \left (x\right ) - 5 \, \log \left (3\right )}{x^{9} + 2 \, x^{6} - x}\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=-\log \left (-x^{9} \log \left (x\right ) - 2 \, x^{6} \log \left (x\right ) + x \log \left (x\right ) + 5 \, \log \left (3\right )\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 38.39 (sec) , antiderivative size = 186, normalized size of antiderivative = 5.31 \[ \int \frac {-5 \log (3)+\left (x-12 x^6-9 x^9\right ) \log ^2(x)}{-5 x \log (3) \log (x)+\left (-x^2+2 x^7+x^{10}\right ) \log ^2(x)} \, dx=\ln \left (8\,x^6\,\ln \left (x\right )+4\,x^9\,\ln \left (x\right )-8\,x^{11}\,\ln \left (x\right )-8\,x^{14}\,\ln \left (x\right )-2\,x^{17}\,\ln \left (x\right )+10\,\ln \left (3\right )\,\ln \left (x\right )-2\,x\,\ln \left (x\right )-120\,x^5\,\ln \left (3\right )\,\ln \left (x\right )-90\,x^8\,\ln \left (3\right )\,\ln \left (x\right )\right )-\ln \left (10\,\ln \left (3\right )-8\,x^6\,\ln \left (x\right )-4\,x^9\,\ln \left (x\right )+8\,x^{11}\,\ln \left (x\right )+8\,x^{14}\,\ln \left (x\right )+2\,x^{17}\,\ln \left (x\right )-20\,x^5\,\ln \left (3\right )-10\,x^8\,\ln \left (3\right )+2\,x\,\ln \left (x\right )\right )-\ln \left (x^{17}+4\,x^{14}+4\,x^{11}-2\,x^9+45\,\ln \left (3\right )\,x^8-4\,x^6+60\,\ln \left (3\right )\,x^5+x-5\,\ln \left (3\right )\right )+\ln \left (x^8+2\,x^5-1\right ) \]
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