Integrand size = 91, antiderivative size = 27 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x}{(1-x) \left (x+81 x^2-\log (x)-x \log (x)\right )} \]
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\[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=\int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {1-81 x^2+162 x^3-\left (1+x^2\right ) \log (x)}{(1-x)^2 (x (1+81 x)-(1+x) \log (x))^2} \, dx \\ & = \int \left (\frac {-1-x+161 x^2+81 x^3}{\left (-1+x^2\right ) \left (x+81 x^2-\log (x)-x \log (x)\right )^2}+\frac {1+x^2}{(-1+x)^2 (1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )}\right ) \, dx \\ & = \int \frac {-1-x+161 x^2+81 x^3}{\left (-1+x^2\right ) \left (x+81 x^2-\log (x)-x \log (x)\right )^2} \, dx+\int \frac {1+x^2}{(-1+x)^2 (1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )} \, dx \\ & = \int \left (\frac {161}{\left (x+81 x^2-\log (x)-x \log (x)\right )^2}+\frac {120}{(-1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )^2}+\frac {81 x}{\left (x+81 x^2-\log (x)-x \log (x)\right )^2}-\frac {40}{(1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )^2}\right ) \, dx+\int \left (\frac {1}{(-1+x)^2 \left (x+81 x^2-\log (x)-x \log (x)\right )}+\frac {1}{2 (-1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )}+\frac {1}{2 (1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {1}{(-1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )} \, dx+\frac {1}{2} \int \frac {1}{(1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )} \, dx-40 \int \frac {1}{(1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )^2} \, dx+81 \int \frac {x}{\left (x+81 x^2-\log (x)-x \log (x)\right )^2} \, dx+120 \int \frac {1}{(-1+x) \left (x+81 x^2-\log (x)-x \log (x)\right )^2} \, dx+161 \int \frac {1}{\left (x+81 x^2-\log (x)-x \log (x)\right )^2} \, dx+\int \frac {1}{(-1+x)^2 \left (x+81 x^2-\log (x)-x \log (x)\right )} \, dx \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x}{(-1+x) \left (-x-81 x^2+\log (x)+x \log (x)\right )} \]
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Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {x}{\left (-1+x \right ) \left (x \ln \left (x \right )-81 x^{2}+\ln \left (x \right )-x \right )}\) | \(25\) |
risch | \(-\frac {x}{\left (-1+x \right ) \left (81 x^{2}-\ln \left (x \right )-x \ln \left (x \right )+x \right )}\) | \(27\) |
norman | \(-\frac {x}{81 x^{3}-x^{2} \ln \left (x \right )-80 x^{2}-x +\ln \left (x \right )}\) | \(29\) |
parallelrisch | \(-\frac {x}{81 x^{3}-x^{2} \ln \left (x \right )-80 x^{2}-x +\ln \left (x \right )}\) | \(29\) |
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Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=-\frac {x}{81 \, x^{3} - 80 \, x^{2} - {\left (x^{2} - 1\right )} \log \left (x\right ) - x} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x}{- 81 x^{3} + 80 x^{2} + x + \left (x^{2} - 1\right ) \log {\left (x \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=-\frac {x}{81 \, x^{3} - 80 \, x^{2} - {\left (x^{2} - 1\right )} \log \left (x\right ) - x} \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=-\frac {x}{81 \, x^{3} - x^{2} \log \left (x\right ) - 80 \, x^{2} - x + \log \left (x\right )} \]
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Timed out. \[ \int \frac {1-81 x^2+162 x^3+\left (-1-x^2\right ) \log (x)}{x^2+160 x^3+6238 x^4-12960 x^5+6561 x^6+\left (-2 x-160 x^2+164 x^3+160 x^4-162 x^5\right ) \log (x)+\left (1-2 x^2+x^4\right ) \log ^2(x)} \, dx=\int -\frac {\ln \left (x\right )\,\left (x^2+1\right )+81\,x^2-162\,x^3-1}{{\ln \left (x\right )}^2\,\left (x^4-2\,x^2+1\right )-\ln \left (x\right )\,\left (162\,x^5-160\,x^4-164\,x^3+160\,x^2+2\,x\right )+x^2+160\,x^3+6238\,x^4-12960\,x^5+6561\,x^6} \,d x \]
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