Integrand size = 72, antiderivative size = 24 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=x-\log ^2\left (x \left (1-e^2-x+16 x^5\right )\right ) \]
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\[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=\int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{\left (-1+e^2\right ) x+x^2-16 x^6} \, dx \\ & = \int \frac {\left (-1+e^2\right ) x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{\left (-1+e^2\right ) x+x^2-16 x^6} \, dx \\ & = \int \frac {\left (-1+e^2\right ) x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{x \left (-1+e^2+x-16 x^5\right )} \, dx \\ & = \int \left (1-\frac {2 \left (1-e^2-2 x+96 x^5\right ) \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{x \left (1-e^2-x+16 x^5\right )}\right ) \, dx \\ & = x-2 \int \frac {\left (1-e^2-2 x+96 x^5\right ) \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{x \left (1-e^2-x+16 x^5\right )} \, dx \\ & = x-2 \int \left (\frac {\log \left (x \left (1-e^2-x+16 x^5\right )\right )}{x}+\frac {\left (1-80 x^4\right ) \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5}\right ) \, dx \\ & = x-2 \int \frac {\log \left (x \left (1-e^2-x+16 x^5\right )\right )}{x} \, dx-2 \int \frac {\left (1-80 x^4\right ) \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx \\ & = x-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \frac {\left (1-e^2-x+16 x^5+x \left (-1+80 x^4\right )\right ) \log (x)}{x \left (1-e^2-x+16 x^5\right )} \, dx-2 \int \left (\frac {\log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5}-\frac {80 x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5}\right ) \, dx \\ & = x-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \left (\frac {\log (x)}{x}+\frac {\left (1-80 x^4\right ) \log (x)}{-1+e^2+x-16 x^5}\right ) \, dx-2 \int \frac {\log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx \\ & = x-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \frac {\log (x)}{x} \, dx+2 \int \frac {\left (1-80 x^4\right ) \log (x)}{-1+e^2+x-16 x^5} \, dx+2 \int \frac {\left (1-e^2-2 x+96 x^5\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx}{x \left (1-e^2-x+16 x^5\right )} \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx-\left (2 \log \left (x \left (1-e^2-x+16 x^5\right )\right )\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx \\ & = x+\log ^2(x)-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \left (\frac {\log (x)}{-1+e^2+x-16 x^5}-\frac {80 x^4 \log (x)}{-1+e^2+x-16 x^5}\right ) \, dx+2 \int \left (\frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{x}+\frac {\left (-1+80 x^4\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx}{1-e^2-x+16 x^5}\right ) \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx-\left (2 \log \left (x \left (1-e^2-x+16 x^5\right )\right )\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx \\ & = x+\log ^2(x)-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \frac {\log (x)}{-1+e^2+x-16 x^5} \, dx+2 \int \frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{x} \, dx+2 \int \frac {\left (-1+80 x^4\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx}{1-e^2-x+16 x^5} \, dx-160 \int \frac {x^4 \log (x)}{-1+e^2+x-16 x^5} \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx-\left (2 \log \left (x \left (1-e^2-x+16 x^5\right )\right )\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx \\ & = x+\log ^2(x)-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \frac {\log (x)}{-1+e^2+x-16 x^5} \, dx+2 \int \frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{x} \, dx+2 \int \left (\frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{-1+e^2+x-16 x^5}-\frac {80 x^4 \int \frac {1}{-1+e^2+x-16 x^5} \, dx}{-1+e^2+x-16 x^5}\right ) \, dx-160 \int \frac {x^4 \log (x)}{-1+e^2+x-16 x^5} \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx-\left (2 \log \left (x \left (1-e^2-x+16 x^5\right )\right )\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx \\ & = x+\log ^2(x)-2 \log (x) \log \left (x \left (1-e^2-x+16 x^5\right )\right )+2 \int \frac {\log (x)}{-1+e^2+x-16 x^5} \, dx+2 \int \frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{x} \, dx+2 \int \frac {\int \frac {1}{-1+e^2+x-16 x^5} \, dx}{-1+e^2+x-16 x^5} \, dx-160 \int \frac {x^4 \log (x)}{-1+e^2+x-16 x^5} \, dx+160 \int \frac {x^4 \log \left (x \left (1-e^2-x+16 x^5\right )\right )}{-1+e^2+x-16 x^5} \, dx-160 \int \frac {x^4 \int \frac {1}{-1+e^2+x-16 x^5} \, dx}{-1+e^2+x-16 x^5} \, dx-\left (2 \log \left (x \left (1-e^2-x+16 x^5\right )\right )\right ) \int \frac {1}{-1+e^2+x-16 x^5} \, dx \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=x-\log ^2\left (x \left (1-e^2-x+16 x^5\right )\right ) \]
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Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
risch | \(x -\ln \left (-{\mathrm e}^{2} x +16 x^{6}-x^{2}+x \right )^{2}\) | \(25\) |
default | \(x -\ln \left (-{\mathrm e}^{2} x +16 x^{6}-x^{2}+x \right )^{2}\) | \(27\) |
norman | \(x -\ln \left (-{\mathrm e}^{2} x +16 x^{6}-x^{2}+x \right )^{2}\) | \(27\) |
parts | \(x -\ln \left (-{\mathrm e}^{2} x +16 x^{6}-x^{2}+x \right )^{2}\) | \(27\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=-\log \left (16 \, x^{6} - x^{2} - x e^{2} + x\right )^{2} + x \]
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Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=x - \log {\left (16 x^{6} - x^{2} - x e^{2} + x \right )}^{2} \]
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Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=-\log \left (16 \, x^{5} - x - e^{2} + 1\right )^{2} - 2 \, \log \left (16 \, x^{5} - x - e^{2} + 1\right ) \log \left (x\right ) - \log \left (x\right )^{2} + x \]
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\[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=\int { \frac {16 \, x^{6} - x^{2} - x e^{2} - 2 \, {\left (96 \, x^{5} - 2 \, x - e^{2} + 1\right )} \log \left (16 \, x^{6} - x^{2} - x e^{2} + x\right ) + x}{16 \, x^{6} - x^{2} - x e^{2} + x} \,d x } \]
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Time = 13.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-x+e^2 x+x^2-16 x^6+\left (2-2 e^2-4 x+192 x^5\right ) \log \left (x-e^2 x-x^2+16 x^6\right )}{-x+e^2 x+x^2-16 x^6} \, dx=x-{\ln \left (x-x\,{\mathrm {e}}^2-x^2+16\,x^6\right )}^2 \]
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