Integrand size = 106, antiderivative size = 27 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=e^{\frac {4+x+\frac {x^2}{\log (x)}}{-1+\frac {1}{16 x}}}+x \]
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\[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=\int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+\exp \left (\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}\right ) \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+\exp \left (\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}\right ) \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{(-1+16 x)^2 \log ^2(x)} \, dx \\ & = \int \left (1+\frac {16 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-x^2+16 x^3+3 x^2 \log (x)-32 x^3 \log (x)+4 \log ^2(x)+2 x \log ^2(x)-16 x^2 \log ^2(x)\right )}{(1-16 x)^2 \log ^2(x)}\right ) \, dx \\ & = x+16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-x^2+16 x^3+3 x^2 \log (x)-32 x^3 \log (x)+4 \log ^2(x)+2 x \log ^2(x)-16 x^2 \log ^2(x)\right )}{(1-16 x)^2 \log ^2(x)} \, dx \\ & = x+16 \int \left (-\frac {2 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-2-x+8 x^2\right )}{(-1+16 x)^2}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2}{(-1+16 x) \log ^2(x)}-\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2 (-3+32 x)}{(-1+16 x)^2 \log (x)}\right ) \, dx \\ & = x+16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2}{(-1+16 x) \log ^2(x)} \, dx-16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2 (-3+32 x)}{(-1+16 x)^2 \log (x)} \, dx-32 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-2-x+8 x^2\right )}{(-1+16 x)^2} \, dx \\ & = x+16 \int \left (\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 \log ^2(x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{16 \log ^2(x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 (-1+16 x) \log ^2(x)}\right ) \, dx-16 \int \left (\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 \log (x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{8 \log (x)}-\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 (-1+16 x)^2 \log (x)}\right ) \, dx-32 \int \left (\frac {1}{32} \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )-\frac {65 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{32 (-1+16 x)^2}\right ) \, dx \\ & = x+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{\log ^2(x)} \, dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x) \log ^2(x)} \, dx-\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{\log (x)} \, dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x)^2 \log (x)} \, dx-2 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{\log (x)} \, dx+65 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x)^2} \, dx-\int \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \, dx+\int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{\log ^2(x)} \, dx \\ \end{align*}
Time = 4.42 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=e^{-\frac {65}{16}-x-\frac {65}{16 (-1+16 x)}-\frac {16 x^3}{(-1+16 x) \log (x)}}+x \]
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Time = 0.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {16 x \left (x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )\right )}{\left (16 x -1\right ) \ln \left (x \right )}}\) | \(30\) |
parallelrisch | \(x +{\mathrm e}^{\frac {\left (-16 x^{2}-64 x \right ) \ln \left (x \right )-16 x^{3}}{\left (16 x -1\right ) \ln \left (x \right )}}+\frac {3}{32}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\left (-\frac {16 \, {\left (x^{3} + {\left (x^{2} + 4 \, x\right )} \log \left (x\right )\right )}}{{\left (16 \, x - 1\right )} \log \left (x\right )}\right )} \]
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Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\frac {- 16 x^{3} + \left (- 16 x^{2} - 64 x\right ) \log {\left (x \right )}}{\left (16 x - 1\right ) \log {\left (x \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (23) = 46\).
Time = 0.36 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.30 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx={\left (x e^{\left (x + \frac {x^{2}}{\log \left (x\right )} + \frac {x}{16 \, \log \left (x\right )} + \frac {65}{16 \, {\left (16 \, x - 1\right )}} + \frac {1}{256 \, \log \left (x\right )} + \frac {65}{16}\right )} + e^{\left (-\frac {1}{256 \, {\left (16 \, x - 1\right )} \log \left (x\right )}\right )}\right )} e^{\left (-x - \frac {x^{2}}{\log \left (x\right )} - \frac {x}{16 \, \log \left (x\right )} - \frac {65}{16 \, {\left (16 \, x - 1\right )}} - \frac {1}{256 \, \log \left (x\right )} - \frac {65}{16}\right )} \]
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Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\left (-\frac {16 \, {\left (x^{3} + x^{2} \log \left (x\right ) + 4 \, x \log \left (x\right )\right )}}{16 \, x \log \left (x\right ) - \log \left (x\right )}\right )} \]
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Time = 13.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x+{\mathrm {e}}^{\frac {64\,x\,\ln \left (x\right )}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {16\,x^2\,\ln \left (x\right )}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {16\,x^3}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}} \]
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