Integrand size = 286, antiderivative size = 32 \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=-5+e^{2 (4+x) \left (-\frac {2}{x}+x+\frac {e^x}{\frac {x}{4}+\log (4)+\log (x)}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
\[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=\int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx \]
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Time = 117.57 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75
method | result | size |
risch | \({\mathrm e}^{\frac {2 \left (4+x \right ) \left (4 x^{2} \ln \left (x \right )+8 x^{2} \ln \left (2\right )+x^{3}+4 \,{\mathrm e}^{x} x -8 \ln \left (x \right )-16 \ln \left (2\right )-2 x \right )}{x \left (4 \ln \left (x \right )+8 \ln \left (2\right )+x \right )}}\) | \(56\) |
parallelrisch | \({\mathrm e}^{\frac {\left (8 x^{3}+32 x^{2}-16 x -64\right ) \ln \left (x \right )+\left (8 x^{2}+32 x \right ) {\mathrm e}^{x}+2 \left (8 x^{3}+32 x^{2}-16 x -64\right ) \ln \left (2\right )+2 x^{4}+8 x^{3}-4 x^{2}-16 x}{x \left (4 \ln \left (x \right )+8 \ln \left (2\right )+x \right )}}\) | \(86\) |
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Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (31) = 62\).
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.53 \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=e^{\left (\frac {2 \, {\left (x^{4} + 4 \, x^{3} - 2 \, x^{2} + 4 \, {\left (x^{2} + 4 \, x\right )} e^{x} + 8 \, {\left (x^{3} + 4 \, x^{2} - 2 \, x - 8\right )} \log \left (2\right ) + 4 \, {\left (x^{3} + 4 \, x^{2} - 2 \, x - 8\right )} \log \left (x\right ) - 8 \, x\right )}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (29) = 58\).
Time = 1.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.66 \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=e^{\frac {2 x^{4} + 8 x^{3} - 4 x^{2} - 16 x + \left (8 x^{2} + 32 x\right ) e^{x} + \left (8 x^{3} + 32 x^{2} - 16 x - 64\right ) \log {\left (x \right )} + \left (16 x^{3} + 64 x^{2} - 32 x - 128\right ) \log {\left (2 \right )}}{x^{2} + 4 x \log {\left (x \right )} + 8 x \log {\left (2 \right )}}} \]
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Timed out. \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (31) = 62\).
Time = 1.07 (sec) , antiderivative size = 302, normalized size of antiderivative = 9.44 \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=e^{\left (\frac {2 \, x^{4}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {16 \, x^{3} \log \left (2\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {8 \, x^{3} \log \left (x\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {8 \, x^{3}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {8 \, x^{2} e^{x}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {64 \, x^{2} \log \left (2\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {32 \, x^{2} \log \left (x\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {4 \, x^{2}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} + \frac {32 \, x e^{x}}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {32 \, x \log \left (2\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {16 \, x \log \left (x\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {16 \, x}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {128 \, \log \left (2\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )} - \frac {64 \, \log \left (x\right )}{x^{2} + 8 \, x \log \left (2\right ) + 4 \, x \log \left (x\right )}\right )} \]
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Time = 15.05 (sec) , antiderivative size = 223, normalized size of antiderivative = 6.97 \[ \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx=2^{\frac {64\,\left (x^2-2\right )}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}+\frac {16\,\left (x^2-2\right )}{x+8\,\ln \left (2\right )+4\,\ln \left (x\right )}}\,x^{\frac {8\,\left (x^2+4\,x\right )}{x+8\,\ln \left (2\right )+4\,\ln \left (x\right )}-\frac {16\,\left (x+4\right )}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{\frac {8\,x^2\,{\mathrm {e}}^x}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{-\frac {4\,x^2}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{\frac {2\,x^4}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{\frac {8\,x^3}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{\frac {32\,x\,{\mathrm {e}}^x}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}}\,{\mathrm {e}}^{-\frac {16\,x}{8\,x\,\ln \left (2\right )+4\,x\,\ln \left (x\right )+x^2}} \]
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