Integrand size = 156, antiderivative size = 31 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-1+e^{\left (-e^{5 x^2}+x \log \left (\frac {1}{x+4 x^2}\right )\right )^2}-x \]
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\[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=\int \frac {-1-4 x+\exp \left (e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-1+\frac {2 e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+8 x+10 e^{5 x^2} x+40 e^{5 x^2} x^2-\log \left (\frac {1}{x+4 x^2}\right )-4 x \log \left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}\right ) \, dx \\ & = -x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+8 x+10 e^{5 x^2} x+40 e^{5 x^2} x^2-\log \left (\frac {1}{x+4 x^2}\right )-4 x \log \left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx \\ & = -x+2 \int \left (10 e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}+\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}-\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )+10 x^2 \log \left (\frac {1}{x+4 x^2}\right )+40 x^3 \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}\right ) \, dx \\ & = -x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )+10 x^2 \log \left (\frac {1}{x+4 x^2}\right )+40 x^3 \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx \\ & = -x+2 \int \left (\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{1+4 x}+e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right )\right ) \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\left (1+4 x+10 x^2+40 x^3\right ) \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx \\ & = -x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{1+4 x} \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx-2 \int \left (\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x}+e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+10 x^2\right ) \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx \\ & = -x-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x} \, dx-2 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+10 x^2\right ) \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int \left (\frac {1}{4} e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )+\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{4 (-1-4 x)}-2 e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx \\ & = -x+\frac {1}{2} \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+\frac {1}{2} \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{-1-4 x} \, dx-2 \int \left (-2 e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}+\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx-2 \int \left (e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )+10 e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x^2 \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx-4 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx \\ & = -x+\frac {1}{2} \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+\frac {1}{2} \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{-1-4 x} \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x} \, dx-2 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx+4 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx-4 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx-20 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x^2 \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x+e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x+4 x^2}\right )^{-2 e^{5 x^2} x} \]
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Time = 2.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77
method | result | size |
parallelrisch | \(\frac {1}{8}-x +{\mathrm e}^{x^{2} \ln \left (\frac {1}{\left (1+4 x \right ) x}\right )^{2}-2 x \,{\mathrm e}^{5 x^{2}} \ln \left (\frac {1}{\left (1+4 x \right ) x}\right )+{\mathrm e}^{10 x^{2}}}\) | \(55\) |
risch | \(\text {Expression too large to display}\) | \(863\) |
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Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x + e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} \]
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Time = 0.70 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=- x + e^{x^{2} \log {\left (\frac {1}{4 x^{2} + x} \right )}^{2} - 2 x e^{5 x^{2}} \log {\left (\frac {1}{4 x^{2} + x} \right )} + e^{10 x^{2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (29) = 58\).
Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x + e^{\left (x^{2} \log \left (4 \, x + 1\right )^{2} + 2 \, x^{2} \log \left (4 \, x + 1\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (4 \, x + 1\right ) + 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (x\right ) + e^{\left (10 \, x^{2}\right )}\right )} \]
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\[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{2} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} + 10 \, {\left (4 \, x^{2} + x\right )} e^{\left (10 \, x^{2}\right )} + {\left (8 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} - {\left (8 \, x^{2} + {\left (40 \, x^{3} + 10 \, x^{2} + 4 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )\right )} e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} - 4 \, x - 1}{4 \, x + 1} \,d x } \]
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Time = 12.92 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{10\,x^2}}\,{\mathrm {e}}^{x^2\,{\ln \left (\frac {1}{4\,x^2+x}\right )}^2}}{{\left (\frac {1}{4\,x^2+x}\right )}^{2\,x\,{\mathrm {e}}^{5\,x^2}}}-x \]
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