Integrand size = 106, antiderivative size = 24 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=e^{-e^{4 x (x+\log (-x+\log (3)))}-x} x \]
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\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x}-e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x+4 \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{-1+4 x} \left (2 x^2+x (1-\log (9))+x \log (-x+\log (3))-\log (3) \log (-x+\log (3))\right )\right ) \, dx \\ & = 4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{-1+4 x} \left (2 x^2+x (1-\log (9))+x \log (-x+\log (3))-\log (3) \log (-x+\log (3))\right ) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \left (\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (1+2 x-\log (9))-\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3))\right ) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (1+2 x-\log (9)) \, dx-4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \left (2 \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^3 (-x+\log (3))^{-1+4 x}-\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (-1+\log (9))\right ) \, dx-4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = -\left (4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx\right )+8 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^3 (-x+\log (3))^{-1+4 x} \, dx+(4 (1-\log (9))) \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ \end{align*}
\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx \]
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Time = 1.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(x \,{\mathrm e}^{-x -{\mathrm e}^{4 x \left (\ln \left (\ln \left (3\right )-x \right )+x \right )}}\) | \(21\) |
risch | \(x \,{\mathrm e}^{-\left (\ln \left (3\right )-x \right )^{4 x} {\mathrm e}^{4 x^{2}}-x}\) | \(26\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \]
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Timed out. \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\text {Timed out} \]
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Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \]
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\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int { -\frac {{\left (x^{2} + 4 \, {\left (2 \, x^{3} - 2 \, x^{2} \log \left (3\right ) + x^{2} + {\left (x^{2} - x \log \left (3\right )\right )} \log \left (-x + \log \left (3\right )\right )\right )} e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )} - {\left (x - 1\right )} \log \left (3\right ) - x\right )} e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )}}{x - \log \left (3\right )} \,d x } \]
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Time = 14.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x^2}\,{\left (\ln \left (3\right )-x\right )}^{4\,x}}\,{\mathrm {e}}^{-x} \]
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