Integrand size = 130, antiderivative size = 26 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=1+e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \]
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Time = 0.42 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2494, 6838} \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=\exp \left (x-x^2 \log ^2\left (\frac {1}{\log ^2(x (1-\log (4)))+8 \log (x (1-\log (4)))+16}\right )\right ) \]
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Rule 2494
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right ) \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x (1-\log (4)))} \, dx \\ & = \exp \left (x-x^2 \log ^2\left (\frac {1}{16+8 \log (x (1-\log (4)))+\log ^2(x (1-\log (4)))}\right )\right ) \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \]
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Time = 2.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
method | result | size |
parallelrisch | \({\mathrm e}^{-x^{2} \ln \left (\frac {1}{\ln \left (x -2 x \ln \left (2\right )\right )^{2}+8 \ln \left (x -2 x \ln \left (2\right )\right )+16}\right )^{2}+x}\) | \(36\) |
risch | \(\left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{4 i \pi \,\operatorname {csgn}\left (i \left (\ln \left (-\left (2 \ln \left (2\right )-1\right ) x \right )+4\right )^{2}\right ) x^{2}} \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{-4 i \pi \,\operatorname {csgn}\left (i \left (\ln \left (-x \right )+\ln \left (2 \ln \left (2\right )-1\right )+4\right )\right ) x^{2}} {\mathrm e}^{-\frac {x \left (-x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{6}+4 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{5} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )-6 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{4} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{2}+4 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{3} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{3}-x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{4}+16 x \ln \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}-4\right )}{4}}\) | \(280\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-x^{2} \log \left (\frac {1}{\log \left (-2 \, x \log \left (2\right ) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \left (2\right ) + x\right ) + 16}\right )^{2} + x\right )} \]
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Time = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{- x^{2} \log {\left (\frac {1}{\log {\left (- 2 x \log {\left (2 \right )} + x \right )}^{2} + 8 \log {\left (- 2 x \log {\left (2 \right )} + x \right )} + 16} \right )}^{2} + x} \]
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Time = 0.41 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-4 \, x^{2} \log \left (\log \left (x\right ) + \log \left (-2 \, \log \left (2\right ) + 1\right ) + 4\right )^{2} + x\right )} \]
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Time = 1.44 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-x^{2} \log \left (\log \left (-2 \, x \log \left (2\right ) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \left (2\right ) + x\right ) + 16\right )^{2} + x\right )} \]
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Time = 14.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx={\mathrm {e}}^x\,{\mathrm {e}}^{-x^2\,{\ln \left (\frac {1}{{\ln \left (x-2\,x\,\ln \left (2\right )\right )}^2+8\,\ln \left (x-2\,x\,\ln \left (2\right )\right )+16}\right )}^2} \]
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