Integrand size = 105, antiderivative size = 23 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{-x^2} \log \left (19+x+\left (-e^{2 x}+x\right )^2\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(23)=46\).
Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.43, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2326} \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=\frac {e^{-x^2} \left (x^3-2 e^{2 x} x^2+x^2+e^{4 x} x+19 x\right ) \log \left (x^2-2 e^{2 x} x+x+e^{4 x}+19\right )}{x \left (x^2-2 e^{2 x} x+x+e^{4 x}+19\right )} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{-x^2} \left (19 x+e^{4 x} x+x^2-2 e^{2 x} x^2+x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )}{x \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{-x^2} \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right ) \]
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Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
risch | \({\mathrm e}^{-x^{2}} \ln \left ({\mathrm e}^{4 x}-2 x \,{\mathrm e}^{2 x}+x^{2}+x +19\right )\) | \(26\) |
parallelrisch | \({\mathrm e}^{-x^{2}} \ln \left ({\mathrm e}^{4 x}-2 x \,{\mathrm e}^{2 x}+x^{2}+x +19\right )\) | \(28\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]
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Timed out. \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]
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Time = 0.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]
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Time = 7.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx={\mathrm {e}}^{-x^2}\,\ln \left (x+{\mathrm {e}}^{4\,x}-2\,x\,{\mathrm {e}}^{2\,x}+x^2+19\right ) \]
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