Integrand size = 56, antiderivative size = 23 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}+\log ^2(4)\right )}+x \]
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\[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=\int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (2+\frac {e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )} (1+2 x)}{x^2}\right ) \, dx \\ & = x+\frac {1}{2} \int \frac {e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )} (1+2 x)}{x^2} \, dx \\ & = x+\frac {1}{2} \int \left (\frac {e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )}}{x^2}+\frac {2 e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )}}{x}\right ) \, dx \\ & = x+\frac {1}{2} \int \frac {e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )}}{x^2} \, dx+\int \frac {e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}-4 x+\log ^2(4)\right )}}{x} \, dx \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=e^{-\frac {e^{-2 x}}{2 x}+\frac {\log ^2(4)}{2}}+x \]
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Time = 0.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x +{\mathrm e}^{\frac {4 x \ln \left (2\right )^{2}-{\mathrm e}^{-2 x}}{2 x}}\) | \(23\) |
parts | \(x +{\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}\) | \(28\) |
parallelrisch | \({\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}+\frac {\ln \left ({\mathrm e}^{2 x}\right )}{2}\) | \(34\) |
norman | \(\frac {\left ({\mathrm e}^{2 x} x^{2}+x \,{\mathrm e}^{2 x} {\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}\right ) {\mathrm e}^{-2 x}}{x}\) | \(51\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\left (\frac {{\left (4 \, x e^{\left (2 \, x\right )} \log \left (2\right )^{2} - 1\right )} e^{\left (-2 \, x\right )}}{2 \, x}\right )} \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\frac {\left (2 x e^{2 x} \log {\left (2 \right )}^{2} - \frac {1}{2}\right ) e^{- 2 x}}{x}} \]
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Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\left (2 \, \log \left (2\right )^{2} - \frac {e^{\left (-2 \, x\right )}}{2 \, x}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx={\left (x e^{\left (-2 \, x\right )} + e^{\left (\frac {4 \, x \log \left (2\right )^{2} - 4 \, x^{2} - e^{\left (-2 \, x\right )}}{2 \, x}\right )}\right )} e^{\left (2 \, x\right )} \]
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Time = 9.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x+{\mathrm {e}}^{2\,{\ln \left (2\right )}^2-\frac {{\mathrm {e}}^{-2\,x}}{2\,x}} \]
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