Integrand size = 69, antiderivative size = 32 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} e^{-1-x+x \log (2)+x^2 \log \left (-\frac {1}{(4-x) x}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=2^{x-2} e^{-x-1} \left (\frac {1}{x^2-4 x}\right )^{x^2} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = 2^{-2+x} e^{-1-x} \left (\frac {1}{-4 x+x^2}\right )^{x^2} \\ \end{align*}
Time = 5.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=2^{-2+x} e^{-1-x} \left (\frac {1}{-4 x+x^2}\right )^{x^2} \]
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Time = 0.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\left (\frac {1}{x^{2}-4 x}\right )^{x^{2}} 2^{x} {\mathrm e}^{-1-x}}{4}\) | \(25\) |
default | \(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{x^{2}-4 x}\right )+x \ln \left (2\right )-x -1}}{4}\) | \(27\) |
norman | \(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{x^{2}-4 x}\right )+x \ln \left (2\right )-x -1}}{4}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{\left (x -4\right ) x}\right )+x \ln \left (2\right )-x -1}}{4}\) | \(27\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (x^{2} \log \left (\frac {1}{x^{2} - 4 \, x}\right ) + x \log \left (2\right ) - x - 1\right )} \]
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Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {e^{x^{2} \log {\left (\frac {1}{x^{2} - 4 x} \right )} - x + x \log {\left (2 \right )} - 1}}{4} \]
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Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x - 4\right ) - x^{2} \log \left (x\right ) + x \log \left (2\right ) - x - 1\right )} \]
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Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x^{2} - 4 \, x\right ) + x \log \left (2\right ) - x - 1\right )} \]
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Time = 13.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {2^x\,{\mathrm {e}}^{-x-1}\,{\left (-\frac {1}{4\,x-x^2}\right )}^{x^2}}{4} \]
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