Integrand size = 63, antiderivative size = 25 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{4 \left (-1+\frac {2}{x}+\frac {e^{2 x}}{\frac {1}{16}+\log (2)}\right )} \]
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Time = 2.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 12, 6873, 6838} \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=2^{\frac {64 (2-x)}{x (1+16 \log (2))}} e^{\frac {4 \left (16 e^{2 x} x-x+2\right )}{x (1+16 \log (2))}} \]
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Rule 6
Rule 12
Rule 6838
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}\right ) \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2 (1+16 \log (2))} \, dx \\ & = \frac {\int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}\right ) \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = \frac {\int \frac {8 \exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x (1+16 \log (2))}\right ) \left (-1+16 e^{2 x} x^2-16 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = \frac {8 \int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x (1+16 \log (2))}\right ) \left (-1+16 e^{2 x} x^2-16 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = 2^{\frac {64 (2-x)}{x (1+16 \log (2))}} e^{\frac {4 \left (2-x+16 e^{2 x} x\right )}{x (1+16 \log (2))}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).
Time = 1.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=2^{3+\frac {128-x (67+48 \log (2))}{x+16 x \log (2)}} e^{\frac {8+\left (-4+64 e^{2 x}\right ) x}{x+16 x \log (2)}} \]
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Time = 0.55 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28
method | result | size |
norman | \({\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{16 x \ln \left (2\right )+x}}\) | \(32\) |
risch | \({\mathrm e}^{-\frac {4 \left (-16 x \,{\mathrm e}^{2 x}+16 x \ln \left (2\right )-32 \ln \left (2\right )+x -2\right )}{x \left (16 \ln \left (2\right )+1\right )}}\) | \(34\) |
parallelrisch | \(\frac {16 \,{\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{x \left (16 \ln \left (2\right )+1\right )}} \ln \left (2\right )+{\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{x \left (16 \ln \left (2\right )+1\right )}}}{16 \ln \left (2\right )+1}\) | \(81\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\left (\frac {4 \, {\left (16 \, x e^{\left (2 \, x\right )} - 16 \, {\left (x - 2\right )} \log \left (2\right ) - x + 2\right )}}{16 \, x \log \left (2\right ) + x}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\frac {64 x e^{2 x} - 4 x + \left (128 - 64 x\right ) \log {\left (2 \right )} + 8}{x + 16 x \log {\left (2 \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (22) = 44\).
Time = 0.38 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.76 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=\frac {e^{\left (\frac {64 \, e^{\left (2 \, x\right )}}{16 \, \log \left (2\right ) + 1} - \frac {4}{16 \, \log \left (2\right ) + 1} + \frac {128 \, \log \left (2\right )}{x {\left (16 \, \log \left (2\right ) + 1\right )}} + \frac {8}{x {\left (16 \, \log \left (2\right ) + 1\right )}}\right )}}{2^{\frac {64}{16 \, \log \left (2\right ) + 1}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\left (\frac {64 \, x e^{\left (2 \, x\right )}}{16 \, x \log \left (2\right ) + x} - \frac {64 \, x \log \left (2\right )}{16 \, x \log \left (2\right ) + x} - \frac {4 \, x}{16 \, x \log \left (2\right ) + x} + \frac {128 \, \log \left (2\right )}{16 \, x \log \left (2\right ) + x} + \frac {8}{16 \, x \log \left (2\right ) + x}\right )} \]
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Time = 11.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx={\left (\frac {1}{18446744073709551616}\right )}^{\frac {x-2}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {8}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {4\,x}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {64\,x\,{\mathrm {e}}^{2\,x}}{x+16\,x\,\ln \left (2\right )}} \]
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