Integrand size = 72, antiderivative size = 18 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-25+\frac {1}{64} e^{-\frac {10}{-3+x}}-x \]
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Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 12, 6820, 14, 2240} \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {1}{64} e^{\frac {10}{3-x}}-x \]
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Rule 12
Rule 14
Rule 27
Rule 2240
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}\right ) \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{40 (-3+x)^2} \, dx \\ & = \frac {1}{40} \int \frac {\exp \left (-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}\right ) \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{(-3+x)^2} \, dx \\ & = \frac {1}{40} \int \frac {5 \left (5 e^{-\frac {10}{-3+x}}-32 (-3+x)^2\right )}{4 (3-x)^2} \, dx \\ & = \frac {1}{32} \int \frac {5 e^{-\frac {10}{-3+x}}-32 (-3+x)^2}{(3-x)^2} \, dx \\ & = \frac {1}{32} \text {Subst}\left (\int \frac {5 e^{-10/x}-32 x^2}{x^2} \, dx,x,-3+x\right ) \\ & = \frac {1}{32} \text {Subst}\left (\int \left (-32+\frac {5 e^{-10/x}}{x^2}\right ) \, dx,x,-3+x\right ) \\ & = -x+\frac {5}{32} \text {Subst}\left (\int \frac {e^{-10/x}}{x^2} \, dx,x,-3+x\right ) \\ & = \frac {1}{64} e^{\frac {10}{3-x}}-x \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {1}{32} \left (\frac {1}{2} e^{\frac {10}{3-x}}-32 x\right ) \]
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Time = 0.91 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78
method | result | size |
risch | \(-x +\frac {9 x^{2} {\mathrm e}^{-\frac {2 \left (\ln \left (\frac {6 x}{5}\right ) x -3 \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{400}\) | \(32\) |
default | \(-x +\frac {\left (27 x^{3}-81 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-3600+1200 x}\) | \(42\) |
parts | \(-x +\frac {\left (27 x^{3}-81 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-3600+1200 x}\) | \(42\) |
parallelrisch | \(\frac {\left (-10800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}} x^{2}+243 x^{3}+10800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}} x -729 x^{2}+64800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-32400+10800 x}\) | \(105\) |
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Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.61 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {1}{400} \, {\left (9 \, x^{2} - 400 \, x e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {9 x^{2} e^{- \frac {2 \left (\left (x - 3\right ) \log {\left (\frac {6 x}{5} \right )} + 5\right )}{x - 3}}}{400} - x \]
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Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-x + \frac {1}{64} \, e^{\left (-\frac {10}{x - 3}\right )} \]
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\[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\int { \frac {{\left (9 \, x^{2} - 40 \, {\left (x^{2} - 6 \, x + 9\right )} e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}}{40 \, {\left (x^{2} - 6 \, x + 9\right )}} \,d x } \]
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Timed out. \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-\int \frac {{\mathrm {e}}^{-\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left ({\mathrm {e}}^{\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left (40\,x^2-240\,x+360\right )-9\,x^2\right )}{40\,x^2-240\,x+360} \,d x \]
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