Optimal. Leaf size=18 \[ -25+\frac {1}{64} e^{-\frac {10}{-3+x}}-x \]
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Rubi [A]
time = 0.32, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps
used = 8, number of rules used = 5, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 12, 6820,
14, 2240} \begin {gather*} \frac {1}{64} e^{\frac {10}{3-x}}-x \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 27
Rule 2240
Rule 6820
Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [A]
time = 0.06, size = 23, normalized size = 1.28 \begin {gather*} \frac {1}{32} \left (\frac {1}{2} e^{\frac {10}{3-x}}-32 x\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.77, size = 42, normalized size = 2.33
method | result | size |
risch | \(-x +\frac {9 x^{2} {\mathrm e}^{-\frac {2 \left (\ln \left (\frac {6 x}{5}\right ) x -3 \ln \left (\frac {6 x}{5}\right )+5\right )}{x -3}}}{400}\) | \(32\) |
default | \(-x +\frac {\left (-\frac {27}{10} x^{2}+\frac {9}{10} x^{3}\right ) {\mathrm e}^{-\frac {2 \left (\left (x -3\right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{x -3}}}{40 x -120}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 14, normalized size = 0.78 \begin {gather*} -x + \frac {1}{64} \, e^{\left (-\frac {10}{x - 3}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 47, normalized size = 2.61 \begin {gather*} \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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs.
\(2 (12) = 24\).
time = 0.17, size = 26, normalized size = 1.44 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} -\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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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