Optimal. Leaf size=22 \[ \frac {2}{-5+e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \]
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Rubi [A] time = 1.56, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6741, 6708, 32} \begin {gather*} -\frac {2}{5-e^{\frac {1}{12} e^{2 x-5}} \log (x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 32
Rule 6708
Rule 6741
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{x \left (5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)\right )^2} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{(5-x)^2} \, dx,x,e^{\frac {1}{12} e^{-5+2 x}} \log (x)\right )\right )\\ &=-\frac {2}{5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.31, size = 23, normalized size = 1.05 \begin {gather*} -\frac {2}{5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 20, normalized size = 0.91 \begin {gather*} \frac {2}{e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \relax (x) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 65, normalized size = 2.95 \begin {gather*} \frac {2 \, {\left (x e^{\left (2 \, x\right )} \log \relax (x) + 6 \, e^{5}\right )}}{x e^{\left (2 \, x + \frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x)^{2} - 5 \, x e^{\left (2 \, x\right )} \log \relax (x) + 6 \, e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )} + 5\right )} \log \relax (x) - 30 \, e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 19, normalized size = 0.86
method | result | size |
risch | \(\frac {2}{\ln \relax (x ) {\mathrm e}^{\frac {{\mathrm e}^{2 x -5}}{12}}-5}\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (x e^{\left (2 \, x - 5\right )} \log \relax (x) + 6\right )} e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )}}{6 \, {\left (x e^{\left (\frac {1}{6} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x)^{2} - 10 \, x e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x) + 25 \, x\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.16, size = 18, normalized size = 0.82 \begin {gather*} \frac {2}{{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-5}}{12}}\,\ln \relax (x)-5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.30, size = 15, normalized size = 0.68 \begin {gather*} \frac {2}{e^{\frac {e^{2 x - 5}}{12}} \log {\relax (x )} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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