Optimal. Leaf size=31 \[ \frac {e^{e^{\left (-x+\frac {3 \log (x)}{x}\right )^2}}}{\frac {1}{3 x}-x} \]
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Rubi [B] time = 1.01, antiderivative size = 125, normalized size of antiderivative = 4.03, number of steps used = 3, number of rules used = 3, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1594, 28, 2288} \begin {gather*} \frac {3 e^{\frac {e^{\frac {x^4+9 \log ^2(x)}{x^2}}}{x^6}} \left (3 x^6-10 x^4+3 x^2+9 \left (1-3 x^2\right ) \log ^2(x)-9 \left (1-3 x^2\right ) \log (x)\right )}{x^2 \left (1-3 x^2\right )^2 \left (\frac {-2 x^3+3 x+6 x \log (x)-\frac {9 \log (x)}{x}}{x^2}+\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^3}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 28
Rule 1594
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}}} \left (3 x^2+9 x^4+e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}} \left (-18 x^2+60 x^4-18 x^6+\left (54-162 x^2\right ) \log (x)+\left (-54+162 x^2\right ) \log ^2(x)\right )\right )}{x^2 \left (1-6 x^2+9 x^4\right )} \, dx\\ &=9 \int \frac {e^{e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}}} \left (3 x^2+9 x^4+e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}} \left (-18 x^2+60 x^4-18 x^6+\left (54-162 x^2\right ) \log (x)+\left (-54+162 x^2\right ) \log ^2(x)\right )\right )}{x^2 \left (-3+9 x^2\right )^2} \, dx\\ &=\frac {3 e^{\frac {e^{\frac {x^4+9 \log ^2(x)}{x^2}}}{x^6}} \left (3 x^2-10 x^4+3 x^6-9 \left (1-3 x^2\right ) \log (x)+9 \left (1-3 x^2\right ) \log ^2(x)\right )}{x^2 \left (1-3 x^2\right )^2 \left (\frac {3 x-2 x^3-\frac {9 \log (x)}{x}+6 x \log (x)}{x^2}+\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^3}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 33, normalized size = 1.06 \begin {gather*} -\frac {3 e^{\frac {e^{x^2+\frac {9 \log ^2(x)}{x^2}}}{x^6}} x}{-1+3 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 35, normalized size = 1.13 \begin {gather*} -\frac {3 \, x e^{\left (e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )}}{3 \, x^{2} - 1} \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*} \int \frac {3 \, {\left (3 \, x^{4} + x^{2} - 2 \, {\left (3 \, x^{6} - 10 \, x^{4} - 9 \, {\left (3 \, x^{2} - 1\right )} \log \relax (x)^{2} + 3 \, x^{2} + 9 \, {\left (3 \, x^{2} - 1\right )} \log \relax (x)\right )} e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )} e^{\left (e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )}}{9 \, x^{6} - 6 \, x^{4} + x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 31, normalized size = 1.00
method | result | size |
risch | \(-\frac {3 x \,{\mathrm e}^{{\mathrm e}^{\frac {\left (-x^{2}+3 \ln \relax (x )\right )^{2}}{x^{2}}}}}{3 x^{2}-1}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3 \, x e^{\left (\frac {e^{\left (x^{2} + \frac {9 \, \log \relax (x)^{2}}{x^{2}}\right )}}{x^{6}}\right )}}{3 \, x^{2} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.44, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {9\,{\ln \relax (x)}^2}{x^2}}}{x^6}}}{x^2-\frac {1}{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.76, size = 36, normalized size = 1.16 \begin {gather*} - \frac {3 x e^{e^{\frac {x^{4} - 6 x^{2} \log {\relax (x )} + 9 \log {\relax (x )}^{2}}{x^{2}}}}}{3 x^{2} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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