Optimal. Leaf size=22 \[ -3+e^{e^{-1+\frac {5 (2-\log (\log (5)))}{4 x^2}}} \]
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Rubi [F] time = 0.78, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}\right ) (-10+5 \log (\log (5)))}{2 x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=-\left (\frac {1}{2} (5 (2-\log (\log (5)))) \int \frac {\exp \left (e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}\right )}{x^3} \, dx\right )\\ &=-\left (\frac {1}{4} (5 (2-\log (\log (5)))) \operatorname {Subst}\left (\int \frac {\exp \left (-1+e^{-1+\frac {5}{2 x}} \log ^{-\frac {5}{4 x}}(5)-\frac {5 (-2+\log (\log (5)))}{4 x}\right )}{x^2} \, dx,x,x^2\right )\right )\\ &=\frac {1}{4} (5 (2-\log (\log (5)))) \operatorname {Subst}\left (\int \exp \left (-1+e^{-1+\frac {5 x}{2}} \log ^{-\frac {5 x}{4}}(5)-\frac {5}{4} x (-2+\log (\log (5)))\right ) \, dx,x,\frac {1}{x^2}\right )\\ &=(2-\log (\log (5))) \operatorname {Subst}\left (\int \exp \left (-1+e^{-1+2 x} \log ^{-x}(5)-x (-2+\log (\log (5)))\right ) \, dx,x,\frac {5}{4 x^2}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 24, normalized size = 1.09 \begin {gather*} e^{e^{-1+\frac {5}{2 x^2}} \log ^{-\frac {5}{4 x^2}}(5)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.73, size = 59, normalized size = 2.68 \begin {gather*} e^{\left (\frac {4 \, x^{2} e^{\left (-\frac {4 \, x^{2} + 5 \, \log \left (\log \relax (5)\right ) - 10}{4 \, x^{2}}\right )} - 4 \, x^{2} - 5 \, \log \left (\log \relax (5)\right ) + 10}{4 \, x^{2}} + \frac {4 \, x^{2} + 5 \, \log \left (\log \relax (5)\right ) - 10}{4 \, x^{2}}\right )} \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*} \mathit {undef} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 20, normalized size = 0.91
method | result | size |
norman | \({\mathrm e}^{{\mathrm e}^{\frac {-5 \ln \left (\ln \relax (5)\right )-4 x^{2}+10}{4 x^{2}}}}\) | \(20\) |
risch | \(\frac {{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \relax (5)\right )-10}{4 x^{2}}}} \ln \left (\ln \relax (5)\right )}{\ln \left (\ln \relax (5)\right )-2}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \relax (5)\right )-10}{4 x^{2}}}}}{\ln \left (\ln \relax (5)\right )-2}\) | \(60\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 17, normalized size = 0.77 \begin {gather*} e^{\left (e^{\left (-\frac {5 \, \log \left (\log \relax (5)\right )}{4 \, x^{2}} + \frac {5}{2 \, x^{2}} - 1\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.44, size = 20, normalized size = 0.91 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{-1}\,{\mathrm {e}}^{\frac {5}{2\,x^2}}}{{\ln \relax (5)}^{\frac {5}{4\,x^2}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.35, size = 20, normalized size = 0.91 \begin {gather*} e^{e^{\frac {- x^{2} - \frac {5 \log {\left (\log {\relax (5 )} \right )}}{4} + \frac {5}{2}}{x^{2}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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