∫ ee10−4x2−5 log(log(5)) ____________________________ 4x2 +10−4x2−5 log(log(5)) ____________________________ 4x2 (−10+5 log(log(5))) _________________________________________________________________________________

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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*}

Int[(E^(E^((10 - 4*x^2 - 5*Log[Log[5]])/(4*x^2)) + (10 - 4*x^2 - 5*Log[Log[5]])/(4*x^2))*(-10 + 5*Log[Log[5]])

(2 - Log[Log[5]])*Defer[Subst][Defer[Int][E^(-1 + E^(-1 + 2*x)/Log[5]^x - x*(-2 + Log[Log[5]])), x], x, 5/(4*x

\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*}

Integrate[(E^(E^((10 - 4*x^2 - 5*Log[Log[5]])/(4*x^2)) + (10 - 4*x^2 - 5*Log[Log[5]])/(4*x^2))*(-10 + 5*Log[Lo

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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*}

integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*exp(exp(1/4*(-5*log(log(5))-4*x^2+10)/

e^(1/4*(4*x^2*e^(-1/4*(4*x^2 + 5*log(log(5)) - 10)/x^2) - 4*x^2 - 5*log(log(5)) + 10)/x^2 + 1/4*(4*x^2 + 5*log

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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*exp(exp(1/4*(-5*log(log(5))-4*x^2+10)/

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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	\(\frac {{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \relax (5)\right )-10}{4 x^{2}}}} \ln \left (\ln \relax (5)\right )}{\ln \left (\ln \relax (5)\right )-2}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \relax (5)\right )-10}{4 x^{2}}}}}{\ln \left (\ln \relax (5)\right )-2}\)	\(60\)

int(1/2*(5*ln(ln(5))-10)*exp(1/4*(-5*ln(ln(5))-4*x^2+10)/x^2)*exp(exp(1/4*(-5*ln(ln(5))-4*x^2+10)/x^2))/x^3,x,

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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*}

integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*exp(exp(1/4*(-5*log(log(5))-4*x^2+10)/

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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*}

int((exp(-((5*log(log(5)))/4 + x^2 - 5/2)/x^2)*exp(exp(-((5*log(log(5)))/4 + x^2 - 5/2)/x^2))*(5*log(log(5)) -

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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*}

integrate(1/2*(5*ln(ln(5))-10)*exp(1/4*(-5*ln(ln(5))-4*x**2+10)/x**2)*exp(exp(1/4*(-5*ln(ln(5))-4*x**2+10)/x**

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3.46.6 \(\int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx\)