Integrand size = 57, antiderivative size = 22 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=-3+e^{e^{-1+\frac {5 (2-\log (\log (5)))}{4 x^2}}} \]
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=e^{e^{-1+\frac {5}{2 x^2}} \log ^{-\frac {5}{4 x^2}}(5)} \]
Integrate[(E^(E^((10 - 4*x^2 - 5*Log[Log[5]])/(4*x^2)) + (10 - 4*x^2 - 5*L og[Log[5]])/(4*x^2))*(-10 + 5*Log[Log[5]]))/(2*x^3),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(5 \log (\log (5))-10) \exp \left (\frac {-4 x^2+10-5 \log (\log (5))}{4 x^2}+e^{\frac {-4 x^2+10-5 \log (\log (5))}{4 x^2}}\right )}{2 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{2} (2-\log (\log (5))) \int \frac {\exp \left (e^{\frac {5-2 x^2}{2 x^2}} \log ^{-\frac {5}{4 x^2}}(5)-\frac {4 x^2-5 (2-\log (\log (5)))}{4 x^2}\right )}{x^3}dx\) |
\(\Big \downarrow \) 7283 |
\(\displaystyle -\frac {5}{4} (2-\log (\log (5))) \int \frac {\exp \left (-1+\frac {5}{2 x^2}+e^{\frac {5}{2 x^2}-1} \log ^{-\frac {5}{4 x^2}}(5)\right ) \log ^{-\frac {5}{4 x^2}}(5)}{x^4}dx^2\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {5}{4} (2-\log (\log (5))) \int \exp \left (-1+\frac {5}{2 x^2}+e^{\frac {5}{2 x^2}-1} \log ^{-\frac {5}{4 x^2}}(5)\right ) \log ^{-\frac {5}{4 x^2}}(5)d\frac {1}{x^2}\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle 5 (2-\log (\log (5))) \int \exp \left (-1+\frac {5}{2 x^2}+e^{\frac {5}{2 x^2}-1} \log ^{-\frac {5}{4 x^2}}(5)\right ) \log ^{-\frac {5}{4 x^2}}(5)d\frac {1}{4 x^2}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle 5 (2-\log (\log (5))) \int \exp \left (-1+\frac {5}{2 x^2}+e^{\frac {5}{2 x^2}-1} \log ^{-\frac {5}{4 x^2}}(5)\right ) \log ^{-\frac {5}{4 x^2}}(5)d\frac {1}{4 x^2}\) |
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*x^3),x]
3.16.6.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, x ]}, Simp[1/lst[[2]] Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x] , x], x, (lst[[3]]*x)^lst[[2]]], x] /; !FalseQ[lst] && NeQ[lst[[2]], m + 1 ]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] || !AlgebraicF unctionQ[u, x])
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \({\mathrm e}^{{\mathrm e}^{\frac {-5 \ln \left (\ln \left (5\right )\right )-4 x^{2}+10}{4 x^{2}}}}\) | \(20\) |
norman | \({\mathrm e}^{{\mathrm e}^{\frac {-5 \ln \left (\ln \left (5\right )\right )-4 x^{2}+10}{4 x^{2}}}}\) | \(20\) |
default | \(\frac {2 \left (\frac {5 \ln \left (\ln \left (5\right )\right )}{2}-5\right ) {\mathrm e}^{{\mathrm e}^{\frac {-5 \ln \left (\ln \left (5\right )\right )-4 x^{2}+10}{4 x^{2}}}}}{5 \left (\ln \left (\ln \left (5\right )\right )-2\right )}\) | \(36\) |
parallelrisch | \(\frac {2 \left (\frac {5 \ln \left (\ln \left (5\right )\right )}{2}-5\right ) {\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \left (5\right )\right )-10}{4 x^{2}}}}}{5 \left (\ln \left (\ln \left (5\right )\right )-2\right )}\) | \(36\) |
risch | \(\frac {{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \left (5\right )\right )-10}{4 x^{2}}}} \ln \left (\ln \left (5\right )\right )}{\ln \left (\ln \left (5\right )\right )-2}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{-\frac {4 x^{2}+5 \ln \left (\ln \left (5\right )\right )-10}{4 x^{2}}}}}{\ln \left (\ln \left (5\right )\right )-2}\) | \(60\) |
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,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (16) = 32\).
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=e^{\left (\frac {4 \, x^{2} e^{\left (-\frac {4 \, x^{2} + 5 \, \log \left (\log \left (5\right )\right ) - 10}{4 \, x^{2}}\right )} - 4 \, x^{2} - 5 \, \log \left (\log \left (5\right )\right ) + 10}{4 \, x^{2}} + \frac {4 \, x^{2} + 5 \, \log \left (\log \left (5\right )\right ) - 10}{4 \, x^{2}}\right )} \]
integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*ex p(exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2))/x^3,x, algorithm=\
e^(1/4*(4*x^2*e^(-1/4*(4*x^2 + 5*log(log(5)) - 10)/x^2) - 4*x^2 - 5*log(lo g(5)) + 10)/x^2 + 1/4*(4*x^2 + 5*log(log(5)) - 10)/x^2)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=e^{e^{\frac {- x^{2} - \frac {5 \log {\left (\log {\left (5 \right )} \right )}}{4} + \frac {5}{2}}{x^{2}}}} \]
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**2))/x**3,x)
Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=e^{\left (e^{\left (-\frac {5 \, \log \left (\log \left (5\right )\right )}{4 \, x^{2}} + \frac {5}{2 \, x^{2}} - 1\right )}\right )} \]
integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*ex p(exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2))/x^3,x, algorithm=\
\[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx=\int { \frac {5 \, {\left (\log \left (\log \left (5\right )\right ) - 2\right )} e^{\left (-\frac {4 \, x^{2} + 5 \, \log \left (\log \left (5\right )\right ) - 10}{4 \, x^{2}} + e^{\left (-\frac {4 \, x^{2} + 5 \, \log \left (\log \left (5\right )\right ) - 10}{4 \, x^{2}}\right )}\right )}}{2 \, x^{3}} \,d x } \]
integrate(1/2*(5*log(log(5))-10)*exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2)*ex p(exp(1/4*(-5*log(log(5))-4*x^2+10)/x^2))/x^3,x, algorithm=\
Time = 12.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{e^{\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}}+\frac {10-4 x^2-5 \log (\log (5))}{4 x^2}} (-10+5 \log (\log (5)))}{2 x^3} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-1}\,{\mathrm {e}}^{\frac {5}{2\,x^2}}}{{\ln \left (5\right )}^{\frac {5}{4\,x^2}}}} \]