Integrand size = 185, antiderivative size = 32 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{e^{e^{2 \left (-x+\log \left (\frac {4}{x}\right )\right )^2}}+\frac {8}{-3+x}}}{x} \]
\[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx \]
Integrate[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2) + 2*x^2 - 4*x*L og[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4) + E^(8/(-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4),x]
Integrate[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2) + 2*x^2 - 4*x*L og[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4) + E^(8/(-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4), x]
Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(32)=64\).
Time = 6.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 4.22, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2026, 2726}
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.
\(\displaystyle \int \frac {e^{e^{e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )}}} \left (\left (e^{\frac {8}{x-3}} \left (-4 x^3+20 x^2-12 x-36\right ) \log \left (\frac {4}{x}\right )+e^{\frac {8}{x-3}} \left (4 x^4-20 x^3+12 x^2+36 x\right )\right ) \exp \left (2 x^2+e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )}+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )\right )+e^{\frac {8}{x-3}} \left (-x^2-2 x-9\right )\right )}{x^4-6 x^3+9 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{e^{e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )}}} \left (\left (e^{\frac {8}{x-3}} \left (-4 x^3+20 x^2-12 x-36\right ) \log \left (\frac {4}{x}\right )+e^{\frac {8}{x-3}} \left (4 x^4-20 x^3+12 x^2+36 x\right )\right ) \exp \left (2 x^2+e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )}+2 \log ^2\left (\frac {4}{x}\right )-4 x \log \left (\frac {4}{x}\right )\right )+e^{\frac {8}{x-3}} \left (-x^2-2 x-9\right )\right )}{x^2 \left (x^2-6 x+9\right )}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {\left (e^{-\frac {8}{3-x}} \left (x^4-5 x^3+3 x^2+9 x\right )-e^{-\frac {8}{3-x}} \left (x^3-5 x^2+3 x+9\right ) \log \left (\frac {4}{x}\right )\right ) \exp \left (\exp \left (4^{-4 x} \left (\frac {1}{x}\right )^{-4 x} e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )}\right )\right )}{x^2 \left (x^2-6 x+9\right ) \left (x-\log \left (\frac {4}{x}\right )-\frac {\log \left (\frac {4}{x}\right )}{x}+1\right )}\) |
Int[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2) + 2*x^2 - 4*x*Log[4/x ] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4) + E^(8/ (-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4), x]
(E^E^(E^(2*x^2 + 2*Log[4/x]^2)/(4^(4*x)*(x^(-1))^(4*x)))*((9*x + 3*x^2 - 5 *x^3 + x^4)/E^(8/(3 - x)) - ((9 + 3*x - 5*x^2 + x^3)*Log[4/x])/E^(8/(3 - x ))))/(x^2*(9 - 6*x + x^2)*(1 + x - Log[4/x] - Log[4/x]/x))
3.6.89.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Timed out.
\[\int \frac {\left (\left (\left (-4 x^{3}+20 x^{2}-12 x -36\right ) {\mathrm e}^{\frac {8}{-3+x}} \ln \left (\frac {4}{x}\right )+\left (4 x^{4}-20 x^{3}+12 x^{2}+36 x \right ) {\mathrm e}^{\frac {8}{-3+x}}\right ) {\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}} {\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}+\left (-x^{2}-2 x -9\right ) {\mathrm e}^{\frac {8}{-3+x}}\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}}}{x^{4}-6 x^{3}+9 x^{2}}d x\]
int((((-4*x^3+20*x^2-12*x-36)*exp(4/(-3+x))^2*ln(4/x)+(4*x^4-20*x^3+12*x^2 +36*x)*exp(4/(-3+x))^2)*exp(ln(4/x)^2-2*x*ln(4/x)+x^2)^2*exp(exp(ln(4/x)^2 -2*x*ln(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(-3+x))^2)*exp(exp(exp(ln(4/x)^2-2 *x*ln(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x)
int((((-4*x^3+20*x^2-12*x-36)*exp(4/(-3+x))^2*ln(4/x)+(4*x^4-20*x^3+12*x^2 +36*x)*exp(4/(-3+x))^2)*exp(ln(4/x)^2-2*x*ln(4/x)+x^2)^2*exp(exp(ln(4/x)^2 -2*x*ln(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(-3+x))^2)*exp(exp(exp(ln(4/x)^2-2 *x*ln(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x} \]
integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(-3+x))^2*log(4/x)+(4*x^4-20*x^3 +12*x^2+36*x)*exp(4/(-3+x))^2)*exp(log(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(-3+x))^2)*exp(exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm=\
Time = 95.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\frac {8}{x - 3}} e^{e^{e^{2 x^{2} - 4 x \log {\left (\frac {4}{x} \right )} + 2 \log {\left (\frac {4}{x} \right )}^{2}}}}}{x} \]
integrate((((-4*x**3+20*x**2-12*x-36)*exp(4/(-3+x))**2*ln(4/x)+(4*x**4-20* x**3+12*x**2+36*x)*exp(4/(-3+x))**2)*exp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2*e xp(exp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2)+(-x**2-2*x-9)*exp(4/(-3+x))**2)*ex p(exp(exp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2))/(x**4-6*x**3+9*x**2),x)
Time = 0.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 8 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} + 4 \, x \log \left (x\right ) - 8 \, \log \left (2\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2}\right )}\right )}\right )}}{x} \]
integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(-3+x))^2*log(4/x)+(4*x^4-20*x^3 +12*x^2+36*x)*exp(4/(-3+x))^2)*exp(log(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(-3+x))^2)*exp(exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm=\
e^(8/(x - 3) + e^(e^(2*x^2 - 8*x*log(2) + 8*log(2)^2 + 4*x*log(x) - 8*log( 2)*log(x) + 2*log(x)^2)))/x
\[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\int { -\frac {{\left (4 \, {\left ({\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )} \log \left (\frac {4}{x}\right ) - {\left (x^{4} - 5 \, x^{3} + 3 \, x^{2} + 9 \, x\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2} + e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )} + {\left (x^{2} + 2 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x^{4} - 6 \, x^{3} + 9 \, x^{2}} \,d x } \]
integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(-3+x))^2*log(4/x)+(4*x^4-20*x^3 +12*x^2+36*x)*exp(4/(-3+x))^2)*exp(log(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(-3+x))^2)*exp(exp(exp( log(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm=\
integrate(-(4*((x^3 - 5*x^2 + 3*x + 9)*e^(8/(x - 3))*log(4/x) - (x^4 - 5*x ^3 + 3*x^2 + 9*x)*e^(8/(x - 3)))*e^(2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2 + e^(2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2)) + (x^2 + 2*x + 9)*e^(8/(x - 3)))* e^(e^(e^(2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2)))/(x^4 - 6*x^3 + 9*x^2), x)
Time = 11.69 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{{\left (\frac {1}{256}\right )}^x\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x}\right )}^2}\,{\mathrm {e}}^{8\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{8\,\ln \left (2\right )-4\,x}}}\,{\mathrm {e}}^{\frac {8}{x-3}}}{x} \]
int(-(exp(exp(exp(2*log(4/x)^2 - 4*x*log(4/x) + 2*x^2)))*(exp(8/(x - 3))*( 2*x + x^2 + 9) - exp(exp(2*log(4/x)^2 - 4*x*log(4/x) + 2*x^2))*exp(2*log(4 /x)^2 - 4*x*log(4/x) + 2*x^2)*(exp(8/(x - 3))*(36*x + 12*x^2 - 20*x^3 + 4* x^4) - exp(8/(x - 3))*log(4/x)*(12*x - 20*x^2 + 4*x^3 + 36))))/(9*x^2 - 6* x^3 + x^4),x)