Integrand size = 136, antiderivative size = 29 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=5 e^{-x \log (4)-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}}+\log (x) \]
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=5\ 4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}}+\log (x) \]
Integrate[(E^(((2*x + x^2)*Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4 + 4*x + x^2)*Log[x] + (-20*x - 20*x^2 - 5*x^3)*Log[4]*Log[x] + (-20 - 10*x)* Log[3]^2*Log[Log[x]] + 5*x*Log[3]^2*Log[x]*Log[Log[x]]^2)/(E^(((2*x + x^2) *Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4*x + 4*x^2 + x^3)*Log[x]),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.
| \(\displaystyle \int \frac {\exp \left (-\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right ) \left (\left (x^2+4 x+4\right ) \log (x) \exp \left (\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right )+\left (-5 x^3-20 x^2-20 x\right ) \log (4) \log (x)+5 x \log ^2(3) \log (x) \log ^2(\log (x))+(-10 x-20) \log ^2(3) \log (\log (x))\right )}{\left (x^3+4 x^2+4 x\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\exp \left (-\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right ) \left (\left (x^2+4 x+4\right ) \log (x) \exp \left (\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right )+\left (-5 x^3-20 x^2-20 x\right ) \log (4) \log (x)+5 x \log ^2(3) \log (x) \log ^2(\log (x))+(-10 x-20) \log ^2(3) \log (\log (x))\right )}{x \left (x^2+4 x+4\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {\exp \left (-\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right ) \left (\left (x^2+4 x+4\right ) \log (x) \exp \left (\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right )+\left (-5 x^3-20 x^2-20 x\right ) \log (4) \log (x)+5 x \log ^2(3) \log (x) \log ^2(\log (x))+(-10 x-20) \log ^2(3) \log (\log (x))\right )}{x (x+2)^2 \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4^x \exp \left (\frac {\log ^2(3) \log ^2(\log (x))}{x+2}-\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right )}{x}-\frac {5 \left (x^3 \log (4) \log (x)+4 x^2 \log (4) \log (x)-x \log ^2(3) \log (x) \log ^2(\log (x))+2 x \log ^2(3) \log (\log (x))+4 \log ^2(3) \log (\log (x))+4 x \log (4) \log (x)\right ) \exp \left (-\frac {\left (x^2+2 x\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{x+2}\right )}{x (x+2)^2 \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -5 \log (4) \int \exp \left (-\frac {\log ^2(3) \log ^2(\log (x))+\left (x^2+2 x\right ) \log (4)}{x+2}\right )dx-5 \log ^2(3) \int \frac {\exp \left (-\frac {\log ^2(3) \log ^2(\log (x))+\left (x^2+2 x\right ) \log (4)}{x+2}\right ) \log (\log (x))}{x \log (x)}dx+5 \log ^2(3) \int \frac {\exp \left (-\frac {\log ^2(3) \log ^2(\log (x))+\left (x^2+2 x\right ) \log (4)}{x+2}\right ) \log (\log (x))}{(x+2) \log (x)}dx+5 \log ^2(3) \int \frac {\exp \left (-\frac {\log ^2(3) \log ^2(\log (x))+\left (x^2+2 x\right ) \log (4)}{x+2}\right ) \log ^2(\log (x))}{(x+2)^2}dx+\log (x)\) |
Int[(E^(((2*x + x^2)*Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4 + 4*x + x^2)*Log[x] + (-20*x - 20*x^2 - 5*x^3)*Log[4]*Log[x] + (-20 - 10*x)*Log[3] ^2*Log[Log[x]] + 5*x*Log[3]^2*Log[x]*Log[Log[x]]^2)/(E^(((2*x + x^2)*Log[4 ] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4*x + 4*x^2 + x^3)*Log[x]),x]
3.26.37.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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])
Time = 28.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62
| method | result | size |
| risch | \(\ln \left (x \right )+5 \,16^{-\frac {x}{2+x}} 4^{-\frac {x^{2}}{2+x}} {\mathrm e}^{-\frac {\ln \left (3\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}}{2+x}}\) | \(47\) |
| parallelrisch | \(\frac {\left (10+{\mathrm e}^{\frac {\ln \left (3\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}+2 \left (x^{2}+2 x \right ) \ln \left (2\right )}{2+x}} \ln \left (x \right ) x +2 \ln \left (x \right ) {\mathrm e}^{\frac {\ln \left (3\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}+2 \left (x^{2}+2 x \right ) \ln \left (2\right )}{2+x}}+5 x \right ) {\mathrm e}^{-\frac {\ln \left (3\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}+\left (2 x^{2}+4 x \right ) \ln \left (2\right )}{2+x}}}{2+x}\) | \(109\) |
int(((x^2+4*x+4)*ln(x)*exp((ln(3)^2*ln(ln(x))^2+2*(x^2+2*x)*ln(2))/(2+x))+ 5*x*ln(3)^2*ln(x)*ln(ln(x))^2+(-10*x-20)*ln(3)^2*ln(ln(x))+2*(-5*x^3-20*x^ 2-20*x)*ln(2)*ln(x))/(x^3+4*x^2+4*x)/ln(x)/exp((ln(3)^2*ln(ln(x))^2+2*(x^2 +2*x)*ln(2))/(2+x)),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx={\left (e^{\left (\frac {\log \left (3\right )^{2} \log \left (\log \left (x\right )\right )^{2} + 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (2\right )}{x + 2}\right )} \log \left (x\right ) + 5\right )} e^{\left (-\frac {\log \left (3\right )^{2} \log \left (\log \left (x\right )\right )^{2} + 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (2\right )}{x + 2}\right )} \]
integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log( 2))/(2+x))+5*x*log(3)^2*log(x)*log(log(x))^2+(-10*x-20)*log(3)^2*log(log(x ))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log(3 )^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm=\
(e^((log(3)^2*log(log(x))^2 + 2*(x^2 + 2*x)*log(2))/(x + 2))*log(x) + 5)*e ^(-(log(3)^2*log(log(x))^2 + 2*(x^2 + 2*x)*log(2))/(x + 2))
Time = 0.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=\log {\left (x \right )} + 5 e^{- \frac {\left (2 x^{2} + 4 x\right ) \log {\left (2 \right )} + \log {\left (3 \right )}^{2} \log {\left (\log {\left (x \right )} \right )}^{2}}{x + 2}} \]
integrate(((x**2+4*x+4)*ln(x)*exp((ln(3)**2*ln(ln(x))**2+2*(x**2+2*x)*ln(2 ))/(2+x))+5*x*ln(3)**2*ln(x)*ln(ln(x))**2+(-10*x-20)*ln(3)**2*ln(ln(x))+2* (-5*x**3-20*x**2-20*x)*ln(2)*ln(x))/(x**3+4*x**2+4*x)/ln(x)/exp((ln(3)**2* ln(ln(x))**2+2*(x**2+2*x)*ln(2))/(2+x)),x)
Time = 0.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=5 \, e^{\left (-\frac {\log \left (3\right )^{2} \log \left (\log \left (x\right )\right )^{2}}{x + 2} - 2 \, x \log \left (2\right )\right )} + \log \left (x\right ) \]
integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log( 2))/(2+x))+5*x*log(3)^2*log(x)*log(log(x))^2+(-10*x-20)*log(3)^2*log(log(x ))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log(3 )^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm=\
Time = 0.82 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=5 \, e^{\left (-\frac {\log \left (3\right )^{2} \log \left (\log \left (x\right )\right )^{2}}{x + 2} - \frac {2 \, x^{2} \log \left (2\right )}{x + 2} - \frac {4 \, x \log \left (2\right )}{x + 2}\right )} + \log \left (x\right ) \]
integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log( 2))/(2+x))+5*x*log(3)^2*log(x)*log(log(x))^2+(-10*x-20)*log(3)^2*log(log(x ))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log(3 )^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm=\
Time = 11.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {e^{-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (e^{\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{\left (4 x+4 x^2+x^3\right ) \log (x)} \, dx=\ln \left (x\right )+\frac {5\,{\mathrm {e}}^{-\frac {{\ln \left (\ln \left (x\right )\right )}^2\,{\ln \left (3\right )}^2}{x+2}}}{2^{\frac {4\,x}{x+2}}\,2^{\frac {2\,x^2}{x+2}}} \]
int(-(exp(-(2*log(2)*(2*x + x^2) + log(log(x))^2*log(3)^2)/(x + 2))*(2*log (2)*log(x)*(20*x + 20*x^2 + 5*x^3) - exp((2*log(2)*(2*x + x^2) + log(log(x ))^2*log(3)^2)/(x + 2))*log(x)*(4*x + x^2 + 4) + log(log(x))*log(3)^2*(10* x + 20) - 5*x*log(log(x))^2*log(3)^2*log(x)))/(log(x)*(4*x + 4*x^2 + x^3)) ,x)