Integrand size = 83, antiderivative size = 28 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2 e^{x-\left (3+\log \left (e^{2 e^{4+\frac {x}{\log (4)}}}\right )\right )^2} x \]
Time = 0.87 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2 e^{-9-12 e^{4+\frac {x}{\log (4)}}-4 e^{8+\frac {2 x}{\log (4)}}+x} x \]
Integrate[(E^(-9 - 12*E^((x + 4*Log[4])/Log[4]) - 4*E^((2*(x + 4*Log[4]))/ Log[4]) + x)*(-24*E^((x + 4*Log[4])/Log[4])*x - 16*E^((2*(x + 4*Log[4]))/L og[4])*x + (2 + 2*x)*Log[4]))/Log[4],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.
\(\displaystyle \int \frac {\left (-24 x e^{\frac {x+4 \log (4)}{\log (4)}}-16 x e^{\frac {2 (x+4 \log (4))}{\log (4)}}+(2 x+2) \log (4)\right ) \exp \left (x-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}-9\right )}{\log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -2 e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9} \left (12 e^{\frac {x}{\log (4)}+4} x+8 e^{\frac {2 x}{\log (4)}+8} x-(x+1) \log (4)\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \int e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9} \left (12 e^{\frac {x}{\log (4)}+4} x+8 e^{\frac {2 x}{\log (4)}+8} x-(x+1) \log (4)\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \int \left (12 \exp \left (\frac {x}{\log (4)}+x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-5\right ) x+8 \exp \left (\frac {2 x}{\log (4)}+x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-1\right ) x-e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9} (x+1) \log (4)\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (12 \int \exp \left (\left (1+\frac {1}{\log (4)}\right ) x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-5\right ) xdx+8 \int \exp \left (\left (1+\frac {2}{\log (4)}\right ) x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-1\right ) xdx-\log (4) \int e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9}dx-\log (4) \int e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9} xdx\right )}{\log (4)}\) |
Int[(E^(-9 - 12*E^((x + 4*Log[4])/Log[4]) - 4*E^((2*(x + 4*Log[4]))/Log[4] ) + x)*(-24*E^((x + 4*Log[4])/Log[4])*x - 16*E^((2*(x + 4*Log[4]))/Log[4]) *x + (2 + 2*x)*Log[4]))/Log[4],x]
3.11.19.3.1 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]]
Time = 0.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
risch | \(2 x \,{\mathrm e}^{-4 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{\ln \left (2\right )}}-12 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{2 \ln \left (2\right )}}+x -9}\) | \(37\) |
norman | \(2 x \,{\mathrm e}^{-4 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{\ln \left (2\right )}}-12 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{2 \ln \left (2\right )}}+x -9}\) | \(40\) |
parallelrisch | \(2 x \,{\mathrm e}^{-4 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{\ln \left (2\right )}}-12 \,{\mathrm e}^{\frac {8 \ln \left (2\right )+x}{2 \ln \left (2\right )}}+x -9}\) | \(40\) |
int(1/2*(-16*x*exp(1/2*(8*ln(2)+x)/ln(2))^2-24*x*exp(1/2*(8*ln(2)+x)/ln(2) )+2*(2+2*x)*ln(2))*exp(-4*exp(1/2*(8*ln(2)+x)/ln(2))^2-12*exp(1/2*(8*ln(2) +x)/ln(2))+x-9)/ln(2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2 \, x e^{\left (x - 4 \, e^{\left (\frac {x + 8 \, \log \left (2\right )}{\log \left (2\right )}\right )} - 12 \, e^{\left (\frac {x + 8 \, \log \left (2\right )}{2 \, \log \left (2\right )}\right )} - 9\right )} \]
integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2) +x)/log(2))+2*(2+2*x)*log(2))*exp(-4*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp (1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2 x e^{x - 4 e^{\frac {2 \left (\frac {x}{2} + 4 \log {\left (2 \right )}\right )}{\log {\left (2 \right )}}} - 12 e^{\frac {\frac {x}{2} + 4 \log {\left (2 \right )}}{\log {\left (2 \right )}}} - 9} \]
integrate(1/2*(-16*x*exp(1/2*(8*ln(2)+x)/ln(2))**2-24*x*exp(1/2*(8*ln(2)+x )/ln(2))+2*(2+2*x)*ln(2))*exp(-4*exp(1/2*(8*ln(2)+x)/ln(2))**2-12*exp(1/2* (8*ln(2)+x)/ln(2))+x-9)/ln(2),x)
Time = 0.41 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2 \, x e^{\left (x - 4 \, e^{\left (\frac {x}{\log \left (2\right )} + 8\right )} - 12 \, e^{\left (\frac {x}{2 \, \log \left (2\right )} + 4\right )} - 9\right )} \]
integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2) +x)/log(2))+2*(2+2*x)*log(2))*exp(-4*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp (1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm=\
\[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=\int { -\frac {2 \, {\left (4 \, x e^{\left (\frac {x + 8 \, \log \left (2\right )}{\log \left (2\right )}\right )} + 6 \, x e^{\left (\frac {x + 8 \, \log \left (2\right )}{2 \, \log \left (2\right )}\right )} - {\left (x + 1\right )} \log \left (2\right )\right )} e^{\left (x - 4 \, e^{\left (\frac {x + 8 \, \log \left (2\right )}{\log \left (2\right )}\right )} - 12 \, e^{\left (\frac {x + 8 \, \log \left (2\right )}{2 \, \log \left (2\right )}\right )} - 9\right )}}{\log \left (2\right )} \,d x } \]
integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2) +x)/log(2))+2*(2+2*x)*log(2))*exp(-4*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp (1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm=\
integrate(-2*(4*x*e^((x + 8*log(2))/log(2)) + 6*x*e^(1/2*(x + 8*log(2))/lo g(2)) - (x + 1)*log(2))*e^(x - 4*e^((x + 8*log(2))/log(2)) - 12*e^(1/2*(x + 8*log(2))/log(2)) - 9)/log(2), x)
Time = 18.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right )}{\log (4)} \, dx=2\,x\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{\frac {x}{\ln \left (2\right )}}\,{\mathrm {e}}^8}\,{\mathrm {e}}^{-12\,{\mathrm {e}}^{\frac {x}{2\,\ln \left (2\right )}}\,{\mathrm {e}}^4}\,{\mathrm {e}}^{-9}\,{\mathrm {e}}^x \]