Integrand size = 80, antiderivative size = 26 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=e^{e^{-1+\frac {1}{3} e^{5 \left (e^{e^x}-x\right )}+2 x}} \]
Time = 1.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=e^{e^{-1+\frac {1}{3} e^{5 e^{e^x}-5 x}+2 x}} \]
Integrate[(E^(E^((-3 + E^(5*E^E^x - 5*x) + 6*x)/3) + (-3 + E^(5*E^E^x - 5* x) + 6*x)/3)*(6 + E^(5*E^E^x - 5*x)*(-5 + 5*E^(E^x + 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.
\(\displaystyle \int \frac {1}{3} \left (e^{5 e^{e^x}-5 x} \left (5 e^{x+e^x}-5\right )+6\right ) \exp \left (\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \exp \left (\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}\right ) \left (6-5 e^{5 e^{e^x}-5 x} \left (1-e^{x+e^x}\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (5 \exp \left (5 \left (e^{e^x}-x\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}+\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )\right ) \left (-1+e^{x+e^x}\right )+6 \exp \left (\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (6 \int \exp \left (\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}\right )dx-5 \int \exp \left (5 \left (e^{e^x}-x\right )+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}+\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )\right )dx+5 \int \exp \left (5 \left (e^{e^x}-x\right )+e^x+e^{\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )}+x+\frac {1}{3} \left (6 x+e^{5 e^{e^x}-5 x}-3\right )\right )dx\right )\) |
Int[(E^(E^((-3 + E^(5*E^E^x - 5*x) + 6*x)/3) + (-3 + E^(5*E^E^x - 5*x) + 6 *x)/3)*(6 + E^(5*E^E^x - 5*x)*(-5 + 5*E^(E^x + x))))/3,x]
3.25.42.3.1 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 = 0.69 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{{\mathrm e}^{x}}-5 x}}{3}+2 x -1}}\) | \(20\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{{\mathrm e}^{x}}-5 x}}{3}+2 x -1}}\) | \(20\) |
int(1/3*((5*exp(x)*exp(exp(x))-5)*exp(5*exp(exp(x))-5*x)+6)*exp(1/3*exp(5* exp(exp(x))-5*x)+2*x-1)*exp(exp(1/3*exp(5*exp(exp(x))-5*x)+2*x-1)),x,metho d=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=e^{\left (e^{\left (2 \, x + \frac {1}{3} \, e^{\left (-5 \, {\left (x e^{x} - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )} - 1\right )}\right )} \]
integrate(1/3*((5*exp(x)*exp(exp(x))-5)*exp(5*exp(exp(x))-5*x)+6)*exp(1/3* exp(5*exp(exp(x))-5*x)+2*x-1)*exp(exp(1/3*exp(5*exp(exp(x))-5*x)+2*x-1)),x , algorithm=\
Time = 6.80 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=e^{e^{2 x + \frac {e^{- 5 x + 5 e^{e^{x}}}}{3} - 1}} \]
integrate(1/3*((5*exp(x)*exp(exp(x))-5)*exp(5*exp(exp(x))-5*x)+6)*exp(1/3* exp(5*exp(exp(x))-5*x)+2*x-1)*exp(exp(1/3*exp(5*exp(exp(x))-5*x)+2*x-1)),x )
Time = 0.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=e^{\left (e^{\left (2 \, x + \frac {1}{3} \, e^{\left (-5 \, x + 5 \, e^{\left (e^{x}\right )}\right )} - 1\right )}\right )} \]
integrate(1/3*((5*exp(x)*exp(exp(x))-5)*exp(5*exp(exp(x))-5*x)+6)*exp(1/3* exp(5*exp(exp(x))-5*x)+2*x-1)*exp(exp(1/3*exp(5*exp(exp(x))-5*x)+2*x-1)),x , algorithm=\
\[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx=\int { \frac {1}{3} \, {\left (5 \, {\left (e^{\left (x + e^{x}\right )} - 1\right )} e^{\left (-5 \, x + 5 \, e^{\left (e^{x}\right )}\right )} + 6\right )} e^{\left (2 \, x + e^{\left (2 \, x + \frac {1}{3} \, e^{\left (-5 \, x + 5 \, e^{\left (e^{x}\right )}\right )} - 1\right )} + \frac {1}{3} \, e^{\left (-5 \, x + 5 \, e^{\left (e^{x}\right )}\right )} - 1\right )} \,d x } \]
integrate(1/3*((5*exp(x)*exp(exp(x))-5)*exp(5*exp(exp(x))-5*x)+6)*exp(1/3* exp(5*exp(exp(x))-5*x)+2*x-1)*exp(exp(1/3*exp(5*exp(exp(x))-5*x)+2*x-1)),x , algorithm=\
integrate(1/3*(5*(e^(x + e^x) - 1)*e^(-5*x + 5*e^(e^x)) + 6)*e^(2*x + e^(2 *x + 1/3*e^(-5*x + 5*e^(e^x)) - 1) + 1/3*e^(-5*x + 5*e^(e^x)) - 1), x)
Time = 9.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{3} e^{e^{\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )}+\frac {1}{3} \left (-3+e^{5 e^{e^x}-5 x}+6 x\right )} \left (6+e^{5 e^{e^x}-5 x} \left (-5+5 e^{e^x+x}\right )\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{5\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^{-5\,x}}{3}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-1}} \]