Integrand size = 72, antiderivative size = 36 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x-\frac {e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (x+\frac {e^{-x} x}{3}\right )}{x} \] Output:
x-exp(5*exp(1/2*exp(x^2)+1/2*x))/x*(x+1/3*x/exp(x))
Time = 2.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\frac {1}{6} e^{5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}} \left (-6-2 e^{-x}\right )+x \] Input:
Integrate[(6*E^x + E^(5*E^((E^x^2 + x)/2))*(2 + E^((E^x^2 + x)/2)*(-5 - 15 *E^x + E^x^2*(-10*x - 30*E^x*x))))/(6*E^x),x]
Output:
(E^(5*E^(E^x^2/2 + x/2))*(-6 - 2/E^x))/6 + 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.
\(\displaystyle \int \frac {1}{6} e^{-x} \left (e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (e^{x^2} \left (-30 e^x x-10 x\right )-15 e^x-5\right )+2\right )+6 e^x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int e^{-x} \left (e^{5 e^{\frac {1}{2} \left (x+e^{x^2}\right )}} \left (2-5 e^{\frac {1}{2} \left (x+e^{x^2}\right )} \left (2 e^{x^2} \left (3 e^x x+x\right )+3 e^x+1\right )\right )+6 e^x\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{6} \int \left (e^{5 e^{\frac {1}{2} \left (x+e^{x^2}\right )}-x} \left (-10 e^{x^2+\frac {x}{2}+\frac {e^{x^2}}{2}} x-30 e^{x^2+\frac {3 x}{2}+\frac {e^{x^2}}{2}} x-5 e^{\frac {x}{2}+\frac {e^{x^2}}{2}}-15 e^{\frac {3 x}{2}+\frac {e^{x^2}}{2}}+2\right )+6\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (-10 \text {Subst}\left (\int \exp \left (\frac {1}{2} \left (-2 x+e^{4 x^2}+10 e^{x+\frac {e^{4 x^2}}{2}}\right )\right )dx,x,\frac {x}{2}\right )-30 \text {Subst}\left (\int e^{\frac {1}{2} \left (2 x+e^{4 x^2}+10 e^{x+\frac {e^{4 x^2}}{2}}\right )}dx,x,\frac {x}{2}\right )+2 \int e^{5 e^{\frac {1}{2} \left (x+e^{x^2}\right )}-x}dx-10 \int e^{\frac {1}{2} \left (2 x^2-x+e^{x^2}+10 e^{\frac {x}{2}+\frac {e^{x^2}}{2}}\right )} xdx-30 \int e^{\frac {1}{2} \left (2 x^2+x+e^{x^2}+10 e^{\frac {x}{2}+\frac {e^{x^2}}{2}}\right )} xdx+6 x\right )\) |
Input:
Int[(6*E^x + E^(5*E^((E^x^2 + x)/2))*(2 + E^((E^x^2 + x)/2)*(-5 - 15*E^x + E^x^2*(-10*x - 30*E^x*x))))/(6*E^x),x]
Output:
$Aborted
Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81
method | result | size |
risch | \(x -\frac {\left (3 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-x +5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}}}{3}\) | \(29\) |
parallelrisch | \(-\frac {\left (-6 \ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x}+6 \,{\mathrm e}^{5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}} {\mathrm e}^{x}+2 \,{\mathrm e}^{5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}}\right ) {\mathrm e}^{-x}}{6}\) | \(49\) |
Input:
int(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2* x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))+6*exp(x))/exp(x),x,method=_RETURNVERB OSE)
Output:
x-1/3*(3*exp(x)+1)*exp(-x+5*exp(1/2*exp(x^2)+1/2*x))
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\frac {1}{3} \, {\left (3 \, x e^{x} - {\left (3 \, e^{x} + 1\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2 )+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))+6*exp(x))/exp(x),x, algorithm=" fricas")
Output:
1/3*(3*x*e^x - (3*e^x + 1)*e^(5*e^(1/2*x + 1/2*e^(x^2))))*e^(-x)
Time = 2.67 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x + \frac {\left (-3 - e^{- x}\right ) e^{5 e^{\frac {x}{2} + \frac {e^{x^{2}}}{2}}}}{3} \] Input:
integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x**2)-15*exp(x)-5)*exp(1/2*exp(x* *2)+1/2*x)+2)*exp(5*exp(1/2*exp(x**2)+1/2*x))+6*exp(x))/exp(x),x)
Output:
x + (-3 - exp(-x))*exp(5*exp(x/2 + exp(x**2)/2))/3
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=-\frac {1}{3} \, {\left (3 \, e^{x} + 1\right )} e^{\left (-x + 5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} + x \] Input:
integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2 )+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))+6*exp(x))/exp(x),x, algorithm=" maxima")
Output:
-1/3*(3*e^x + 1)*e^(-x + 5*e^(1/2*x + 1/2*e^(x^2))) + x
\[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\int { -\frac {1}{6} \, {\left ({\left (5 \, {\left (2 \, {\left (3 \, x e^{x} + x\right )} e^{\left (x^{2}\right )} + 3 \, e^{x} + 1\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )} - 2\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} - 6 \, e^{x}\right )} e^{\left (-x\right )} \,d x } \] Input:
integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2 )+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))+6*exp(x))/exp(x),x, algorithm=" giac")
Output:
integrate(-1/6*((5*(2*(3*x*e^x + x)*e^(x^2) + 3*e^x + 1)*e^(1/2*x + 1/2*e^ (x^2)) - 2)*e^(5*e^(1/2*x + 1/2*e^(x^2))) - 6*e^x)*e^(-x), x)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x-{\mathrm {e}}^{5\,\sqrt {{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}\,\sqrt {{\mathrm {e}}^x}-x}\,\left ({\mathrm {e}}^x+\frac {1}{3}\right ) \] Input:
int(exp(-x)*(exp(x) - (exp(5*exp(x/2 + exp(x^2)/2))*(exp(x/2 + exp(x^2)/2) *(15*exp(x) + exp(x^2)*(10*x + 30*x*exp(x)) + 5) - 2))/6),x)
Output:
x - exp(5*exp(exp(x^2))^(1/2)*exp(x)^(1/2) - x)*(exp(x) + 1/3)
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.50 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\frac {-3 e^{5 e^{\frac {e^{x^{2}}}{2}+\frac {x}{2}}+x}-e^{5 e^{\frac {e^{x^{2}}}{2}+\frac {x}{2}}}+3 e^{x} x}{3 e^{x}} \] Input:
int(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2* x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))+6*exp(x))/exp(x),x)
Output:
( - 3*e**(5*e**((e**(x**2) + x)/2) + x) - e**(5*e**((e**(x**2) + x)/2)) + 3*e**x*x)/(3*e**x)