Integrand size = 57, antiderivative size = 30 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=-3+\frac {1}{5} e^{-\frac {e^{16+x}}{2}} \left (x+x \left (e^{e^x}+2 x\right )\right ) \] Output:
1/5*(x*(2*x+exp(exp(x)))+x)/exp(1/2*exp(x+16))-3
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\frac {1}{5} e^{-\frac {e^{16+x}}{2}} x \left (1+e^{e^x}+2 x\right ) \] Input:
Integrate[(2 + 8*x + E^E^x*(2 + 2*E^x*x - E^(16 + x)*x) + E^(16 + x)*(-x - 2*x^2))/(10*E^(E^(16 + x)/2)),x]
Output:
(x*(1 + E^E^x + 2*x))/(5*E^(E^(16 + x)/2))
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.
\(\displaystyle \int \frac {1}{10} e^{-\frac {e^{x+16}}{2}} \left (e^{x+16} \left (-2 x^2-x\right )+8 x+e^{e^x} \left (2 e^x x-e^{x+16} x+2\right )+2\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int e^{-\frac {e^{x+16}}{2}} \left (8 x+e^{e^x} \left (2 e^x x-e^{x+16} x+2\right )-e^{x+16} \left (2 x^2+x\right )+2\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{10} \int \left (8 e^{-\frac {e^{x+16}}{2}} x-e^{x-\frac {e^{x+16}}{2}+16} (2 x+1) x+2 e^{-\frac {e^{x+16}}{2}}+e^{e^x-\frac {e^{x+16}}{2}} \left (2 e^x \left (1-\frac {e^{16}}{2}\right ) x+2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (-2 \int e^{x-\frac {e^{x+16}}{2}+16} x^2dx+8 \int e^{-\frac {e^{x+16}}{2}} xdx-\int e^{x-\frac {e^{x+16}}{2}+16} xdx+2 \operatorname {ExpIntegralEi}\left (-\frac {e^{x+16}}{2}\right )+\frac {2 \left (2-e^{16}\right ) e^{x+e^x-\frac {e^{x+16}}{2}} x}{2 e^x-e^{x+16}}\right )\) |
Input:
Int[(2 + 8*x + E^E^x*(2 + 2*E^x*x - E^(16 + x)*x) + E^(16 + x)*(-x - 2*x^2 ))/(10*E^(E^(16 + x)/2)),x]
Output:
$Aborted
Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63
method | result | size |
risch | \(\frac {\left (1+2 x +{\mathrm e}^{{\mathrm e}^{x}}\right ) x \,{\mathrm e}^{-\frac {{\mathrm e}^{x +16}}{2}}}{5}\) | \(19\) |
norman | \(\left (\frac {x}{5}+\frac {2 x^{2}}{5}+\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{5}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{x +16}}{2}}\) | \(26\) |
parallelrisch | \(-\frac {\left (-4 x^{2}-2 x \,{\mathrm e}^{{\mathrm e}^{x}}-2 x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{x +16}}{2}}}{10}\) | \(27\) |
Input:
int(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x +2)/exp(1/2*exp(x+16)),x,method=_RETURNVERBOSE)
Output:
1/5*(1+2*x+exp(exp(x)))*x*exp(-1/2*exp(x+16))
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\frac {1}{5} \, {\left (2 \, x^{2} + x\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )} + \frac {1}{5} \, x e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )} + e^{x}\right )} \] Input:
integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+1 6)+8*x+2)/exp(1/2*exp(x+16)),x, algorithm="fricas")
Output:
1/5*(2*x^2 + x)*e^(-1/2*e^(x + 16)) + 1/5*x*e^(-1/2*e^(x + 16) + e^x)
\[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=- \frac {\int \left (- 8 x e^{- \frac {e^{16} e^{x}}{2}}\right )\, dx + \int \left (- 2 e^{- \frac {e^{16} e^{x}}{2}} e^{e^{x}}\right )\, dx + \int x e^{16} e^{x} e^{- \frac {e^{16} e^{x}}{2}}\, dx + \int \left (- 2 x e^{x} e^{- \frac {e^{16} e^{x}}{2}} e^{e^{x}}\right )\, dx + \int 2 x^{2} e^{16} e^{x} e^{- \frac {e^{16} e^{x}}{2}}\, dx + \int x e^{16} e^{x} e^{- \frac {e^{16} e^{x}}{2}} e^{e^{x}}\, dx + \int \left (- 2 e^{- \frac {e^{16} e^{x}}{2}}\right )\, dx}{10} \] Input:
integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x**2-x)*exp(x+ 16)+8*x+2)/exp(1/2*exp(x+16)),x)
Output:
-(Integral(-8*x*exp(-exp(16)*exp(x)/2), x) + Integral(-2*exp(-exp(16)*exp( x)/2)*exp(exp(x)), x) + Integral(x*exp(16)*exp(x)*exp(-exp(16)*exp(x)/2), x) + Integral(-2*x*exp(x)*exp(-exp(16)*exp(x)/2)*exp(exp(x)), x) + Integra l(2*x**2*exp(16)*exp(x)*exp(-exp(16)*exp(x)/2), x) + Integral(x*exp(16)*ex p(x)*exp(-exp(16)*exp(x)/2)*exp(exp(x)), x) + Integral(-2*exp(-exp(16)*exp (x)/2), x))/10
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\frac {1}{5} \, {\left (2 \, x^{2} + x e^{\left (e^{x}\right )} + x\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )} \] Input:
integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+1 6)+8*x+2)/exp(1/2*exp(x+16)),x, algorithm="maxima")
Output:
1/5*(2*x^2 + x*e^(e^x) + x)*e^(-1/2*e^(x + 16))
\[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\int { -\frac {1}{10} \, {\left ({\left (2 \, x^{2} + x\right )} e^{\left (x + 16\right )} + {\left (x e^{\left (x + 16\right )} - 2 \, x e^{x} - 2\right )} e^{\left (e^{x}\right )} - 8 \, x - 2\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )} \,d x } \] Input:
integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+1 6)+8*x+2)/exp(1/2*exp(x+16)),x, algorithm="giac")
Output:
integrate(-1/10*((2*x^2 + x)*e^(x + 16) + (x*e^(x + 16) - 2*x*e^x - 2)*e^( e^x) - 8*x - 2)*e^(-1/2*e^(x + 16)), x)
Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}\,{\mathrm {e}}^x}{2}}\,\left (2\,x+{\mathrm {e}}^{{\mathrm {e}}^x}+1\right )}{5} \] Input:
int(exp(-exp(x + 16)/2)*((4*x)/5 - (exp(x + 16)*(x + 2*x^2))/10 + (exp(exp (x))*(2*x*exp(x) - x*exp(x + 16) + 2))/10 + 1/5),x)
Output:
(x*exp(-(exp(16)*exp(x))/2)*(2*x + exp(exp(x)) + 1))/5
Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx=\frac {x \left (e^{e^{x}}+2 x +1\right )}{5 e^{\frac {e^{x} e^{16}}{2}}} \] Input:
int(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x +2)/exp(1/2*exp(x+16)),x)
Output:
(x*(e**(e**x) + 2*x + 1))/(5*e**((e**x*e**16)/2))