Integrand size = 70, antiderivative size = 26 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2 \] Output:
4*x^2/exp(x+exp(x)-4/5*x/exp(x))^2
Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2 \] Input:
Integrate[(E^(-x - (2*(5*E^(2*x) - 4*x + 5*E^x*x))/(5*E^x))*(32*x^2 - 40*E ^(2*x)*x^2 - 32*x^3 + E^x*(40*x - 40*x^2)))/5,x]
Output:
4*E^(-2*E^x - 2*x + (8*x)/(5*E^x))*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.
\(\displaystyle \int \frac {1}{5} \left (-32 x^3-40 e^{2 x} x^2+32 x^2+e^x \left (40 x-40 x^2\right )\right ) \exp \left (-x-\frac {2}{5} e^{-x} \left (5 e^x x-4 x+5 e^{2 x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int 8 \exp \left (-x-\frac {2}{5} e^{-x} \left (5 e^x x-4 x+5 e^{2 x}\right )\right ) \left (-4 x^3-5 e^{2 x} x^2+4 x^2+5 e^x \left (x-x^2\right )\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8}{5} \int \exp \left (-x-\frac {2}{5} e^{-x} \left (5 e^x x-4 x+5 e^{2 x}\right )\right ) \left (-4 x^3-5 e^{2 x} x^2+4 x^2+5 e^x \left (x-x^2\right )\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {8}{5} \int e^{-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} \left (-4 x^3-5 e^{2 x} x^2+4 x^2+5 e^x \left (x-x^2\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {8}{5} \int \left (-4 e^{-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} x^3+4 e^{-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} x^2-5 \exp \left (2 x-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )\right ) x^2-5 e^{x-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} (x-1) x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{5} \left (-4 \int e^{-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} x^3dx-5 \int e^{-\frac {1}{5} e^{-x} \left (5 e^x x-8 x+10 e^{2 x}\right )} x^2dx-5 \int e^{-\frac {2}{5} e^{-x} \left (5 e^x x-4 x+5 e^{2 x}\right )} x^2dx+4 \int e^{-\frac {1}{5} e^{-x} \left (15 e^x x-8 x+10 e^{2 x}\right )} x^2dx+5 \int e^{-\frac {2}{5} e^{-x} \left (5 e^x x-4 x+5 e^{2 x}\right )} xdx\right )\) |
Input:
Int[(E^(-x - (2*(5*E^(2*x) - 4*x + 5*E^x*x))/(5*E^x))*(32*x^2 - 40*E^(2*x) *x^2 - 32*x^3 + E^x*(40*x - 40*x^2)))/5,x]
Output:
$Aborted
Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(4 x^{2} {\mathrm e}^{-\frac {2 \left (5 \,{\mathrm e}^{2 x}+5 \,{\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{-x}}{5}}\) | \(28\) |
parallelrisch | \(4 x^{2} {\mathrm e}^{-\frac {2 \left (5 \,{\mathrm e}^{2 x}+5 \,{\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{-x}}{5}}\) | \(30\) |
Input:
int(1/5*(-40*exp(x)^2*x^2+(-40*x^2+40*x)*exp(x)-32*x^3+32*x^2)/exp(x)/exp( 1/5*(5*exp(x)^2+5*exp(x)*x-4*x)/exp(x))^2,x,method=_RETURNVERBOSE)
Output:
4*x^2*exp(-2/5*(5*exp(2*x)+5*exp(x)*x-4*x)*exp(-x))
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 \, x^{2} e^{\left (-\frac {1}{5} \, {\left (15 \, x e^{x} - 8 \, x + 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + x\right )} \] Input:
integrate(1/5*(-40*exp(x)^2*x^2+(-40*x^2+40*x)*exp(x)-32*x^3+32*x^2)/exp(x )/exp(1/5*(5*exp(x)^2+5*exp(x)*x-4*x)/exp(x))^2,x, algorithm="fricas")
Output:
4*x^2*e^(-1/5*(15*x*e^x - 8*x + 10*e^(2*x))*e^(-x) + x)
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 x^{2} e^{- 2 \left (x e^{x} - \frac {4 x}{5} + e^{2 x}\right ) e^{- x}} \] Input:
integrate(1/5*(-40*exp(x)**2*x**2+(-40*x**2+40*x)*exp(x)-32*x**3+32*x**2)/ exp(x)/exp(1/5*(5*exp(x)**2+5*exp(x)*x-4*x)/exp(x))**2,x)
Output:
4*x**2*exp(-2*(x*exp(x) - 4*x/5 + exp(2*x))*exp(-x))
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 \, x^{2} e^{\left (\frac {8}{5} \, x e^{\left (-x\right )} - 2 \, x - 2 \, e^{x}\right )} \] Input:
integrate(1/5*(-40*exp(x)^2*x^2+(-40*x^2+40*x)*exp(x)-32*x^3+32*x^2)/exp(x )/exp(1/5*(5*exp(x)^2+5*exp(x)*x-4*x)/exp(x))^2,x, algorithm="maxima")
Output:
4*x^2*e^(8/5*x*e^(-x) - 2*x - 2*e^x)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4 \, x^{2} e^{\left (-\frac {1}{5} \, {\left (5 \, x e^{x} - 8 \, x + 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} - x\right )} \] Input:
integrate(1/5*(-40*exp(x)^2*x^2+(-40*x^2+40*x)*exp(x)-32*x^3+32*x^2)/exp(x )/exp(1/5*(5*exp(x)^2+5*exp(x)*x-4*x)/exp(x))^2,x, algorithm="giac")
Output:
4*x^2*e^(-1/5*(5*x*e^x - 8*x + 10*e^(2*x))*e^(-x) - x)
Time = 3.97 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=4\,x^2\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {8\,x\,{\mathrm {e}}^{-x}}{5}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x} \] Input:
int(-exp(-2*exp(-x)*(exp(2*x) - (4*x)/5 + x*exp(x)))*exp(-x)*(8*x^2*exp(2* x) - (exp(x)*(40*x - 40*x^2))/5 - (32*x^2)/5 + (32*x^3)/5),x)
Output:
4*x^2*exp(-2*x)*exp((8*x*exp(-x))/5)*exp(-2*exp(x))
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {1}{5} e^{-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )} \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx=\frac {4 e^{\frac {8 x}{5 e^{x}}} x^{2}}{e^{2 e^{x}+2 x}} \] Input:
int(1/5*(-40*exp(x)^2*x^2+(-40*x^2+40*x)*exp(x)-32*x^3+32*x^2)/exp(x)/exp( 1/5*(5*exp(x)^2+5*exp(x)*x-4*x)/exp(x))^2,x)
Output:
(4*e**((8*x)/(5*e**x))*x**2)/e**(2*e**x + 2*x)