Integrand size = 83, antiderivative size = 29 \[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\frac {e^{-e^{-2 x} \left (e^{-e^x+x}+\frac {x}{5}\right ) x}}{x} \]
Time = 1.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\frac {e^{-\frac {1}{5} e^{-2 x} x \left (5 e^{-e^x+x}+x\right )}}{x} \]
Integrate[(E^(-2*x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x)))*(-5*E^(2*x) - 2 *x^2 + 2*x^3 + E^(-E^x + x)*(-5*x + 5*x^2 + 5*E^x*x^2)))/(5*x^2),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.
\(\displaystyle \int \frac {\left (2 x^3-2 x^2+e^{x-e^x} \left (5 e^x x^2+5 x^2-5 x\right )-5 e^{2 x}\right ) \exp \left (-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )-2 x\right )}{5 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {\exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right ) \left (-2 x^3+2 x^2+5 e^{2 x}+5 e^{x-e^x} \left (-e^x x^2-x^2+x\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {\exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right ) \left (-2 x^3+2 x^2+5 e^{2 x}+5 e^{x-e^x} \left (-e^x x^2-x^2+x\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (-2 \exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right ) (x-1)-\frac {5 \exp \left (-x-e^x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right ) (x-1)}{x}+\frac {5 \exp \left (-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )-e^x\right ) \left (e^{e^x}-x^2\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 \int \exp \left (-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )-e^x\right )dx-2 \int \exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right )dx+5 \int \exp \left (-x-e^x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right )dx-5 \int \frac {\exp \left (-x-e^x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right )}{x}dx+2 \int \exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )\right ) xdx-5 \int \frac {e^{-\frac {1}{5} e^{-2 x} \left (x^2+5 e^{x-e^x} x\right )}}{x^2}dx\right )\) |
Int[(E^(-2*x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x)))*(-5*E^(2*x) - 2*x^2 + 2*x^3 + E^(-E^x + x)*(-5*x + 5*x^2 + 5*E^x*x^2)))/(5*x^2),x]
3.16.100.3.1 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 = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {{\mathrm e}^{-\frac {x \left (5 \,{\mathrm e}^{x -{\mathrm e}^{x}}+x \right ) {\mathrm e}^{-2 x}}{5}}}{x}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {x \left (5 \,{\mathrm e}^{x -{\mathrm e}^{x}}+x \right ) {\mathrm e}^{-2 x}}{5}}}{x}\) | \(26\) |
int(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x^2)/x^ 2/exp(x)^2/exp(1/5*(5*x*exp(x-exp(x))+x^2)/exp(x)^2),x,method=_RETURNVERBO SE)
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\frac {e^{\left (-\frac {1}{5} \, {\left (x^{2} + 10 \, x e^{\left (2 \, x\right )} + 5 \, x e^{\left (x - e^{x}\right )}\right )} e^{\left (-2 \, x\right )} + 2 \, x\right )}}{x} \]
integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x ^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-exp(x))+x^2)/exp(x)^2),x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\frac {e^{- \left (\frac {x^{2}}{5} + x e^{x - e^{x}}\right ) e^{- 2 x}}}{x} \]
integrate(1/5*((5*exp(x)*x**2+5*x**2-5*x)*exp(x-exp(x))-5*exp(x)**2+2*x**3 -2*x**2)/x**2/exp(x)**2/exp(1/5*(5*x*exp(x-exp(x))+x**2)/exp(x)**2),x)
\[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\int { \frac {{\left (2 \, x^{3} - 2 \, x^{2} + 5 \, {\left (x^{2} e^{x} + x^{2} - x\right )} e^{\left (x - e^{x}\right )} - 5 \, e^{\left (2 \, x\right )}\right )} e^{\left (-\frac {1}{5} \, {\left (x^{2} + 5 \, x e^{\left (x - e^{x}\right )}\right )} e^{\left (-2 \, x\right )} - 2 \, x\right )}}{5 \, x^{2}} \,d x } \]
integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x ^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-exp(x))+x^2)/exp(x)^2),x, algorithm=\
1/5*integrate((2*x^3 - 2*x^2 + 5*(x^2*e^x + x^2 - x)*e^(x - e^x) - 5*e^(2* x))*e^(-1/5*(x^2 + 5*x*e^(x - e^x))*e^(-2*x) - 2*x)/x^2, x)
\[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\int { \frac {{\left (2 \, x^{3} - 2 \, x^{2} + 5 \, {\left (x^{2} e^{x} + x^{2} - x\right )} e^{\left (x - e^{x}\right )} - 5 \, e^{\left (2 \, x\right )}\right )} e^{\left (-\frac {1}{5} \, {\left (x^{2} + 5 \, x e^{\left (x - e^{x}\right )}\right )} e^{\left (-2 \, x\right )} - 2 \, x\right )}}{5 \, x^{2}} \,d x } \]
integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x ^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-exp(x))+x^2)/exp(x)^2),x, algorithm=\
integrate(1/5*(2*x^3 - 2*x^2 + 5*(x^2*e^x + x^2 - x)*e^(x - e^x) - 5*e^(2* x))*e^(-1/5*(x^2 + 5*x*e^(x - e^x))*e^(-2*x) - 2*x)/x^2, x)
Time = 10.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )} \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx=\frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{-2\,x}}{5}}}{x} \]