Integrand size = 117, antiderivative size = 34 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=e^{5+\frac {4 e^{2 e^{-x} x} x^2}{5 x+\frac {x}{-1+x}}} x \]
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=e^{5+\frac {4 e^{2 e^{-x} x} (-1+x) x}{-4+5 x}} x \]
Integrate[(E^(-x + (-20 + 25*x + E^((2*x)/E^x)*(-4*x + 4*x^2))/(-4 + 5*x)) *(E^x*(16 - 40*x + 25*x^2) + E^((2*x)/E^x)*(32*x^2 - 104*x^3 + 112*x^4 - 4 0*x^5 + E^x*(16*x - 32*x^2 + 20*x^3))))/(16 - 40*x + 25*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.
\(\displaystyle \int \frac {\left (e^x \left (25 x^2-40 x+16\right )+e^{2 e^{-x} x} \left (-40 x^5+112 x^4-104 x^3+32 x^2+e^x \left (20 x^3-32 x^2+16 x\right )\right )\right ) \exp \left (\frac {e^{2 e^{-x} x} \left (4 x^2-4 x\right )+25 x-20}{5 x-4}-x\right )}{25 x^2-40 x+16} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 100 \int \frac {\exp \left (\frac {-25 x+4 e^{2 e^{-x} x} \left (x-x^2\right )+20}{4-5 x}-x\right ) \left (e^x \left (25 x^2-40 x+16\right )+4 e^{2 e^{-x} x} \left (-10 x^5+28 x^4-26 x^3+8 x^2+e^x \left (5 x^3-8 x^2+4 x\right )\right )\right )}{100 (4-5 x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (e^x \left (25 x^2-40 x+16\right )+4 e^{2 e^{-x} x} \left (-10 x^5+28 x^4-26 x^3+8 x^2+e^x \left (5 x^3-8 x^2+4 x\right )\right )\right ) \exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}-x\right )}{(4-5 x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}+2 e^{-x} x-x\right )}{(5 x-4)^2}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} x^2-4 e^{2 e^{-x} x} x+25 x-20}{5 x-4}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {5 x^2}{4-5 x}+2 e^{-x} x-\frac {4 e^{2 e^{-x} x} (x-1) x}{4-5 x}-\frac {29 x}{4-5 x}+\frac {20}{4-5 x}\right )}{(5 x-4)^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} x^2-4 e^{2 e^{-x} x} x+25 x-20}{5 x-4}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {5 x^2}{4-5 x}+2 e^{-x} x-\frac {4 e^{2 e^{-x} x} (x-1) x}{4-5 x}-\frac {29 x}{4-5 x}+\frac {20}{4-5 x}\right )}{(5 x-4)^2}\right )dx\) |
Int[(E^(-x + (-20 + 25*x + E^((2*x)/E^x)*(-4*x + 4*x^2))/(-4 + 5*x))*(E^x* (16 - 40*x + 25*x^2) + E^((2*x)/E^x)*(32*x^2 - 104*x^3 + 112*x^4 - 40*x^5 + E^x*(16*x - 32*x^2 + 20*x^3))))/(16 - 40*x + 25*x^2),x]
3.30.66.3.1 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]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 10.55 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(x \,{\mathrm e}^{\frac {\left (4 x^{2}-4 x \right ) {\mathrm e}^{2 x \,{\mathrm e}^{-x}}+25 x -20}{5 x -4}}\) | \(36\) |
risch | \(x \,{\mathrm e}^{\frac {4 \,{\mathrm e}^{2 x \,{\mathrm e}^{-x}} x^{2}-4 \,{\mathrm e}^{2 x \,{\mathrm e}^{-x}} x +25 x -20}{5 x -4}}\) | \(41\) |
int((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp(x/exp (x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25*x-20)/ (5*x-4))/(25*x^2-40*x+16)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\left (x - \frac {5 \, x^{2} - 4 \, {\left (x^{2} - x\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} - 29 \, x + 20}{5 \, x - 4}\right )} \]
integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp (x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm=\
Time = 100.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\frac {25 x + \left (4 x^{2} - 4 x\right ) e^{2 x e^{- x}} - 20}{5 x - 4}} \]
integrate((((20*x**3-32*x**2+16*x)*exp(x)-40*x**5+112*x**4-104*x**3+32*x** 2)*exp(x/exp(x))**2+(25*x**2-40*x+16)*exp(x))*exp(((4*x**2-4*x)*exp(x/exp( x))**2+25*x-20)/(5*x-4))/(25*x**2-40*x+16)/exp(x),x)
Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\left (\frac {4}{5} \, x e^{\left (2 \, x e^{\left (-x\right )}\right )} - \frac {16 \, e^{\left (2 \, x e^{\left (-x\right )}\right )}}{25 \, {\left (5 \, x - 4\right )}} - \frac {4}{25} \, e^{\left (2 \, x e^{\left (-x\right )}\right )} + 5\right )} \]
integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp (x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm=\
\[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=\int { -\frac {{\left (4 \, {\left (10 \, x^{5} - 28 \, x^{4} + 26 \, x^{3} - 8 \, x^{2} - {\left (5 \, x^{3} - 8 \, x^{2} + 4 \, x\right )} e^{x}\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} - {\left (25 \, x^{2} - 40 \, x + 16\right )} e^{x}\right )} e^{\left (-x + \frac {4 \, {\left (x^{2} - x\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} + 25 \, x - 20}{5 \, x - 4}\right )}}{25 \, x^{2} - 40 \, x + 16} \,d x } \]
integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp (x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm=\
integrate(-(4*(10*x^5 - 28*x^4 + 26*x^3 - 8*x^2 - (5*x^3 - 8*x^2 + 4*x)*e^ x)*e^(2*x*e^(-x)) - (25*x^2 - 40*x + 16)*e^x)*e^(-x + (4*(x^2 - x)*e^(2*x* e^(-x)) + 25*x - 20)/(5*x - 4))/(25*x^2 - 40*x + 16), x)
Time = 9.77 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x\,{\mathrm {e}}^{-\frac {4\,x\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{-x}}}{5\,x-4}}\,{\mathrm {e}}^{-\frac {20}{5\,x-4}}\,{\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{-x}}}{5\,x-4}}\,{\mathrm {e}}^{\frac {25\,x}{5\,x-4}} \]
int((exp(-x)*exp(-(exp(2*x*exp(-x))*(4*x - 4*x^2) - 25*x + 20)/(5*x - 4))* (exp(x)*(25*x^2 - 40*x + 16) + exp(2*x*exp(-x))*(32*x^2 - 104*x^3 + 112*x^ 4 - 40*x^5 + exp(x)*(16*x - 32*x^2 + 20*x^3))))/(25*x^2 - 40*x + 16),x)