Integrand size = 55, antiderivative size = 25 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 e^{10 \left (-x+\frac {5 \left (2 e^2+x\right )}{-4+x}\right )} x \]
Time = 0.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 e^{\frac {100 e^2-10 (-9+x) x}{-4+x}} x \]
Integrate[(E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))*(1600 - 36800*x - 1000 0*E^2*x + 8100*x^2 - 1000*x^3))/(16 - 8*x + 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.
| \(\displaystyle \int \frac {e^{\frac {2 \left (-5 x^2+45 x+50 e^2\right )}{x-4}} \left (-1000 x^3+8100 x^2-10000 e^2 x-36800 x+1600\right )}{x^2-8 x+16} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{\frac {2 \left (-5 x^2+45 x+50 e^2\right )}{x-4}} \left (-1000 x^3+8100 x^2+\left (-36800-10000 e^2\right ) x+1600\right )}{x^2-8 x+16}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int \frac {25 e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} \left (-10 x^3+81 x^2-4 \left (92+25 e^2\right ) x+16\right )}{(4-x)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 100 \int \frac {e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} \left (-10 x^3+81 x^2-4 \left (92+25 e^2\right ) x+16\right )}{(4-x)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 100 \int \left (-10 e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} x+e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}}-\frac {100 e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} \left (2+e^2\right )}{x-4}-\frac {400 e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} \left (2+e^2\right )}{(x-4)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 100 \left (\int e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}}dx-400 \left (2+e^2\right ) \int \frac {e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}}}{(x-4)^2}dx-100 \left (2+e^2\right ) \int \frac {e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}}}{x-4}dx-10 \int e^{-\frac {10 \left (-x^2+9 x+10 e^2\right )}{4-x}} xdx\right )\) |
Int[(E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))*(1600 - 36800*x - 10000*E^2* x + 8100*x^2 - 1000*x^3))/(16 - 8*x + x^2),x]
3.21.59.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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 = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(100 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}\) | \(25\) |
| gosper | \(100 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}\) | \(27\) |
| parallelrisch | \(100 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}\) | \(27\) |
| norman | \(\frac {-400 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}+100 x^{2} {\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}}{x -4}\) | \(60\) |
int((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)-5*x^2+ 45*x)/(x-4))^2/(x^2-8*x+16),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 \, x e^{\left (-\frac {10 \, {\left (x^{2} - 9 \, x - 10 \, e^{2}\right )}}{x - 4}\right )} \]
integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)- 5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 x e^{\frac {2 \left (- 5 x^{2} + 45 x + 50 e^{2}\right )}{x - 4}} \]
integrate((-10000*exp(2)*x-1000*x**3+8100*x**2-36800*x+1600)*exp((50*exp(2 )-5*x**2+45*x)/(x-4))**2/(x**2-8*x+16),x)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 \, x e^{\left (-10 \, x + \frac {100 \, e^{2}}{x - 4} + \frac {200}{x - 4} + 50\right )} \]
integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)- 5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100 \, x e^{\left (-\frac {5 \, {\left (2 \, x^{2} - 5 \, x e^{2} - 18 \, x\right )}}{x - 4} - 25 \, e^{2}\right )} \]
integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)- 5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx=100\,x\,{\mathrm {e}}^{\frac {-10\,x^2+90\,x+100\,{\mathrm {e}}^2}{x-4}} \]