Integrand size = 106, antiderivative size = 28 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=\left (5-5 e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}+2 x\right )^2 \]
Time = 0.62 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=\left (-13+5 e^{\frac {1}{5} e^{20} \left (4+\frac {20}{-4+x}+x\right )}-2 (-4+x)\right )^2 \]
Integrate[(320 - 32*x - 44*x^2 + 8*x^3 + E^(20 + (2*E^20*(4 + x^2))/(-20 + 5*x))*(-40 - 80*x + 10*x^2) + E^((E^20*(4 + x^2))/(-20 + 5*x))*(-320 + 16 0*x - 20*x^2 + E^20*(40 + 96*x + 22*x^2 - 4*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.
\(\displaystyle \int \frac {8 x^3-44 x^2+e^{\frac {2 e^{20} \left (x^2+4\right )}{5 x-20}+20} \left (10 x^2-80 x-40\right )+e^{\frac {e^{20} \left (x^2+4\right )}{5 x-20}} \left (-20 x^2+e^{20} \left (-4 x^3+22 x^2+96 x+40\right )+160 x-320\right )-32 x+320}{x^2-8 x+16} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {4 x^3-22 x^2-16 x-5 e^{20-\frac {2 e^{20} \left (x^2+4\right )}{5 (4-x)}} \left (-x^2+8 x+4\right )-e^{-\frac {e^{20} \left (x^2+4\right )}{5 (4-x)}} \left (10 x^2-80 x-e^{20} \left (-2 x^3+11 x^2+48 x+20\right )+160\right )+160}{2 (4-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {4 x^3-22 x^2-16 x-5 e^{20-\frac {2 e^{20} \left (x^2+4\right )}{5 (4-x)}} \left (-x^2+8 x+4\right )-e^{-\frac {e^{20} \left (x^2+4\right )}{5 (4-x)}} \left (10 x^2-80 x-e^{20} \left (-2 x^3+11 x^2+48 x+20\right )+160\right )+160}{(4-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {4 x^3}{(x-4)^2}-\frac {22 x^2}{(x-4)^2}-\frac {16 x}{(x-4)^2}+\frac {5 e^{\frac {2 e^{20} \left (x^2+4\right )}{5 (x-4)}+20} \left (x^2-8 x-4\right )}{(x-4)^2}+\frac {e^{\frac {e^{20} \left (x^2+4\right )}{5 (x-4)}} \left (-2 e^{20} x^3-\left (10-11 e^{20}\right ) x^2+16 \left (5+3 e^{20}\right ) x-20 \left (8-e^{20}\right )\right )}{(4-x)^2}+\frac {160}{(x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-5 \left (2+e^{20}\right ) \int e^{\frac {e^{20} \left (x^2+4\right )}{5 (x-4)}}dx+5 \int e^{\frac {2 e^{20} \left (x^2+4\right )}{5 (x-4)}+20}dx+260 \int \frac {e^{\frac {e^{20} \left (x^2+4\right )}{5 (x-4)}+20}}{(x-4)^2}dx-100 \int \frac {e^{\frac {2 e^{20} \left (x^2+4\right )}{5 (x-4)}+20}}{(x-4)^2}dx+40 \int \frac {e^{\frac {e^{20} \left (x^2+4\right )}{5 (x-4)}+20}}{x-4}dx-2 \int e^{\frac {e^{20} \left (x^2+4\right )}{5 (x-4)}+20} xdx+2 x^2+10 x\right )\) |
Int[(320 - 32*x - 44*x^2 + 8*x^3 + E^(20 + (2*E^20*(4 + x^2))/(-20 + 5*x)) *(-40 - 80*x + 10*x^2) + E^((E^20*(4 + x^2))/(-20 + 5*x))*(-320 + 160*x - 20*x^2 + E^20*(40 + 96*x + 22*x^2 - 4*x^3)))/(16 - 8*x + x^2),x]
3.12.6.3.1 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 = 1.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
risch | \(4 x^{2}+25 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 \left (x -4\right )}}+20 x +\left (-50-20 x \right ) {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}\) | \(48\) |
parallelrisch | \(4 x^{2}-20 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+25 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 \left (x -4\right )}}+20 x -50 \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-64\) | \(65\) |
norman | \(\frac {4 x^{2}+4 x^{3}-100 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+30 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+25 x \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-20 x^{2} {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+200 \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-320}{x -4}\) | \(118\) |
parts | \(4 x^{2}+20 x +\frac {30 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-20 x^{2} {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+200 \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}}{x -4}+\frac {-100 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+25 x \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}}{x -4}\) | \(123\) |
int(((10*x^2-80*x-40)*exp(20)*exp((x^2+4)*exp(20)/(5*x-20))^2+((-4*x^3+22* x^2+96*x+40)*exp(20)-20*x^2+160*x-320)*exp((x^2+4)*exp(20)/(5*x-20))+8*x^3 -44*x^2-32*x+320)/(x^2-8*x+16),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=4 \, x^{2} - 10 \, {\left (2 \, x + 5\right )} e^{\left (\frac {{\left (x^{2} + 4\right )} e^{20}}{5 \, {\left (x - 4\right )}}\right )} + 20 \, x + 25 \, e^{\left (\frac {2 \, {\left (x^{2} + 4\right )} e^{20}}{5 \, {\left (x - 4\right )}}\right )} \]
integrate(((10*x^2-80*x-40)*exp(20)*exp((x^2+4)*exp(20)/(5*x-20))^2+((-4*x ^3+22*x^2+96*x+40)*exp(20)-20*x^2+160*x-320)*exp((x^2+4)*exp(20)/(5*x-20)) +8*x^3-44*x^2-32*x+320)/(x^2-8*x+16),x, algorithm=\
4*x^2 - 10*(2*x + 5)*e^(1/5*(x^2 + 4)*e^20/(x - 4)) + 20*x + 25*e^(2/5*(x^ 2 + 4)*e^20/(x - 4))
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=4 x^{2} + 20 x + \left (- 20 x - 50\right ) e^{\frac {\left (x^{2} + 4\right ) e^{20}}{5 x - 20}} + 25 e^{\frac {2 \left (x^{2} + 4\right ) e^{20}}{5 x - 20}} \]
integrate(((10*x**2-80*x-40)*exp(20)*exp((x**2+4)*exp(20)/(5*x-20))**2+((- 4*x**3+22*x**2+96*x+40)*exp(20)-20*x**2+160*x-320)*exp((x**2+4)*exp(20)/(5 *x-20))+8*x**3-44*x**2-32*x+320)/(x**2-8*x+16),x)
4*x**2 + 20*x + (-20*x - 50)*exp((x**2 + 4)*exp(20)/(5*x - 20)) + 25*exp(2 *(x**2 + 4)*exp(20)/(5*x - 20))
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.32 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=4 \, x^{2} - 10 \, {\left (2 \, x e^{\left (\frac {4}{5} \, e^{20}\right )} + 5 \, e^{\left (\frac {4}{5} \, e^{20}\right )}\right )} e^{\left (\frac {1}{5} \, x e^{20} + \frac {4 \, e^{20}}{x - 4}\right )} + 20 \, x + 25 \, e^{\left (\frac {2}{5} \, x e^{20} + \frac {8 \, e^{20}}{x - 4} + \frac {8}{5} \, e^{20}\right )} \]
integrate(((10*x^2-80*x-40)*exp(20)*exp((x^2+4)*exp(20)/(5*x-20))^2+((-4*x ^3+22*x^2+96*x+40)*exp(20)-20*x^2+160*x-320)*exp((x^2+4)*exp(20)/(5*x-20)) +8*x^3-44*x^2-32*x+320)/(x^2-8*x+16),x, algorithm=\
4*x^2 - 10*(2*x*e^(4/5*e^20) + 5*e^(4/5*e^20))*e^(1/5*x*e^20 + 4*e^20/(x - 4)) + 20*x + 25*e^(2/5*x*e^20 + 8*e^20/(x - 4) + 8/5*e^20)
Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (24) = 48\).
Time = 0.38 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.50 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx={\left (4 \, x^{2} e^{20} + 20 \, x e^{20} - 20 \, x e^{\left (\frac {x^{2} e^{20} + x e^{20}}{5 \, {\left (x - 4\right )}} - \frac {1}{5} \, e^{20} + 20\right )} + 25 \, e^{\left (\frac {2 \, {\left (x^{2} e^{20} + x e^{20}\right )}}{5 \, {\left (x - 4\right )}} - \frac {2}{5} \, e^{20} + 20\right )} - 50 \, e^{\left (\frac {x^{2} e^{20} + x e^{20}}{5 \, {\left (x - 4\right )}} - \frac {1}{5} \, e^{20} + 20\right )}\right )} e^{\left (-20\right )} \]
integrate(((10*x^2-80*x-40)*exp(20)*exp((x^2+4)*exp(20)/(5*x-20))^2+((-4*x ^3+22*x^2+96*x+40)*exp(20)-20*x^2+160*x-320)*exp((x^2+4)*exp(20)/(5*x-20)) +8*x^3-44*x^2-32*x+320)/(x^2-8*x+16),x, algorithm=\
(4*x^2*e^20 + 20*x*e^20 - 20*x*e^(1/5*(x^2*e^20 + x*e^20)/(x - 4) - 1/5*e^ 20 + 20) + 25*e^(2/5*(x^2*e^20 + x*e^20)/(x - 4) - 2/5*e^20 + 20) - 50*e^( 1/5*(x^2*e^20 + x*e^20)/(x - 4) - 1/5*e^20 + 20))*e^(-20)
Time = 0.40 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.43 \[ \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx=25\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {2\,x^2\,{\mathrm {e}}^{20}}{5\,x-20}}-50\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {x^2\,{\mathrm {e}}^{20}}{5\,x-20}}+4\,x^2-x\,\left (20\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {x^2\,{\mathrm {e}}^{20}}{5\,x-20}}-20\right ) \]
int(-(32*x - exp((exp(20)*(x^2 + 4))/(5*x - 20))*(160*x + exp(20)*(96*x + 22*x^2 - 4*x^3 + 40) - 20*x^2 - 320) + 44*x^2 - 8*x^3 + exp(20)*exp((2*exp (20)*(x^2 + 4))/(5*x - 20))*(80*x - 10*x^2 + 40) - 320)/(x^2 - 8*x + 16),x )