Integrand size = 131, antiderivative size = 30 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=\frac {5}{2 \left (4-e^{2 \left (x+2 \left (e^{1-x}+x\right )\right )^2}+x\right )} \]
Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=-\frac {5}{2 \left (-4+e^{2 e^{-2 x} \left (2 e+3 e^x x\right )^2}-x\right )} \]
Integrate[(-5 + E^(8*E^(2 - 2*x) + 24*E^(1 - x)*x + 18*x^2)*(-80*E^(2 - 2* x) + E^(1 - x)*(120 - 120*x) + 180*x))/(32 + 2*E^(16*E^(2 - 2*x) + 48*E^(1 - x)*x + 36*x^2) + E^(8*E^(2 - 2*x) + 24*E^(1 - x)*x + 18*x^2)*(-16 - 4*x ) + 16*x + 2*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.
\(\displaystyle \int \frac {e^{18 x^2+24 e^{1-x} x+8 e^{2-2 x}} \left (e^{1-x} (120-120 x)-80 e^{2-2 x}+180 x\right )-5}{2 x^2+2 e^{36 x^2+48 e^{1-x} x+16 e^{2-2 x}}+e^{18 x^2+24 e^{1-x} x+8 e^{2-2 x}} (-4 x-16)+16 x+32} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{18 x^2+24 e^{1-x} x+8 e^{2-2 x}} \left (e^{1-x} (120-120 x)-80 e^{2-2 x}+180 x\right )-5}{2 \left (x-e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {5 \left (4 e^{18 x^2+24 e^{1-x} x+8 e^{2-2 x}} \left (-6 e^{1-x} (1-x)+4 e^{2-2 x}-9 x\right )+1\right )}{\left (x-e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{2} \int \frac {4 e^{18 x^2+24 e^{1-x} x+8 e^{2-2 x}} \left (-6 e^{1-x} (1-x)+4 e^{2-2 x}-9 x\right )+1}{\left (x-e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{2} \int \left (-\frac {36 e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2} x}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}+\frac {24 e^{18 x^2+24 e^{1-x} x-x+8 e^{2-2 x}+1} x}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}+\frac {16 e^{18 x^2+24 e^{1-x} x-2 x+8 e^{2-2 x}+2}}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}-\frac {24 e^{18 x^2+24 e^{1-x} x-x+8 e^{2-2 x}+1}}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}+\frac {1}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{2} \left (16 \int \frac {e^{18 x^2+24 e^{1-x} x-2 x+8 e^{2-2 x}+2}}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}dx-24 \int \frac {e^{18 x^2+24 e^{1-x} x-x+8 e^{2-2 x}+1}}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}dx+24 \int \frac {e^{18 x^2+24 e^{1-x} x-x+8 e^{2-2 x}+1} x}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}dx+\int \frac {1}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}dx-36 \int \frac {e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2} x}{\left (-x+e^{2 e^{-2 x} \left (3 e^x x+2 e\right )^2}-4\right )^2}dx\right )\) |
Int[(-5 + E^(8*E^(2 - 2*x) + 24*E^(1 - x)*x + 18*x^2)*(-80*E^(2 - 2*x) + E ^(1 - x)*(120 - 120*x) + 180*x))/(32 + 2*E^(16*E^(2 - 2*x) + 48*E^(1 - x)* x + 36*x^2) + E^(8*E^(2 - 2*x) + 24*E^(1 - x)*x + 18*x^2)*(-16 - 4*x) + 16 *x + 2*x^2),x]
3.14.13.3.1 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 = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {5}{2 \left (x -{\mathrm e}^{8 \,{\mathrm e}^{2-2 x}+24 x \,{\mathrm e}^{1-x}+18 x^{2}}+4\right )}\) | \(34\) |
norman | \(\frac {5}{2 \left (x -{\mathrm e}^{8 \,{\mathrm e}^{2-2 x}+24 x \,{\mathrm e}^{1-x}+18 x^{2}}+4\right )}\) | \(36\) |
parallelrisch | \(\frac {5}{2 \left (x -{\mathrm e}^{8 \,{\mathrm e}^{2-2 x}+24 x \,{\mathrm e}^{1-x}+18 x^{2}}+4\right )}\) | \(36\) |
int(((-80*exp(1-x)^2+(-120*x+120)*exp(1-x)+180*x)*exp(8*exp(1-x)^2+24*x*ex p(1-x)+18*x^2)-5)/(2*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)^2+(-16-4*x)*ex p(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)+2*x^2+16*x+32),x,method=_RETURNVERBOS E)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=\frac {5}{2 \, {\left (x - e^{\left (18 \, x^{2} + 24 \, x e^{\left (-x + 1\right )} + 8 \, e^{\left (-2 \, x + 2\right )}\right )} + 4\right )}} \]
integrate(((-80*exp(1-x)^2+(-120*x+120)*exp(1-x)+180*x)*exp(8*exp(1-x)^2+2 4*x*exp(1-x)+18*x^2)-5)/(2*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)^2+(-16-4 *x)*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)+2*x^2+16*x+32),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=- \frac {5}{- 2 x + 2 e^{18 x^{2} + 24 x e^{1 - x} + 8 e^{2 - 2 x}} - 8} \]
integrate(((-80*exp(1-x)**2+(-120*x+120)*exp(1-x)+180*x)*exp(8*exp(1-x)**2 +24*x*exp(1-x)+18*x**2)-5)/(2*exp(8*exp(1-x)**2+24*x*exp(1-x)+18*x**2)**2+ (-16-4*x)*exp(8*exp(1-x)**2+24*x*exp(1-x)+18*x**2)+2*x**2+16*x+32),x)
Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=\frac {5}{2 \, {\left (x - e^{\left (18 \, x^{2} + 24 \, x e^{\left (-x + 1\right )} + 8 \, e^{\left (-2 \, x + 2\right )}\right )} + 4\right )}} \]
integrate(((-80*exp(1-x)^2+(-120*x+120)*exp(1-x)+180*x)*exp(8*exp(1-x)^2+2 4*x*exp(1-x)+18*x^2)-5)/(2*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)^2+(-16-4 *x)*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)+2*x^2+16*x+32),x, algorithm=\
Time = 0.46 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=\frac {5}{2 \, {\left (x - e^{\left (18 \, x^{2} + 24 \, x e^{\left (-x + 1\right )} + 8 \, e^{\left (-2 \, x + 2\right )}\right )} + 4\right )}} \]
integrate(((-80*exp(1-x)^2+(-120*x+120)*exp(1-x)+180*x)*exp(8*exp(1-x)^2+2 4*x*exp(1-x)+18*x^2)-5)/(2*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)^2+(-16-4 *x)*exp(8*exp(1-x)^2+24*x*exp(1-x)+18*x^2)+2*x^2+16*x+32),x, algorithm=\
Time = 10.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-5+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} \left (-80 e^{2-2 x}+e^{1-x} (120-120 x)+180 x\right )}{32+2 e^{16 e^{2-2 x}+48 e^{1-x} x+36 x^2}+e^{8 e^{2-2 x}+24 e^{1-x} x+18 x^2} (-16-4 x)+16 x+2 x^2} \, dx=\frac {5}{2\,\left (x-{\mathrm {e}}^{8\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2}\,{\mathrm {e}}^{18\,x^2}\,{\mathrm {e}}^{24\,x\,{\mathrm {e}}^{-x}\,\mathrm {e}}+4\right )} \]