Integrand size = 109, antiderivative size = 32 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=4 e^{2+e^{-e^{2 x}+x}+x} \left (5-\frac {3}{x (1+4 x)}\right ) \]
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=-\frac {4 e^{2+e^{-e^{2 x}+x}+x} \left (3-5 x-20 x^2\right )}{x (1+4 x)} \]
Integrate[(E^(2 + E^(-E^(2*x) + x) + x)*(12 + 84*x - 28*x^2 + 160*x^3 + 32 0*x^4 + E^(-E^(2*x) + x)*(-12*x - 28*x^2 + 160*x^3 + 320*x^4 + E^(2*x)*(24 *x + 56*x^2 - 320*x^3 - 640*x^4))))/(x^2 + 8*x^3 + 16*x^4),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.
\(\displaystyle \int \frac {e^{x+e^{x-e^{2 x}}+2} \left (320 x^4+160 x^3-28 x^2+e^{x-e^{2 x}} \left (320 x^4+160 x^3-28 x^2+e^{2 x} \left (-640 x^4-320 x^3+56 x^2+24 x\right )-12 x\right )+84 x+12\right )}{16 x^4+8 x^3+x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{x+e^{x-e^{2 x}}+2} \left (320 x^4+160 x^3-28 x^2+e^{x-e^{2 x}} \left (320 x^4+160 x^3-28 x^2+e^{2 x} \left (-640 x^4-320 x^3+56 x^2+24 x\right )-12 x\right )+84 x+12\right )}{x^2 \left (16 x^2+8 x+1\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{x+e^{x-e^{2 x}}+2} \left (320 x^4+160 x^3-28 x^2+e^{x-e^{2 x}} \left (320 x^4+160 x^3-28 x^2+e^{2 x} \left (-640 x^4-320 x^3+56 x^2+24 x\right )-12 x\right )+84 x+12\right )}{x^2 (4 x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {320 e^{x+e^{x-e^{2 x}}+2} x^2}{(4 x+1)^2}+\frac {4 e^{2 x-e^{2 x}+e^{x-e^{2 x}}+2} \left (20 x^2+5 x-3\right )}{(4 x+1) x}-\frac {8 e^{4 x-e^{2 x}+e^{x-e^{2 x}}+2} \left (20 x^2+5 x-3\right )}{(4 x+1) x}+\frac {12 e^{x+e^{x-e^{2 x}}+2}}{(4 x+1)^2 x^2}+\frac {160 e^{x+e^{x-e^{2 x}}+2} x}{(4 x+1)^2}-\frac {28 e^{x+e^{x-e^{2 x}}+2}}{(4 x+1)^2}+\frac {84 e^{x+e^{x-e^{2 x}}+2}}{(4 x+1)^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 20 \text {Subst}\left (\int e^{e^{-x^2} x+2}dx,x,e^x\right )+20 \text {Subst}\left (\int e^{-x^2+e^{-x^2} x+2} xdx,x,e^x\right )-40 \text {Subst}\left (\int e^{-x^2+e^{-x^2} x+2} x^3dx,x,e^x\right )+12 \int \frac {e^{x+e^{x-e^{2 x}}+2}}{x^2}dx-12 \int \frac {e^{x+e^{x-e^{2 x}}+2}}{x}dx-12 \int \frac {e^{2 x-e^{2 x}+e^{x-e^{2 x}}+2}}{x}dx+24 \int \frac {e^{4 x-e^{2 x}+e^{x-e^{2 x}}+2}}{x}dx-192 \int \frac {e^{x+e^{x-e^{2 x}}+2}}{(4 x+1)^2}dx+48 \int \frac {e^{x+e^{x-e^{2 x}}+2}}{4 x+1}dx+48 \int \frac {e^{2 x-e^{2 x}+e^{x-e^{2 x}}+2}}{4 x+1}dx-96 \int \frac {e^{4 x-e^{2 x}+e^{x-e^{2 x}}+2}}{4 x+1}dx\) |
Int[(E^(2 + E^(-E^(2*x) + x) + x)*(12 + 84*x - 28*x^2 + 160*x^3 + 320*x^4 + E^(-E^(2*x) + x)*(-12*x - 28*x^2 + 160*x^3 + 320*x^4 + E^(2*x)*(24*x + 5 6*x^2 - 320*x^3 - 640*x^4))))/(x^2 + 8*x^3 + 16*x^4),x]
3.28.59.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 1.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {4 \left (20 x^{2}+5 x -3\right ) {\mathrm e}^{{\mathrm e}^{x -{\mathrm e}^{2 x}}+2+x}}{x \left (1+4 x \right )}\) | \(36\) |
parallelrisch | \(\frac {320 x^{2} {\mathrm e}^{{\mathrm e}^{x -{\mathrm e}^{2 x}}+2+x}+80 \,{\mathrm e}^{{\mathrm e}^{x -{\mathrm e}^{2 x}}+2+x} x -48 \,{\mathrm e}^{{\mathrm e}^{x -{\mathrm e}^{2 x}}+2+x}}{4 x \left (1+4 x \right )}\) | \(63\) |
int((((-640*x^4-320*x^3+56*x^2+24*x)*exp(x)^2+320*x^4+160*x^3-28*x^2-12*x) *exp(-exp(x)^2+x)+320*x^4+160*x^3-28*x^2+84*x+12)*exp(exp(-exp(x)^2+x)+2+x )/(16*x^4+8*x^3+x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=\frac {4 \, {\left (20 \, x^{2} + 5 \, x - 3\right )} e^{\left (x + e^{\left (x - e^{\left (2 \, x\right )}\right )} + 2\right )}}{4 \, x^{2} + x} \]
integrate((((-640*x^4-320*x^3+56*x^2+24*x)*exp(x)^2+320*x^4+160*x^3-28*x^2 -12*x)*exp(-exp(x)^2+x)+320*x^4+160*x^3-28*x^2+84*x+12)*exp(exp(-exp(x)^2+ x)+2+x)/(16*x^4+8*x^3+x^2),x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=\frac {\left (80 x^{2} + 20 x - 12\right ) e^{x + e^{x - e^{2 x}} + 2}}{4 x^{2} + x} \]
integrate((((-640*x**4-320*x**3+56*x**2+24*x)*exp(x)**2+320*x**4+160*x**3- 28*x**2-12*x)*exp(-exp(x)**2+x)+320*x**4+160*x**3-28*x**2+84*x+12)*exp(exp (-exp(x)**2+x)+2+x)/(16*x**4+8*x**3+x**2),x)
Time = 0.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=\frac {4 \, {\left (20 \, x^{2} e^{2} + 5 \, x e^{2} - 3 \, e^{2}\right )} e^{\left (x + e^{\left (x - e^{\left (2 \, x\right )}\right )}\right )}}{4 \, x^{2} + x} \]
integrate((((-640*x^4-320*x^3+56*x^2+24*x)*exp(x)^2+320*x^4+160*x^3-28*x^2 -12*x)*exp(-exp(x)^2+x)+320*x^4+160*x^3-28*x^2+84*x+12)*exp(exp(-exp(x)^2+ x)+2+x)/(16*x^4+8*x^3+x^2),x, algorithm=\
\[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=\int { \frac {4 \, {\left (80 \, x^{4} + 40 \, x^{3} - 7 \, x^{2} + {\left (80 \, x^{4} + 40 \, x^{3} - 7 \, x^{2} - 2 \, {\left (80 \, x^{4} + 40 \, x^{3} - 7 \, x^{2} - 3 \, x\right )} e^{\left (2 \, x\right )} - 3 \, x\right )} e^{\left (x - e^{\left (2 \, x\right )}\right )} + 21 \, x + 3\right )} e^{\left (x + e^{\left (x - e^{\left (2 \, x\right )}\right )} + 2\right )}}{16 \, x^{4} + 8 \, x^{3} + x^{2}} \,d x } \]
integrate((((-640*x^4-320*x^3+56*x^2+24*x)*exp(x)^2+320*x^4+160*x^3-28*x^2 -12*x)*exp(-exp(x)^2+x)+320*x^4+160*x^3-28*x^2+84*x+12)*exp(exp(-exp(x)^2+ x)+2+x)/(16*x^4+8*x^3+x^2),x, algorithm=\
integrate(4*(80*x^4 + 40*x^3 - 7*x^2 + (80*x^4 + 40*x^3 - 7*x^2 - 2*(80*x^ 4 + 40*x^3 - 7*x^2 - 3*x)*e^(2*x) - 3*x)*e^(x - e^(2*x)) + 21*x + 3)*e^(x + e^(x - e^(2*x)) + 2)/(16*x^4 + 8*x^3 + x^2), x)
Time = 12.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2+e^{-e^{2 x}+x}+x} \left (12+84 x-28 x^2+160 x^3+320 x^4+e^{-e^{2 x}+x} \left (-12 x-28 x^2+160 x^3+320 x^4+e^{2 x} \left (24 x+56 x^2-320 x^3-640 x^4\right )\right )\right )}{x^2+8 x^3+16 x^4} \, dx=20\,{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^x-\frac {3\,{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^x}{x^2+\frac {x}{4}} \]
int((exp(x + exp(x - exp(2*x)) + 2)*(84*x + exp(x - exp(2*x))*(exp(2*x)*(2 4*x + 56*x^2 - 320*x^3 - 640*x^4) - 12*x - 28*x^2 + 160*x^3 + 320*x^4) - 2 8*x^2 + 160*x^3 + 320*x^4 + 12))/(x^2 + 8*x^3 + 16*x^4),x)