Integrand size = 82, antiderivative size = 28 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx=-e^4+e^x+e^{\frac {e^{3-x^2}}{x+x^4}} \end {dmath*}
Time = 2.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx=e^x+e^{\frac {e^{3-x^2}}{x \left (1+x^3\right )}} \end {dmath*}
Integrate[(E^(3 + E^(3 - x^2)/(x + x^4))*(-1 - 2*x^2 - 4*x^3 - 2*x^5) + E^ (x + x^2)*(x^2 + 2*x^5 + x^8))/(E^x^2*(x^2 + 2*x^5 + x^8)),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.
\(\displaystyle \int \frac {e^{-x^2} \left (e^{x^2+x} \left (x^8+2 x^5+x^2\right )+e^{\frac {e^{3-x^2}}{x^4+x}+3} \left (-2 x^5-4 x^3-2 x^2-1\right )\right )}{x^8+2 x^5+x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x^2} \left (e^{x^2+x} \left (x^8+2 x^5+x^2\right )+e^{\frac {e^{3-x^2}}{x^4+x}+3} \left (-2 x^5-4 x^3-2 x^2-1\right )\right )}{x^2 \left (x^6+2 x^3+1\right )}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {e^{-x^2} \left (e^{\frac {e^{3-x^2}}{x^4+x}+3} \left (2 x^5+4 x^3+2 x^2+1\right )-e^{x^2+x} \left (x^8+2 x^5+x^2\right )\right )}{x^2 \left (x^3+1\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x^4+x}} \left (2 x^5+4 x^3+2 x^2+1\right )-e^{x^2+x} \left (x^8+2 x^5+x^2\right )\right )}{x^2 \left (x^3+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3} \left (2 x^5+4 x^3+2 x^2+1\right )}{x^2 \left (x^3+1\right )^2}-e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{3} \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{\left (-2 x+i \sqrt {3}+1\right )^2}dx-\frac {4 i \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{-2 x+i \sqrt {3}+1}dx}{3 \sqrt {3}}-\int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{x^2}dx+\frac {1}{3} \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{(x+1)^2}dx-\frac {2}{3} \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{x+1}dx+\frac {2}{9} \left (3+i \sqrt {3}\right ) \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{2 x-i \sqrt {3}-1}dx+\frac {4}{3} \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{\left (2 x+i \sqrt {3}-1\right )^2}dx+\frac {2}{9} \left (3-i \sqrt {3}\right ) \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{2 x+i \sqrt {3}-1}dx-\frac {4 i \int \frac {e^{-x^2+\frac {e^{3-x^2}}{x^4+x}+3}}{2 x+i \sqrt {3}-1}dx}{3 \sqrt {3}}+e^x\) |
Int[(E^(3 + E^(3 - x^2)/(x + x^4))*(-1 - 2*x^2 - 4*x^3 - 2*x^5) + E^(x + x ^2)*(x^2 + 2*x^5 + x^8))/(E^x^2*(x^2 + 2*x^5 + x^8)),x]
3.10.33.3.1 Defintions of rubi rules used
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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 = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14
\[{\mathrm e}^{x}+{\mathrm e}^{\frac {{\mathrm e}^{-x^{2}+3}}{x \left (1+x \right ) \left (x^{2}-x +1\right )}}\]
int(((-2*x^5-4*x^3-2*x^2-1)*exp(3)*exp(exp(3)/(x^4+x)/exp(x^2))+(x^8+2*x^5 +x^2)*exp(x)*exp(x^2))/(x^8+2*x^5+x^2)/exp(x^2),x)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx={\left (e^{\left (x + 3\right )} + e^{\left (\frac {3 \, x^{4} + 3 \, x + e^{\left (-x^{2} + 3\right )}}{x^{4} + x}\right )}\right )} e^{\left (-3\right )} \end {dmath*}
integrate(((-2*x^5-4*x^3-2*x^2-1)*exp(3)*exp(exp(3)/(x^4+x)/exp(x^2))+(x^8 +2*x^5+x^2)*exp(x)*exp(x^2))/(x^8+2*x^5+x^2)/exp(x^2),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx=e^{x} + e^{\frac {e^{3} e^{- x^{2}}}{x^{4} + x}} \end {dmath*}
integrate(((-2*x**5-4*x**3-2*x**2-1)*exp(3)*exp(exp(3)/(x**4+x)/exp(x**2)) +(x**8+2*x**5+x**2)*exp(x)*exp(x**2))/(x**8+2*x**5+x**2)/exp(x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (24) = 48\).
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.25 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx={\left (e^{\left (x + \frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x + 1\right )}}\right )} + e^{\left (-\frac {2 \, x e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x^{2} - x + 1\right )}} + \frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x^{2} - x + 1\right )}} + \frac {e^{\left (-x^{2} + 3\right )}}{x}\right )}\right )} e^{\left (-\frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x + 1\right )}}\right )} \end {dmath*}
integrate(((-2*x^5-4*x^3-2*x^2-1)*exp(3)*exp(exp(3)/(x^4+x)/exp(x^2))+(x^8 +2*x^5+x^2)*exp(x)*exp(x^2))/(x^8+2*x^5+x^2)/exp(x^2),x, algorithm=\
(e^(x + 1/3*e^(-x^2 + 3)/(x + 1)) + e^(-2/3*x*e^(-x^2 + 3)/(x^2 - x + 1) + 1/3*e^(-x^2 + 3)/(x^2 - x + 1) + e^(-x^2 + 3)/x))*e^(-1/3*e^(-x^2 + 3)/(x + 1))
\begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx=\int { \frac {{\left ({\left (x^{8} + 2 \, x^{5} + x^{2}\right )} e^{\left (x^{2} + x\right )} - {\left (2 \, x^{5} + 4 \, x^{3} + 2 \, x^{2} + 1\right )} e^{\left (\frac {e^{\left (-x^{2} + 3\right )}}{x^{4} + x} + 3\right )}\right )} e^{\left (-x^{2}\right )}}{x^{8} + 2 \, x^{5} + x^{2}} \,d x } \end {dmath*}
integrate(((-2*x^5-4*x^3-2*x^2-1)*exp(3)*exp(exp(3)/(x^4+x)/exp(x^2))+(x^8 +2*x^5+x^2)*exp(x)*exp(x^2))/(x^8+2*x^5+x^2)/exp(x^2),x, algorithm=\
integrate(((x^8 + 2*x^5 + x^2)*e^(x^2 + x) - (2*x^5 + 4*x^3 + 2*x^2 + 1)*e ^(e^(-x^2 + 3)/(x^4 + x) + 3))*e^(-x^2)/(x^8 + 2*x^5 + x^2), x)
Time = 15.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \begin {dmath*} \int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{-x^2}}{x^4+x}}+{\mathrm {e}}^x \end {dmath*}