Integrand size = 89, antiderivative size = 29 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{-5+x+2 \left (1+\frac {1-\frac {x}{-5+x}}{x}\right )}}{x}} \]
Time = 5.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{-3-\frac {2}{-5+x}+\frac {2}{x}+x}}{x}} \]
Integrate[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(-50 - 5*x + 35*x^2 - 11*x^3 + x^4))/(2 5*x^3 - 10*x^4 + x^5),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.
\(\displaystyle \int \frac {\left (x^4-11 x^3+35 x^2-5 x-50\right ) \exp \left (\frac {2 x^2-10 x-10}{x^2-5 x}+\frac {e^{\frac {2 x^2-10 x-10}{x^2-5 x}+x-5}}{x}+x-5\right )}{x^5-10 x^4+25 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^4-11 x^3+35 x^2-5 x-50\right ) \exp \left (\frac {2 x^2-10 x-10}{x^2-5 x}+\frac {e^{\frac {2 x^2-10 x-10}{x^2-5 x}+x-5}}{x}+x-5\right )}{x^3 \left (x^2-10 x+25\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{4 (5-x)^2 x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{(5-x)^2 x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (-\frac {27 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 x}+\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 (x-5)}-\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{5 (x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{(x-5)^2}dx-\frac {2}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x-5}dx-\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}dx+\frac {27}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x}dx-2 \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}dx\) |
Int[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(-50 - 5*x + 35*x^2 - 11*x^3 + x^4))/(25*x^3 - 10*x^4 + x^5),x]
3.13.55.3.1 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[(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])
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 = 15.78 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {x^{3}-8 x^{2}+15 x -10}{\left (-5+x \right ) x}}}{x}}\) | \(29\) |
parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {2 x^{2}-10 x -10}{\left (-5+x \right ) x}} {\mathrm e}^{-5+x}}{x}}\) | \(29\) |
int((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)*ex p(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x,method =_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.66 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\left (\frac {x^{3} - 8 \, x^{2} + {\left (x - 5\right )} e^{\left (\frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} + 15 \, x - 10}{x^{2} - 5 \, x} - \frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} \]
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm=\
e^((x^3 - 8*x^2 + (x - 5)*e^((x^3 - 8*x^2 + 15*x - 10)/(x^2 - 5*x)) + 15*x - 10)/(x^2 - 5*x) - (x^3 - 8*x^2 + 15*x - 10)/(x^2 - 5*x))
Time = 0.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{\frac {2 x^{2} - 10 x - 10}{x^{2} - 5 x}} e^{x - 5}}{x}} \]
integrate((x**4-11*x**3+35*x**2-5*x-50)*exp((2*x**2-10*x-10)/(x**2-5*x))*e xp(-5+x)*exp(exp((2*x**2-10*x-10)/(x**2-5*x))*exp(-5+x)/x)/(x**5-10*x**4+2 5*x**3),x)
\[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=\int { \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}} \,d x } \]
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm=\
integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)
\[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=\int { \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}} \,d x } \]
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm=\
integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)
Time = 9.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{\frac {2\,x}{x-5}}\,{\mathrm {e}}^{\frac {10}{5\,x-x^2}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {10}{x-5}}}{x}} \]