Integrand size = 65, antiderivative size = 23 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=e^{\frac {2 e^{1-x} x}{-1+x (1+5 x)}} \]
Time = 1.71 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=e^{\frac {2 e^{1-x} x}{-1+x+5 x^2}} \]
Integrate[(E^(1 - x + (2*E^(1 - x)*x)/(-1 + x + 5*x^2))*(-2 + 2*x - 12*x^2 - 10*x^3))/(1 - 2*x - 9*x^2 + 10*x^3 + 25*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.
\(\displaystyle \int \frac {e^{\frac {2 e^{1-x} x}{5 x^2+x-1}-x+1} \left (-10 x^3-12 x^2+2 x-2\right )}{25 x^4+10 x^3-9 x^2-2 x+1} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {100 e^{\frac {2 e^{1-x} x}{5 x^2+x-1}-x+1} \left (-10 x^3-12 x^2+2 x-2\right )}{21 \sqrt {21} \left (-10 x+\sqrt {21}-1\right )}+\frac {100 e^{\frac {2 e^{1-x} x}{5 x^2+x-1}-x+1} \left (-10 x^3-12 x^2+2 x-2\right )}{21 \sqrt {21} \left (10 x+\sqrt {21}+1\right )}+\frac {100 e^{\frac {2 e^{1-x} x}{5 x^2+x-1}-x+1} \left (-10 x^3-12 x^2+2 x-2\right )}{21 \left (-10 x+\sqrt {21}-1\right )^2}+\frac {100 e^{\frac {2 e^{1-x} x}{5 x^2+x-1}-x+1} \left (-10 x^3-12 x^2+2 x-2\right )}{21 \left (10 x+\sqrt {21}+1\right )^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 e^{x \left (\frac {2 e^{1-x}}{5 x^2+x-1}-1\right )+1} \left (-5 x^3-6 x^2+x-1\right )}{\left (-5 x^2-x+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )} \left (5 x^3+6 x^2-x+1\right )}{\left (-5 x^2-x+1\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )} \left (5 x^3+6 x^2-x+1\right )}{\left (-5 x^2-x+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )} (2-x)}{\left (5 x^2+x-1\right )^2}+\frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )} (x+1)}{5 x^2+x-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\frac {10}{21} \left (1-\sqrt {21}\right ) \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{\left (-10 x+\sqrt {21}-1\right )^2}dx+\frac {200}{21} \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{\left (-10 x+\sqrt {21}-1\right )^2}dx+\frac {10 \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{-10 x+\sqrt {21}-1}dx}{\sqrt {21}}+\frac {1}{7} \left (7+3 \sqrt {21}\right ) \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{10 x-\sqrt {21}+1}dx+\frac {10}{21} \left (1+\sqrt {21}\right ) \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{\left (10 x+\sqrt {21}+1\right )^2}dx+\frac {200}{21} \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{\left (10 x+\sqrt {21}+1\right )^2}dx+\frac {1}{7} \left (7-3 \sqrt {21}\right ) \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{10 x+\sqrt {21}+1}dx+\frac {10 \int \frac {e^{1-x \left (1+\frac {2 e^{1-x}}{-5 x^2-x+1}\right )}}{10 x+\sqrt {21}+1}dx}{\sqrt {21}}\right )\) |
Int[(E^(1 - x + (2*E^(1 - x)*x)/(-1 + x + 5*x^2))*(-2 + 2*x - 12*x^2 - 10* x^3))/(1 - 2*x - 9*x^2 + 10*x^3 + 25*x^4),x]
3.24.65.3.1 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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \({\mathrm e}^{\frac {2 x \,{\mathrm e}^{1-x}}{5 x^{2}+x -1}}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {2 x \,{\mathrm e}^{1-x}}{5 x^{2}+x -1}}\) | \(21\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {2 x \,{\mathrm e}^{1-x}}{5 x^{2}+x -1}}+5 x^{2} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{1-x}}{5 x^{2}+x -1}}-{\mathrm e}^{\frac {2 x \,{\mathrm e}^{1-x}}{5 x^{2}+x -1}}}{5 x^{2}+x -1}\) | \(82\) |
int((-10*x^3-12*x^2+2*x-2)*exp(1-x)*exp(2*x*exp(1-x)/(5*x^2+x-1))/(25*x^4+ 10*x^3-9*x^2-2*x+1),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=e^{\left (x - \frac {5 \, x^{3} - 4 \, x^{2} - 2 \, x e^{\left (-x + 1\right )} - 2 \, x + 1}{5 \, x^{2} + x - 1} - 1\right )} \]
integrate((-10*x^3-12*x^2+2*x-2)*exp(1-x)*exp(2*x*exp(1-x)/(5*x^2+x-1))/(2 5*x^4+10*x^3-9*x^2-2*x+1),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=e^{\frac {2 x e^{1 - x}}{5 x^{2} + x - 1}} \]
integrate((-10*x**3-12*x**2+2*x-2)*exp(1-x)*exp(2*x*exp(1-x)/(5*x**2+x-1)) /(25*x**4+10*x**3-9*x**2-2*x+1),x)
Time = 0.36 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=e^{\left (\frac {2 \, x e^{\left (-x + 1\right )}}{5 \, x^{2} + x - 1}\right )} \]
integrate((-10*x^3-12*x^2+2*x-2)*exp(1-x)*exp(2*x*exp(1-x)/(5*x^2+x-1))/(2 5*x^4+10*x^3-9*x^2-2*x+1),x, algorithm=\
\[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx=\int { -\frac {2 \, {\left (5 \, x^{3} + 6 \, x^{2} - x + 1\right )} e^{\left (-x + \frac {2 \, x e^{\left (-x + 1\right )}}{5 \, x^{2} + x - 1} + 1\right )}}{25 \, x^{4} + 10 \, x^{3} - 9 \, x^{2} - 2 \, x + 1} \,d x } \]
integrate((-10*x^3-12*x^2+2*x-2)*exp(1-x)*exp(2*x*exp(1-x)/(5*x^2+x-1))/(2 5*x^4+10*x^3-9*x^2-2*x+1),x, algorithm=\
integrate(-2*(5*x^3 + 6*x^2 - x + 1)*e^(-x + 2*x*e^(-x + 1)/(5*x^2 + x - 1 ) + 1)/(25*x^4 + 10*x^3 - 9*x^2 - 2*x + 1), x)
Time = 14.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{1-x+\frac {2 e^{1-x} x}{-1+x+5 x^2}} \left (-2+2 x-12 x^2-10 x^3\right )}{1-2 x-9 x^2+10 x^3+25 x^4} \, dx={\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{-x}\,\mathrm {e}}{5\,x^2+x-1}} \]