Integrand size = 94, antiderivative size = 29 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{-5+e^{\frac {2 e^3}{x}}+\frac {\log (2)}{1+x}}}{2 x} \]
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {2^{-\frac {x}{1+x}} e^{-5+e^{\frac {2 e^3}{x}}}}{x} \]
Integrate[(E^(-2 + (-3 - 3*x + E^((2*E^3)/x)*(1 + x) + Log[2])/(1 + x))*(- x - 2*x^2 - x^3 + E^(3 + (2*E^3)/x)*(-2 - 4*x - 2*x^2) - x^2*Log[2]))/(2*x ^3 + 4*x^4 + 2*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.
\(\displaystyle \int \frac {\left (-x^3-2 x^2+e^{\frac {2 e^3}{x}+3} \left (-2 x^2-4 x-2\right )-x^2 \log (2)-x\right ) \exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}-2\right )}{2 x^5+4 x^4+2 x^3} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (-x^3+e^{\frac {2 e^3}{x}+3} \left (-2 x^2-4 x-2\right )+x^2 (-2-\log (2))-x\right ) \exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}-2\right )}{2 x^5+4 x^4+2 x^3}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-x^3+e^{\frac {2 e^3}{x}+3} \left (-2 x^2-4 x-2\right )+x^2 (-2-\log (2))-x\right ) \exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}-2\right )}{x^3 \left (2 x^2+4 x+2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-x^3+e^{\frac {2 e^3}{x}+3} \left (-2 x^2-4 x-2\right )+x^2 (-2-\log (2))-x\right ) \exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}-2\right )}{x^3 \left (\sqrt {2} x+\sqrt {2}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (-x^2-x (2+\log (2))-1\right ) \exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}-2\right )}{2 x^2 (x+1)^2}-\frac {\exp \left (\frac {2 e^3}{x}+\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)-3+\log (2)}{x+1}+1\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)+\log (2)-3}{x+1}+\frac {2 e^3}{x}+1\right )}{x^3}dx-\frac {1}{2} \int \frac {\exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)+\log (2)-3}{x+1}-2\right )}{x^2}dx-\frac {1}{2} \log (2) \int \frac {\exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)+\log (2)-3}{x+1}-2\right )}{x}dx+\frac {1}{2} \log (2) \int \frac {\exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)+\log (2)-3}{x+1}-2\right )}{(x+1)^2}dx+\frac {1}{2} \log (2) \int \frac {\exp \left (\frac {-3 x+e^{\frac {2 e^3}{x}} (x+1)+\log (2)-3}{x+1}-2\right )}{x+1}dx\) |
Int[(E^(-2 + (-3 - 3*x + E^((2*E^3)/x)*(1 + x) + Log[2])/(1 + x))*(-x - 2* x^2 - x^3 + E^(3 + (2*E^3)/x)*(-2 - 4*x - 2*x^2) - x^2*Log[2]))/(2*x^3 + 4 *x^4 + 2*x^5),x]
3.30.14.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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 = 2.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {2^{\frac {1}{1+x}} {\mathrm e}^{-5+{\mathrm e}^{\frac {2 \,{\mathrm e}^{3}}{x}}}}{2 x}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{-2} {\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{3}}{x}}+\ln \left (2\right )-3 x -3}{1+x}}}{2 x}\) | \(37\) |
int(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*ln(2)-x^3-2*x^2-x)*exp(((1+ x)*exp(exp(3)/x)^2+ln(2)-3*x-3)/(1+x))/(2*x^5+4*x^4+2*x^3)/exp(2),x,method =_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\left (-\frac {{\left (5 \, {\left (x + 1\right )} e^{3} - {\left (x + 1\right )} e^{\left (\frac {3 \, x + 2 \, e^{3}}{x}\right )} - e^{3} \log \left (2\right )\right )} e^{\left (-3\right )}}{x + 1}\right )}}{2 \, x} \]
integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*e xp(((1+x)*exp(exp(3)/x)^2+log(2)-3*x-3)/(1+x))/(2*x^5+4*x^4+2*x^3)/exp(2), x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\frac {- 3 x + \left (x + 1\right ) e^{\frac {2 e^{3}}{x}} - 3 + \log {\left (2 \right )}}{x + 1}}}{2 x e^{2}} \]
integrate(((-2*x**2-4*x-2)*exp(3)*exp(exp(3)/x)**2-x**2*ln(2)-x**3-2*x**2- x)*exp(((1+x)*exp(exp(3)/x)**2+ln(2)-3*x-3)/(1+x))/(2*x**5+4*x**4+2*x**3)/ exp(2),x)
Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\left (\frac {\log \left (2\right )}{x + 1} + e^{\left (\frac {2 \, e^{3}}{x}\right )} - 5\right )}}{2 \, x} \]
integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*e xp(((1+x)*exp(exp(3)/x)^2+log(2)-3*x-3)/(1+x))/(2*x^5+4*x^4+2*x^3)/exp(2), x, algorithm=\
\[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\int { -\frac {{\left (x^{3} + x^{2} \log \left (2\right ) + 2 \, x^{2} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (\frac {2 \, e^{3}}{x} + 3\right )} + x\right )} e^{\left (\frac {{\left (x + 1\right )} e^{\left (\frac {2 \, e^{3}}{x}\right )} - 3 \, x + \log \left (2\right ) - 3}{x + 1} - 2\right )}}{2 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} \,d x } \]
integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*e xp(((1+x)*exp(exp(3)/x)^2+log(2)-3*x-3)/(1+x))/(2*x^5+4*x^4+2*x^3)/exp(2), x, algorithm=\
integrate(-1/2*(x^3 + x^2*log(2) + 2*x^2 + 2*(x^2 + 2*x + 1)*e^(2*e^3/x + 3) + x)*e^(((x + 1)*e^(2*e^3/x) - 3*x + log(2) - 3)/(x + 1) - 2)/(x^5 + 2* x^4 + x^3), x)
Time = 13.86 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {2^{\frac {1}{x+1}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {3\,x}{x+1}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^3}{x}}}{x+1}}\,{\mathrm {e}}^{-\frac {3}{x+1}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^3}{x}}}{x+1}}}{2\,x} \]
int(-(exp(-(3*x - log(2) - exp((2*exp(3))/x)*(x + 1) + 3)/(x + 1))*exp(-2) *(x + x^2*log(2) + 2*x^2 + x^3 + exp((2*exp(3))/x)*exp(3)*(4*x + 2*x^2 + 2 )))/(2*x^3 + 4*x^4 + 2*x^5),x)