Integrand size = 82, antiderivative size = 31 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=e+\frac {1}{2} e^{\frac {x}{(2-x) (1+x)}+\frac {4+x}{2}} x \]
Time = 1.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} e^{2+\frac {x}{2}-\frac {x}{-2-x+x^2}} x \]
Integrate[(E^((-8 - 8*x + 3*x^2 + x^3 + (4 + 2*x - 2*x^2)*Log[2])/(-4 - 2* x + 2*x^2))*(8 + 16*x - 2*x^2 - 5*x^3 + x^5))/(8 + 8*x - 6*x^2 - 4*x^3 + 2 *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.
\(\displaystyle \int \frac {\left (x^5-5 x^3-2 x^2+16 x+8\right ) \exp \left (\frac {x^3+3 x^2+\left (-2 x^2+2 x+4\right ) \log (2)-8 x-8}{2 x^2-2 x-4}\right )}{2 x^4-4 x^3-6 x^2+8 x+8} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {\left (x^5-5 x^3-2 x^2+16 x+8\right ) \exp \left (\frac {x^3+3 x^2+\left (-2 x^2+2 x+4\right ) \log (2)-8 x-8}{2 x^2-2 x-4}\right )}{27 (x-2)}+\frac {\left (x^5-5 x^3-2 x^2+16 x+8\right ) \exp \left (\frac {x^3+3 x^2+\left (-2 x^2+2 x+4\right ) \log (2)-8 x-8}{2 x^2-2 x-4}\right )}{27 (x+1)}+\frac {\left (x^5-5 x^3-2 x^2+16 x+8\right ) \exp \left (\frac {x^3+3 x^2+\left (-2 x^2+2 x+4\right ) \log (2)-8 x-8}{2 x^2-2 x-4}\right )}{18 (x-2)^2}+\frac {\left (x^5-5 x^3-2 x^2+16 x+8\right ) \exp \left (\frac {x^3+3 x^2+\left (-2 x^2+2 x+4\right ) \log (2)-8 x-8}{2 x^2-2 x-4}\right )}{18 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{-2 x^2+2 x+4}} \left (x^5-5 x^3-2 x^2+16 x+8\right )}{4 (2-x)^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}} \left (x^5-5 x^3-2 x^2+16 x+8\right )}{(2-x)^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}} x+2 e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}+\frac {4 e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{3 (x-2)}+\frac {2 e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{3 (x+1)}+\frac {8 e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{3 (x-2)^2}-\frac {2 e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{3 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (2 \int e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}dx+\frac {8}{3} \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{(x-2)^2}dx+\frac {4}{3} \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{x-2}dx+\int e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}} xdx-\frac {2}{3} \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{(x+1)^2}dx+\frac {2}{3} \int \frac {e^{\frac {-x^3-3 x^2+8 x+8}{2 \left (-x^2+x+2\right )}}}{x+1}dx\right )\) |
Int[(E^((-8 - 8*x + 3*x^2 + x^3 + (4 + 2*x - 2*x^2)*Log[2])/(-4 - 2*x + 2* x^2))*(8 + 16*x - 2*x^2 - 5*x^3 + x^5))/(8 + 8*x - 6*x^2 - 4*x^3 + 2*x^4), x]
3.7.89.3.1 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 = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
parallelrisch | \(x \,{\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}\) | \(42\) |
gosper | \({\mathrm e}^{-\frac {2 x^{2} \ln \left (2\right )-x^{3}-2 x \ln \left (2\right )-3 x^{2}-4 \ln \left (2\right )+8 x +8}{2 \left (x^{2}-x -2\right )}} x\) | \(47\) |
risch | \(x \,{\mathrm e}^{-\frac {2 x^{2} \ln \left (2\right )-x^{3}-2 x \ln \left (2\right )-3 x^{2}-4 \ln \left (2\right )+8 x +8}{2 \left (1+x \right ) \left (-2+x \right )}}\) | \(47\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-2 x \,{\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}}{x^{2}-x -2}\) | \(145\) |
int((x^5-5*x^3-2*x^2+16*x+8)*exp(((-2*x^2+2*x+4)*ln(2)+x^3+3*x^2-8*x-8)/(2 *x^2-2*x-4))/(2*x^4-4*x^3-6*x^2+8*x+8),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\left (\frac {x^{3} + 3 \, x^{2} - 2 \, {\left (x^{2} - x - 2\right )} \log \left (2\right ) - 8 \, x - 8}{2 \, {\left (x^{2} - x - 2\right )}}\right )} \]
integrate((x^5-5*x^3-2*x^2+16*x+8)*exp(((-2*x^2+2*x+4)*log(2)+x^3+3*x^2-8* x-8)/(2*x^2-2*x-4))/(2*x^4-4*x^3-6*x^2+8*x+8),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\frac {x^{3} + 3 x^{2} - 8 x + \left (- 2 x^{2} + 2 x + 4\right ) \log {\left (2 \right )} - 8}{2 x^{2} - 2 x - 4}} \]
integrate((x**5-5*x**3-2*x**2+16*x+8)*exp(((-2*x**2+2*x+4)*ln(2)+x**3+3*x* *2-8*x-8)/(2*x**2-2*x-4))/(2*x**4-4*x**3-6*x**2+8*x+8),x)
Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {1}{2} \, x - \frac {1}{3 \, {\left (x + 1\right )}} - \frac {2}{3 \, {\left (x - 2\right )}} + 2\right )} \]
integrate((x^5-5*x^3-2*x^2+16*x+8)*exp(((-2*x^2+2*x+4)*log(2)+x^3+3*x^2-8* x-8)/(2*x^2-2*x-4))/(2*x^4-4*x^3-6*x^2+8*x+8),x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {x^{3} - x^{2} - 4 \, x}{2 \, {\left (x^{2} - x - 2\right )}} + 2\right )} \]
integrate((x^5-5*x^3-2*x^2+16*x+8)*exp(((-2*x^2+2*x+4)*log(2)+x^3+3*x^2-8* x-8)/(2*x^2-2*x-4))/(2*x^4-4*x^3-6*x^2+8*x+8),x, algorithm=\
Time = 8.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.32 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=2^{\frac {x^2-2}{-x^2+x+2}-\frac {2\,x}{-2\,x^2+2\,x+4}}\,x\,{\mathrm {e}}^{\frac {8\,x}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {x^3}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {3\,x^2}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{\frac {8}{-2\,x^2+2\,x+4}} \]
int((exp(-(log(2)*(2*x - 2*x^2 + 4) - 8*x + 3*x^2 + x^3 - 8)/(2*x - 2*x^2 + 4))*(16*x - 2*x^2 - 5*x^3 + x^5 + 8))/(8*x - 6*x^2 - 4*x^3 + 2*x^4 + 8), x)