\(\int \frac {e^{\frac {-8-8 x+3 x^2+x^3+(4+2 x-2 x^2) \log (2)}{-4-2 x+2 x^2}} (8+16 x-2 x^2-5 x^3+x^5)}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx\) [689]

Optimal result

Integrand size = 82, antiderivative size = 31 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=e+\frac {1}{2} e^{\frac {x}{(2-x) (1+x)}+\frac {4+x}{2}} x \]

[Out]

exp(-ln(2)+x/(1+x)/(2-x)+2+1/2*x)*x+exp(1)

Rubi [F]

\[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\int \frac {\exp \left (\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}\right ) \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx \]

[In]

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]

[Out]

Defer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 + 2*x - 2*x^2)), x]/2 + (2*Defer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 +
2*x - 2*x^2))/(-2 + x)^2, x])/3 + Defer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 + 2*x - 2*x^2))/(-2 + x), x]/3 + D
efer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 + 2*x - 2*x^2))*x, x]/4 - Defer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 + 2
*x - 2*x^2))/(1 + x)^2, x]/6 + Defer[Int][E^((8 + 8*x - 3*x^2 - x^3)/(4 + 2*x - 2*x^2))/(1 + x), x]/6

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{4 \left (2+x-x^2\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{\left (2+x-x^2\right )^2} \, dx \\ & = \frac {1}{4} \int \left (2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}+\frac {8 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (-2+x)^2}+\frac {4 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (-2+x)}+e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} x-\frac {2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (1+x)^2}+\frac {2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (1+x)}\right ) \, dx \\ & = -\left (\frac {1}{6} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{(1+x)^2} \, dx\right )+\frac {1}{6} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{1+x} \, dx+\frac {1}{4} \int e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} x \, dx+\frac {1}{3} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{-2+x} \, dx+\frac {1}{2} \int e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \, dx+\frac {2}{3} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{(-2+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} e^{2+\frac {x}{2}-\frac {x}{-2-x+x^2}} x \]

[In]

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]

[Out]

(E^(2 + x/2 - x/(-2 - x + x^2))*x)/2

Maple [A] (verified)

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	\(\frac {x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-2 x \,{\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}}{x^{2}-x -2}\)	\(145\)

[In]

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)

[Out]

x*exp(1/2/(x^2-x-2)*((-2*x^2+2*x+4)*ln(2)+x^3+3*x^2-8*x-8))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\left (\frac {x^{3} + 3 \, x^{2} - 2 \, {\left (x^{2} - x - 2\right )} \log \left (2\right ) - 8 \, x - 8}{2 \, {\left (x^{2} - x - 2\right )}}\right )} \]

[In]

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="fricas")

[Out]

x*e^(1/2*(x^3 + 3*x^2 - 2*(x^2 - x - 2)*log(2) - 8*x - 8)/(x^2 - x - 2))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\frac {x^{3} + 3 x^{2} - 8 x + \left (- 2 x^{2} + 2 x + 4\right ) \log {\left (2 \right )} - 8}{2 x^{2} - 2 x - 4}} \]

[In]

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)

[Out]

x*exp((x**3 + 3*x**2 - 8*x + (-2*x**2 + 2*x + 4)*log(2) - 8)/(2*x**2 - 2*x - 4))

Maxima [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {1}{2} \, x - \frac {1}{3 \, {\left (x + 1\right )}} - \frac {2}{3 \, {\left (x - 2\right )}} + 2\right )} \]

[In]

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="maxima")

[Out]

1/2*x*e^(1/2*x - 1/3/(x + 1) - 2/3/(x - 2) + 2)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {x^{3} - x^{2} - 4 \, x}{2 \, {\left (x^{2} - x - 2\right )}} + 2\right )} \]

[In]

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="giac")

[Out]

1/2*x*e^(1/2*(x^3 - x^2 - 4*x)/(x^2 - x - 2) + 2)

Mupad [B] (verification not implemented)

Time = 8.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.32 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=2^{\frac {x^2-2}{-x^2+x+2}-\frac {2\,x}{-2\,x^2+2\,x+4}}\,x\,{\mathrm {e}}^{\frac {8\,x}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {x^3}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {3\,x^2}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{\frac {8}{-2\,x^2+2\,x+4}} \]

[In]

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)

[Out]

2^((x^2 - 2)/(x - x^2 + 2) - (2*x)/(2*x - 2*x^2 + 4))*x*exp((8*x)/(2*x - 2*x^2 + 4))*exp(-x^3/(2*x - 2*x^2 + 4
))*exp(-(3*x^2)/(2*x - 2*x^2 + 4))*exp(8/(2*x - 2*x^2 + 4))