\(\int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} (-2-4 x-2 x^2)-x^2 \log (2))}{2 x^3+4 x^4+2 x^5} \, dx\) [8914]

Optimal result

Integrand size = 94, antiderivative size = 29 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{-5+e^{\frac {2 e^3}{x}}+\frac {\log (2)}{1+x}}}{2 x} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\int \frac {\exp \left (-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}\right ) \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx \]

[In]

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]

[Out]

-Defer[Int][E^(1 + (2*E^3)/x + (-3 - 3*x + E^((2*E^3)/x)*(1 + x) + Log[2])/(1 + x))/x^3, x] - Defer[Int][(2^(1
+ x)^(-1)*E^(-5 + E^((2*E^3)/x)))/x^2, x]/2 - (Log[2]*Defer[Int][(2^(1 + x)^(-1)*E^(-5 + E^((2*E^3)/x)))/x, x
])/2 + (Log[2]*Defer[Int][(2^(1 + x)^(-1)*E^(-5 + E^((2*E^3)/x)))/(1 + x)^2, x])/2 + (Log[2]*Defer[Int][(2^(1
+ x)^(-1)*E^(-5 + E^((2*E^3)/x)))/(1 + x), x])/2

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {2^{-\frac {x}{1+x}} e^{-5+e^{\frac {2 e^3}{x}}}}{x} \]

[In]

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]

[Out]

E^(-5 + E^((2*E^3)/x))/(2^(x/(1 + x))*x)

Maple [A] (verified)

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	\(\frac {{\mathrm e}^{-2} {\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{3}}{x}}+\ln \left (2\right )-3 x -3}{1+x}}}{2 x}\)	\(37\)

[In]

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)

[Out]

1/2*2^(1/(1+x))/x*exp(-5+exp(2*exp(3)/x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\left (-\frac {{\left (5 \, {\left (x + 1\right )} e^{3} - {\left (x + 1\right )} e^{\left (\frac {3 \, x + 2 \, e^{3}}{x}\right )} - e^{3} \log \left (2\right )\right )} e^{\left (-3\right )}}{x + 1}\right )}}{2 \, x} \]

[In]

integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*exp(((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="fricas")

[Out]

1/2*e^(-(5*(x + 1)*e^3 - (x + 1)*e^((3*x + 2*e^3)/x) - e^3*log(2))*e^(-3)/(x + 1))/x

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\frac {- 3 x + \left (x + 1\right ) e^{\frac {2 e^{3}}{x}} - 3 + \log {\left (2 \right )}}{x + 1}}}{2 x e^{2}} \]

[In]

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)

[Out]

exp(-2)*exp((-3*x + (x + 1)*exp(2*exp(3)/x) - 3 + log(2))/(x + 1))/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {e^{\left (\frac {\log \left (2\right )}{x + 1} + e^{\left (\frac {2 \, e^{3}}{x}\right )} - 5\right )}}{2 \, x} \]

[In]

integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*exp(((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="maxima")

[Out]

1/2*e^(log(2)/(x + 1) + e^(2*e^3/x) - 5)/x

Giac [F]

\[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\int { -\frac {{\left (x^{3} + x^{2} \log \left (2\right ) + 2 \, x^{2} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (\frac {2 \, e^{3}}{x} + 3\right )} + x\right )} e^{\left (\frac {{\left (x + 1\right )} e^{\left (\frac {2 \, e^{3}}{x}\right )} - 3 \, x + \log \left (2\right ) - 3}{x + 1} - 2\right )}}{2 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} \,d x } \]

[In]

integrate(((-2*x^2-4*x-2)*exp(3)*exp(exp(3)/x)^2-x^2*log(2)-x^3-2*x^2-x)*exp(((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="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 13.86 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {e^{-2+\frac {-3-3 x+e^{\frac {2 e^3}{x}} (1+x)+\log (2)}{1+x}} \left (-x-2 x^2-x^3+e^{3+\frac {2 e^3}{x}} \left (-2-4 x-2 x^2\right )-x^2 \log (2)\right )}{2 x^3+4 x^4+2 x^5} \, dx=\frac {2^{\frac {1}{x+1}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {3\,x}{x+1}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^3}{x}}}{x+1}}\,{\mathrm {e}}^{-\frac {3}{x+1}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^3}{x}}}{x+1}}}{2\,x} \]

[In]

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)

[Out]

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