\(\int \frac {e^{\frac {2+8 x+2 x^2+(x+2 x^2+x^3) \log (\frac {1}{x})}{2+2 x+(x+x^2) \log (\frac {1}{x})}} (12+12 x+8 x^2+(4 x+4 x^2+4 x^3) \log (\frac {1}{x})+(x^2+2 x^3+x^4) \log ^2(\frac {1}{x}))}{4+8 x+4 x^2+(4 x+8 x^2+4 x^3) \log (\frac {1}{x})+(x^2+2 x^3+x^4) \log ^2(\frac {1}{x})} \, dx\) [7966]

Optimal result

Integrand size = 144, antiderivative size = 24 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{1+x+\frac {4}{(1+x) \left (\frac {2}{x}+\log \left (\frac {1}{x}\right )\right )}} \]

[Out]

exp(x+1+4/(1+x)/(ln(1/x)+2/x))

Rubi [F]

\[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\int \frac {\exp \left (\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}\right ) \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \]

[In]

Int[(E^((2 + 8*x + 2*x^2 + (x + 2*x^2 + x^3)*Log[x^(-1)])/(2 + 2*x + (x + x^2)*Log[x^(-1)]))*(12 + 12*x + 8*x^
2 + (4*x + 4*x^2 + 4*x^3)*Log[x^(-1)] + (x^2 + 2*x^3 + x^4)*Log[x^(-1)]^2))/(4 + 8*x + 4*x^2 + (4*x + 8*x^2 +
4*x^3)*Log[x^(-1)] + (x^2 + 2*x^3 + x^4)*Log[x^(-1)]^2),x]

[Out]

Defer[Int][E^((2*(1 + 4*x + x^2))/((1 + x)*(2 + x*Log[x^(-1)])))*(x^(-1))^((x*(1 + x))/(2 + x*Log[x^(-1)])), x
] + 4*Defer[Int][(E^((2*(1 + 4*x + x^2))/((1 + x)*(2 + x*Log[x^(-1)])))*(x^(-1))^((x*(1 + x))/(2 + x*Log[x^(-1
)])))/(2 + x*Log[x^(-1)])^2, x] + 4*Defer[Int][(E^((2*(1 + 4*x + x^2))/((1 + x)*(2 + x*Log[x^(-1)])))*(x^(-1))
^((x*(1 + x))/(2 + x*Log[x^(-1)])))/((1 + x)*(2 + x*Log[x^(-1)])^2), x] - 4*Defer[Int][(E^((2*(1 + 4*x + x^2))
/((1 + x)*(2 + x*Log[x^(-1)])))*(x^(-1))^(-1 + (x*(1 + x))/(2 + x*Log[x^(-1)])))/((1 + x)^2*(2 + x*Log[x^(-1)]
)), x]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\frac {2 \left (1+4 x+x^2\right )+x (1+x)^2 \log \left (\frac {1}{x}\right )}{(1+x) \left (2+x \log \left (\frac {1}{x}\right )\right )}} \]

[In]

Integrate[(E^((2 + 8*x + 2*x^2 + (x + 2*x^2 + x^3)*Log[x^(-1)])/(2 + 2*x + (x + x^2)*Log[x^(-1)]))*(12 + 12*x
+ 8*x^2 + (4*x + 4*x^2 + 4*x^3)*Log[x^(-1)] + (x^2 + 2*x^3 + x^4)*Log[x^(-1)]^2))/(4 + 8*x + 4*x^2 + (4*x + 8*
x^2 + 4*x^3)*Log[x^(-1)] + (x^2 + 2*x^3 + x^4)*Log[x^(-1)]^2),x]

[Out]

E^((2*(1 + 4*x + x^2) + x*(1 + x)^2*Log[x^(-1)])/((1 + x)*(2 + x*Log[x^(-1)])))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(48\) vs. \(2(23)=46\).

Time = 8.81 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (x^{3}+2 x^{2}+x \right ) \ln \left (\frac {1}{x}\right )+2 x^{2}+8 x +2}{x^{2} \ln \left (\frac {1}{x}\right )+x \ln \left (\frac {1}{x}\right )+2 x +2}}\)	\(49\)
risch	\({\mathrm e}^{\frac {x^{3} \ln \left (\frac {1}{x}\right )+2 x^{2} \ln \left (\frac {1}{x}\right )+x \ln \left (\frac {1}{x}\right )+2 x^{2}+8 x +2}{\left (1+x \right ) \left (x \ln \left (\frac {1}{x}\right )+2\right )}}\)	\(51\)

[In]

int(((x^4+2*x^3+x^2)*ln(1/x)^2+(4*x^3+4*x^2+4*x)*ln(1/x)+8*x^2+12*x+12)*exp(((x^3+2*x^2+x)*ln(1/x)+2*x^2+8*x+2
)/((x^2+x)*ln(1/x)+2*x+2))/((x^4+2*x^3+x^2)*ln(1/x)^2+(4*x^3+8*x^2+4*x)*ln(1/x)+4*x^2+8*x+4),x,method=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (\frac {2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + x\right )} \log \left (\frac {1}{x}\right ) + 8 \, x + 2}{{\left (x^{2} + x\right )} \log \left (\frac {1}{x}\right ) + 2 \, x + 2}\right )} \]

[In]

integrate(((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+4*x^2+4*x)*log(1/x)+8*x^2+12*x+12)*exp(((x^3+2*x^2+x)*log(1/x)+2*
x^2+8*x+2)/((x^2+x)*log(1/x)+2*x+2))/((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+8*x^2+4*x)*log(1/x)+4*x^2+8*x+4),x, al
gorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\frac {2 x^{2} + 8 x + \left (x^{3} + 2 x^{2} + x\right ) \log {\left (\frac {1}{x} \right )} + 2}{2 x + \left (x^{2} + x\right ) \log {\left (\frac {1}{x} \right )} + 2}} \]

[In]

integrate(((x**4+2*x**3+x**2)*ln(1/x)**2+(4*x**3+4*x**2+4*x)*ln(1/x)+8*x**2+12*x+12)*exp(((x**3+2*x**2+x)*ln(1
/x)+2*x**2+8*x+2)/((x**2+x)*ln(1/x)+2*x+2))/((x**4+2*x**3+x**2)*ln(1/x)**2+(4*x**3+8*x**2+4*x)*ln(1/x)+4*x**2+
8*x+4),x)

[Out]

exp((2*x**2 + 8*x + (x**3 + 2*x**2 + x)*log(1/x) + 2)/(2*x + (x**2 + x)*log(1/x) + 2))

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (x - \frac {8}{x \log \left (x\right )^{2} + 2 \, {\left (x - 1\right )} \log \left (x\right ) - 4} - \frac {4}{{\left (x + 1\right )} \log \left (x\right ) + 2 \, x + 2} + 1\right )} \]

[In]

integrate(((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+4*x^2+4*x)*log(1/x)+8*x^2+12*x+12)*exp(((x^3+2*x^2+x)*log(1/x)+2*
x^2+8*x+2)/((x^2+x)*log(1/x)+2*x+2))/((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+8*x^2+4*x)*log(1/x)+4*x^2+8*x+4),x, al
gorithm="maxima")

[Out]

e^(x - 8/(x*log(x)^2 + 2*(x - 1)*log(x) - 4) - 4/((x + 1)*log(x) + 2*x + 2) + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (23) = 46\).

Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 5.46 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (\frac {x^{3} \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} + \frac {2 \, x^{2} \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {2 \, x^{2}}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} + \frac {x \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {8 \, x}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {2}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2}\right )} \]

[In]

integrate(((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+4*x^2+4*x)*log(1/x)+8*x^2+12*x+12)*exp(((x^3+2*x^2+x)*log(1/x)+2*
x^2+8*x+2)/((x^2+x)*log(1/x)+2*x+2))/((x^4+2*x^3+x^2)*log(1/x)^2+(4*x^3+8*x^2+4*x)*log(1/x)+4*x^2+8*x+4),x, al
gorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.04 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx={\mathrm {e}}^{\frac {8\,x}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\mathrm {e}}^{\frac {2\,x^2}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\mathrm {e}}^{\frac {2}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\left (\frac {1}{x}\right )}^{\frac {x^2+x}{x\,\ln \left (\frac {1}{x}\right )+2}} \]

[In]

int((exp((8*x + log(1/x)*(x + 2*x^2 + x^3) + 2*x^2 + 2)/(2*x + log(1/x)*(x + x^2) + 2))*(12*x + log(1/x)*(4*x
+ 4*x^2 + 4*x^3) + log(1/x)^2*(x^2 + 2*x^3 + x^4) + 8*x^2 + 12))/(8*x + log(1/x)*(4*x + 8*x^2 + 4*x^3) + log(1
/x)^2*(x^2 + 2*x^3 + x^4) + 4*x^2 + 4),x)

[Out]

exp((8*x)/(2*x + x*log(1/x) + x^2*log(1/x) + 2))*exp((2*x^2)/(2*x + x*log(1/x) + x^2*log(1/x) + 2))*exp(2/(2*x
+ x*log(1/x) + x^2*log(1/x) + 2))*(1/x)^((x + x^2)/(x*log(1/x) + 2))