\(\int \frac {e^{\frac {e^{5+4 x} (5 x+x^2)+e^5 (x+6 x^2+6 x^3+x^4)}{e^{4 x}+x+x^2}} (e^{5+8 x} (5+2 x)+e^{5+4 x} (1+8 x+10 x^2+4 x^3)+e^5 (5 x^2+12 x^3+9 x^4+2 x^5))}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} (2 x+2 x^2)} \, dx\) [5970]

Optimal result

Integrand size = 147, antiderivative size = 26 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{e^5 x \left (5+x+\frac {1+x}{e^{4 x}+x+x^2}\right )} \]

[Out]

exp(x*exp(5)*(5+x+(1+x)/(exp(4*x)+x^2+x)))

Rubi [F]

\[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=\int \frac {\exp \left (\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}\right ) \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx \]

[In]

Int[(E^((E^(5 + 4*x)*(5*x + x^2) + E^5*(x + 6*x^2 + 6*x^3 + x^4))/(E^(4*x) + x + x^2))*(E^(5 + 8*x)*(5 + 2*x)
+ E^(5 + 4*x)*(1 + 8*x + 10*x^2 + 4*x^3) + E^5*(5*x^2 + 12*x^3 + 9*x^4 + 2*x^5)))/(E^(8*x) + x^2 + 2*x^3 + x^4
+ E^(4*x)*(2*x + 2*x^2)),x]

[Out]

5*Defer[Int][E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2)), x] + 2*Defer
[Int][E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))*x, x] - Defer[Int][(
E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))*x)/(E^(4*x) + x + x^2)^2,
x] + Defer[Int][(E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))*x^2)/(E^(
4*x) + x + x^2)^2, x] + 6*Defer[Int][(E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x)
+ x + x^2))*x^3)/(E^(4*x) + x + x^2)^2, x] + 4*Defer[Int][(E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*
x^2 + x^3))/(E^(4*x) + x + x^2))*x^4)/(E^(4*x) + x + x^2)^2, x] + Defer[Int][E^(5 + (E^5*x*(1 + 5*E^(4*x) + 6*
x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))/(E^(4*x) + x + x^2), x] - 2*Defer[Int][(E^(5 + (E^5*x*(1 +
5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))*x)/(E^(4*x) + x + x^2), x] - 4*Defer[Int][(E^
(5 + (E^5*x*(1 + 5*E^(4*x) + 6*x + E^(4*x)*x + 6*x^2 + x^3))/(E^(4*x) + x + x^2))*x^2)/(E^(4*x) + x + x^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{\left (e^{4 x}+x+x^2\right )^2} \, dx \\ & = \int \left (\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) (5+2 x)-\frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) \left (-1+2 x+4 x^2\right )}{e^{4 x}+x+x^2}+\frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x \left (-1+x+6 x^2+4 x^3\right )}{\left (e^{4 x}+x+x^2\right )^2}\right ) \, dx \\ & = \int \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) (5+2 x) \, dx-\int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) \left (-1+2 x+4 x^2\right )}{e^{4 x}+x+x^2} \, dx+\int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x \left (-1+x+6 x^2+4 x^3\right )}{\left (e^{4 x}+x+x^2\right )^2} \, dx \\ & = \int \left (5 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right )+2 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x\right ) \, dx+\int \left (-\frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x}{\left (e^{4 x}+x+x^2\right )^2}+\frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^2}{\left (e^{4 x}+x+x^2\right )^2}+\frac {6 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^3}{\left (e^{4 x}+x+x^2\right )^2}+\frac {4 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^4}{\left (e^{4 x}+x+x^2\right )^2}\right ) \, dx-\int \left (-\frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right )}{e^{4 x}+x+x^2}+\frac {2 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x}{e^{4 x}+x+x^2}+\frac {4 \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^2}{e^{4 x}+x+x^2}\right ) \, dx \\ & = 2 \int \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x \, dx-2 \int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x}{e^{4 x}+x+x^2} \, dx+4 \int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^4}{\left (e^{4 x}+x+x^2\right )^2} \, dx-4 \int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^2}{e^{4 x}+x+x^2} \, dx+5 \int \exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) \, dx+6 \int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^3}{\left (e^{4 x}+x+x^2\right )^2} \, dx-\int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x}{\left (e^{4 x}+x+x^2\right )^2} \, dx+\int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right ) x^2}{\left (e^{4 x}+x+x^2\right )^2} \, dx+\int \frac {\exp \left (5+\frac {e^5 x \left (1+5 e^{4 x}+6 x+e^{4 x} x+6 x^2+x^3\right )}{e^{4 x}+x+x^2}\right )}{e^{4 x}+x+x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{5 e^5 x+e^5 x^2+\frac {e^5 x+e^5 x^2}{e^{4 x}+x+x^2}} \]

[In]

Integrate[(E^((E^(5 + 4*x)*(5*x + x^2) + E^5*(x + 6*x^2 + 6*x^3 + x^4))/(E^(4*x) + x + x^2))*(E^(5 + 8*x)*(5 +
2*x) + E^(5 + 4*x)*(1 + 8*x + 10*x^2 + 4*x^3) + E^5*(5*x^2 + 12*x^3 + 9*x^4 + 2*x^5)))/(E^(8*x) + x^2 + 2*x^3
+ x^4 + E^(4*x)*(2*x + 2*x^2)),x]

[Out]

E^(5*E^5*x + E^5*x^2 + (E^5*x + E^5*x^2)/(E^(4*x) + x + x^2))

Maple [A] (verified)

Time = 5.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62

method	result	size
parallelrisch	\({\mathrm e}^{\frac {{\mathrm e}^{5} x \left (x^{3}+x \,{\mathrm e}^{4 x}+6 x^{2}+5 \,{\mathrm e}^{4 x}+6 x +1\right )}{{\mathrm e}^{4 x}+x^{2}+x}}\)	\(42\)
risch	\({\mathrm e}^{\frac {x \left (x^{3} {\mathrm e}^{5}+6 x^{2} {\mathrm e}^{5}+{\mathrm e}^{5+4 x} x +6 x \,{\mathrm e}^{5}+5 \,{\mathrm e}^{5+4 x}+{\mathrm e}^{5}\right )}{{\mathrm e}^{4 x}+x^{2}+x}}\)	\(52\)

[In]

int(((5+2*x)*exp(5)*exp(4*x)^2+(4*x^3+10*x^2+8*x+1)*exp(5)*exp(4*x)+(2*x^5+9*x^4+12*x^3+5*x^2)*exp(5))*exp(((x
^2+5*x)*exp(5)*exp(4*x)+(x^4+6*x^3+6*x^2+x)*exp(5))/(exp(4*x)+x^2+x))/(exp(4*x)^2+(2*x^2+2*x)*exp(4*x)+x^4+2*x
^3+x^2),x,method=_RETURNVERBOSE)

[Out]

exp(exp(5)*x*(x^3+x*exp(4*x)+6*x^2+5*exp(4*x)+6*x+1)/(exp(4*x)+x^2+x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{\left (\frac {{\left (x^{4} + 6 \, x^{3} + 6 \, x^{2} + x\right )} e^{10} + {\left (x^{2} + 5 \, x\right )} e^{\left (4 \, x + 10\right )}}{{\left (x^{2} + x\right )} e^{5} + e^{\left (4 \, x + 5\right )}}\right )} \]

[In]

integrate(((5+2*x)*exp(5)*exp(4*x)^2+(4*x^3+10*x^2+8*x+1)*exp(5)*exp(4*x)+(2*x^5+9*x^4+12*x^3+5*x^2)*exp(5))*e
xp(((x^2+5*x)*exp(5)*exp(4*x)+(x^4+6*x^3+6*x^2+x)*exp(5))/(exp(4*x)+x^2+x))/(exp(4*x)^2+(2*x^2+2*x)*exp(4*x)+x
^4+2*x^3+x^2),x, algorithm="fricas")

[Out]

e^(((x^4 + 6*x^3 + 6*x^2 + x)*e^10 + (x^2 + 5*x)*e^(4*x + 10))/((x^2 + x)*e^5 + e^(4*x + 5)))

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{\frac {\left (x^{2} + 5 x\right ) e^{5} e^{4 x} + \left (x^{4} + 6 x^{3} + 6 x^{2} + x\right ) e^{5}}{x^{2} + x + e^{4 x}}} \]

[In]

integrate(((5+2*x)*exp(5)*exp(4*x)**2+(4*x**3+10*x**2+8*x+1)*exp(5)*exp(4*x)+(2*x**5+9*x**4+12*x**3+5*x**2)*ex
p(5))*exp(((x**2+5*x)*exp(5)*exp(4*x)+(x**4+6*x**3+6*x**2+x)*exp(5))/(exp(4*x)+x**2+x))/(exp(4*x)**2+(2*x**2+2
*x)*exp(4*x)+x**4+2*x**3+x**2),x)

[Out]

exp(((x**2 + 5*x)*exp(5)*exp(4*x) + (x**4 + 6*x**3 + 6*x**2 + x)*exp(5))/(x**2 + x + exp(4*x)))

Maxima [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{\left (x^{2} e^{5} + 5 \, x e^{5} - \frac {e^{\left (4 \, x + 5\right )}}{x^{2} + x + e^{\left (4 \, x\right )}} + e^{5}\right )} \]

[In]

integrate(((5+2*x)*exp(5)*exp(4*x)^2+(4*x^3+10*x^2+8*x+1)*exp(5)*exp(4*x)+(2*x^5+9*x^4+12*x^3+5*x^2)*exp(5))*e
xp(((x^2+5*x)*exp(5)*exp(4*x)+(x^4+6*x^3+6*x^2+x)*exp(5))/(exp(4*x)+x^2+x))/(exp(4*x)^2+(2*x^2+2*x)*exp(4*x)+x
^4+2*x^3+x^2),x, algorithm="maxima")

[Out]

e^(x^2*e^5 + 5*x*e^5 - e^(4*x + 5)/(x^2 + x + e^(4*x)) + e^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (23) = 46\).

Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.81 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx=e^{\left (\frac {x^{4} e^{5} + 6 \, x^{3} e^{5} + 6 \, x^{2} e^{5} + x^{2} e^{\left (4 \, x + 5\right )} + 5 \, x^{2} + x e^{5} + 5 \, x e^{\left (4 \, x + 5\right )} + 5 \, x + 5 \, e^{\left (4 \, x\right )}}{x^{2} + x + e^{\left (4 \, x\right )}} - 5\right )} \]

[In]

integrate(((5+2*x)*exp(5)*exp(4*x)^2+(4*x^3+10*x^2+8*x+1)*exp(5)*exp(4*x)+(2*x^5+9*x^4+12*x^3+5*x^2)*exp(5))*e
xp(((x^2+5*x)*exp(5)*exp(4*x)+(x^4+6*x^3+6*x^2+x)*exp(5))/(exp(4*x)+x^2+x))/(exp(4*x)^2+(2*x^2+2*x)*exp(4*x)+x
^4+2*x^3+x^2),x, algorithm="giac")

[Out]

e^((x^4*e^5 + 6*x^3*e^5 + 6*x^2*e^5 + x^2*e^(4*x + 5) + 5*x^2 + x*e^5 + 5*x*e^(4*x + 5) + 5*x + 5*e^(4*x))/(x^
2 + x + e^(4*x)) - 5)

Mupad [B] (verification not implemented)

Time = 11.67 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.46 \[ \int \frac {e^{\frac {e^{5+4 x} \left (5 x+x^2\right )+e^5 \left (x+6 x^2+6 x^3+x^4\right )}{e^{4 x}+x+x^2}} \left (e^{5+8 x} (5+2 x)+e^{5+4 x} \left (1+8 x+10 x^2+4 x^3\right )+e^5 \left (5 x^2+12 x^3+9 x^4+2 x^5\right )\right )}{e^{8 x}+x^2+2 x^3+x^4+e^{4 x} \left (2 x+2 x^2\right )} \, dx={\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}}\,{\mathrm {e}}^{\frac {x^4\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}}\,{\mathrm {e}}^{\frac {6\,x^2\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}}\,{\mathrm {e}}^{\frac {6\,x^3\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}}\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^5}{x+{\mathrm {e}}^{4\,x}+x^2}} \]

[In]

int((exp((exp(5)*(x + 6*x^2 + 6*x^3 + x^4) + exp(4*x)*exp(5)*(5*x + x^2))/(x + exp(4*x) + x^2))*(exp(5)*(5*x^2
+ 12*x^3 + 9*x^4 + 2*x^5) + exp(8*x)*exp(5)*(2*x + 5) + exp(4*x)*exp(5)*(8*x + 10*x^2 + 4*x^3 + 1)))/(exp(8*x
) + exp(4*x)*(2*x + 2*x^2) + x^2 + 2*x^3 + x^4),x)

[Out]

exp((5*x*exp(4*x)*exp(5))/(x + exp(4*x) + x^2))*exp((x^4*exp(5))/(x + exp(4*x) + x^2))*exp((6*x^2*exp(5))/(x +
exp(4*x) + x^2))*exp((6*x^3*exp(5))/(x + exp(4*x) + x^2))*exp((x^2*exp(4*x)*exp(5))/(x + exp(4*x) + x^2))*exp
((x*exp(5))/(x + exp(4*x) + x^2))