If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$ 
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$

My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$
Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.
Any help will be appreciated.
Note: Please don't use root tables and complex numbers too!

I used Sage cell to calculate this.
var("X")
F.<x> = NumberField(X^5 - 2)
y = x/(x^2 + 1) + x^2/(x^4 + 1) + x^3/(x + 1) + x^4/(x^3 + 1)
print(y)
print(y.minpoly())

Output:
-44/765*x^4 + 122/765*x^3 + 79/765*x^2 + 233/765*x + 1336/765
x^5 - 1336/153*x^4 + 1556/51*x^3 - 8182/153*x^2 + 7210/153*x - 12806/765

In the output, the first line is the sum, and the second line is the minimal polynomial of that sum.
None of these is nice enough to be an intended answer. Thus I suspect that the question is slightly wrong.

One possibility is to change $x^5 = 2$ to $x^5 = 1$. Then it is easy to evaluate that the sum is equal to $2$.
