Prove that x is rational
Solution 1:
We may take $x>0$. $\frac{x^5+x}{x^3+x}$ is rational, so $bx^4+b = ax^2 +a$ for some integers $a,b$ and $x = \sqrt{k}$, where $k=c+\sqrt{d}$ is the root of a quadratic equation, with $c,d\in\mathbb{Q}$, $d=0$ or $d>0$ not a square.
If $k$ is rational ($d=0$) but not a square, $x^2-1$ is rational and $x^5+x = (x^3+x)(x^2-1)+2x$ is irrational, a contradiction.
If $k$ is irrational, ($d\neq 0$), then expanding $x^3+x$ we get
$$x^3+x = c^3+3cd+c + (3c^2+d+1)\sqrt{d}$$ is irrational, a contradiction.