What is the inverse of $\sqrt{ax+bx^4}$? I am interested in the inverse function of:
$$
y=\sqrt{ax+bx^4}
$$
Where $a,b,x \in [0,1]$.Furthermore, $a+b=1$ but I’m not so sure that’s as relevant. This came up in the course of (physics) research, and I have to admit it is not immediately clear how to proceed. Wolfram Alpha gives a closed form but it is extremely convoluted and I don’t trust that that form is as simple as it could be. What is the general approach for finding an inverse like this, and does this one have a ‘neat’ form?
 A: The general procedure is: solve for $x$ as a function of $y$. There's a problem, though, which is that it's hard to solve equations sometimes. This one isn't so bad, because first you can square both sides to get $y^2 = ax + b x^4$, and then you can apply the general formula for roots of quartic equations. The complexity of that formula probably accounts for the "extremely convoluted" solution you got from Wolfram Alpha; good luck on finding a simpler formula. 
Just be happy that the expression under the radical wasn't a general quintic polynomial $a + bx + cx^2 + dx^3 + ex^4 + dx^5$: a famous theorem which one learns in a course on Galois Theory says that there is no general formula whatsoever, expressed solely in terms of arithmetic and radicals, for solutions to quintic equations. 
The moral of the story is: there is no "general approach for finding an inverse like this", in the sense that there is no general approach which solves all equations. Every equation to solve can present one with a new mathematical adventure.
A: If this comes from a problem in physics, I suppose that approximations could be sufficient.
Expanding as Taylor series around $x=0$ gives
$$\sqrt{ax+bx^4}=\sqrt{ax} +\frac{b x^{7/2}}{2 \sqrt{a}}-\frac{b^2 x^{13/2}}{8
   a^{3/2}}+\frac{b^3 x^{19/2}}{16 a^{5/2}}+O\left(x^{25/2}\right)$$ Now, using series reversion
$$x=\frac{y^2}{a}-\frac{b y^8}{a^5}+\frac{4 b^2 y^{14}}{a^9}-\frac{22 b^3
   y^{20}}{a^{13}}+O\left(y^{26}\right)$$
Another possible solution could be using Padé approximants. This would give for example, as another  approximation,
$$\frac{6 \sqrt{a}  \left(a^4-b y^6\right)}{6 a^4+5 b y^6}\sqrt{x}+\frac{ y^5
   \left(56 a^4 b-b^2 y^6\right)}{4 a^3 \left(6 a^4+5 b y^6\right)}x-y=0$$ which is a quadratic in $\sqrt{x}$.
