Is there a specific named function that is the inverse of $x+x^a$ for $x$ real? This seems such a simple question that I fear I must have missed some elementary maths.
I am looking for a way to solve $x+x^a = y$ by reference to an already defined function, $a,x,y > 0$ are real.
Failing that a reasonable approximation for $a$ in $(0,1)$.
Many thanks!
 A: The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright $_{1}\Psi_{1}$ function (a generalization of the confluent hypergeometric function $_{1}F_{1}$) providing the inverse function. See a previous answer here for details and references. For this particular case solution is
(Setting $\alpha = a$), for $\alpha>1$
$$x = y\cdot\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})$$ which can be set as $$x=y+\sum_{n=1}^\infty\binom{n\alpha}{n-1}\frac{(-1)^ny^{n(\alpha-1)+1}}{n}$$ whose convergence region is $|y^{\alpha-1}|<|(\alpha-1)^{\alpha-1}\alpha^{-\alpha}|$. For non integer $\alpha>1$ binomials must be set in terms of $\Gamma$ function.
Since Fox-Wright's generalized function can be expressed in terms of Fox-H function we have $$\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H_{1,2}^{1,1}([(0,\alpha)];[(0,1),(-1,\alpha-1)];y^{\alpha-1})$$ $$\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H_{2,1}^{1,1}([(1,1),(2,\alpha-1)];[(1,\alpha)];y^{1-\alpha})$$ for this particular case, Wolfram's Mathematica 12.3 provides an explicit inverse as
$x=y\cdot$FoxH[{{{0,$\alpha$}},{{}}},{{{0,1}},{{-1,$\alpha$-1}}},$y^{\alpha-1}$]
For $0<\alpha<1$ the solution is $$x = y\cdot(1-\,_{1}\Psi_{1}([1,\alpha^{-1}];[2,\alpha^{-1}-1];-y^{\alpha^{-1}-1}))$$
and the above relationships are turned in replacing $\alpha$ by $\alpha^{-1}$ and Fox-Wright function $\,_{1}\Psi_{1}$ by $1-\,_{1}\Psi_{1}$. In this case Mathematica's expression is
$x=y\cdot(1-$FoxH[{{{0,$\alpha^{-1}$}},{{}}},{{{0,1}},{{-1,$\alpha^{-1}$-1}}},$y^{\alpha^{-1}-1}$])
Finally, just to complement this answer, general trinomial equation solutions are developed in section 4 of the following
Reference
Miller A.R., Moskowitz I.S. Reduction of a Class of Fox-Wright Psi Functions for Certain Rational Parameters. Computers Math. Applic. Vol. 30, No. 11, pp. 73-82, (1995). Pergamon
A preprint can be found here. (Document has mis-embedded fonts, isolated commas must be replaced by $\Gamma$ symbol)
A: If $a$ is rational, then the root, say $x_*$, of your equation is algebraic, and (say) Mathematica will find for you with any degree of accuracy. Otherwise, one can approximate $a$ by rational numbers.

Another way to get bounds on $x_*$ is to use a combination of the Newton and secant methods to bracket the root, as follows. For $a\in(0,1)$, using the substitution $u=x^a$, rewrite your equation as
\begin{equation}
    f(u):=u^{1/a}+u=y, \tag{1}
\end{equation}
so that the function $f$ is convex and increasing. Let $u_*$ be the positive root of equation (1), so that $x_*=u_*^{1/a}$ and $f(u_*)=y$.
Note that
\begin{equation}
    u_*\vee u_*^{1/a}<y=f(u_*)<2(u_*\vee u_*^{1/a}),
\end{equation}
where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$. So, letting
\begin{equation}
    u_0:=u_0(y):=\frac y2\wedge\Big(\frac y2\Big)^a,\quad v_0:=v_0(y):=y\wedge y^a,
\end{equation}
we get the initial bracketing of $u_*$:
\begin{equation}
    u_0<u_*<v_0. 
\end{equation}
Use now the following combination of the secant and Newton methods for every natural $n$:
\begin{equation}
    u_n:=u_n(y):=U(u_{n-1},v_{n-1}),\quad v_n:=v_n(y):=V(v_{n-1}),
\end{equation}
where
\begin{equation}
    U(u,v):=u+\frac{y-f(u)}{f(v)-f(u)}\,(v-u),
\end{equation}
\begin{equation}
    V(v):=v-\frac{f(v)-y}{f'(v)}. 
\end{equation}
Then $u_n$ and $v_n$ bracket the root $u_*$ and monotonically converge to it (very fast):
\begin{equation}
    u_{n-1}<u_n<u_*<v_n<v_{n-1},\quad u_n\uparrow u_*,\quad v_n\downarrow u_*. 
\end{equation}
The bracketing $u_*\in[u_1,v_1]$ can be already pretty good, while providing almost digestible explicit lower and upper bounds on $u_*$.
As an illustration, here are the graphs $\{(y,\log_{10}(v_n(y)-u_n(y)))\colon0<y<3\}$ for $a=7/10$ with $n=1$ (blue), $n=2$ (orange), and $n=3$ (green):

