Let us generalize the problem slightly and define:
\begin{equation}
V(x):= \frac{1}{\sqrt{a+(x-b)^2}}
\end{equation}
and now analyze the ODE in question.
The first sensible think that comes into mind is to substitute for $W(x):=1/V(x)$.
In other words we define $f(x):=W^{(-1)}(x)$ and change the abscissa $x\rightarrow f(x)$ and $d/d x \rightarrow 1/f^{'}(x) d/d x$. After having done that we reduce the ODE to a normal form (there is a standard way of doing it) by writing:
\begin{equation}
y(x)= \frac{\sqrt{x}}{(-a+x^2)^{1/4}} v(x)
\end{equation}
This results in a following ODE for the function $v(x)$. We have:
\begin{equation}
v^{''}(x) + \left(\frac{-3 a^2+6 a x^2+4 a B x^3-4 a A x^4-4 B x^5+4 A x^6}{4 x^2\left(x^2-a \right)^2}\right) v(x)=0 \quad (I)
\end{equation}
At the first glance mapping the above onto the hypergeometric equation would be hard because that equation in its normal form has polynomials of order two and six in the numerator and in the denominator respectively. Our equation is different than that so it will be hard to match the respective polynomials.
Let us try a different possibility then and substitute for $V(x)$. To reiterate this means that we define $f(x):=V^{(-1)}(x)$ and change the abscissa as above. Then we reduce the equation to its normal form by writing:
\begin{equation}
y(x)= \frac{1}{x(-1+a x^2)^{1/4}} v(x)
\end{equation}
This results in a following ODE for the function $v(x)$. We have:
\begin{equation}
v^{''}(x) + \left(\frac{4 A-4 B x-4 a A x^2+4 a B x^3+6 a x^4-3 a^2 x^6}{4 x^4 \left(a x^2-1\right)^2}\right) v(x)=0 \quad (II)
\end{equation}
Now the polynomial orders looks nicer and there is hope that this can be mapped onto the hypergeometric equation or its confluent versions.
Update: This is not an answer to this question but we found two ODEs whose solutions are known and which look remarkably similar to the ODEs above.
Firstly let us focus on the ODE $(II)$. We have.
Firstly:
\begin{eqnarray}
\!\!\!\!\!\!\!V^{''}(x) + \left(\frac{A^2-2 \sqrt{a} A^2 x+(-2+a A^2) x^2 - 8 \sqrt{a} x^3 + 4 a^{3/2} x^5 - 2 a^2 x^6}{x^4 (a x^2-1)^2}\right) \cdot V(x)=0 \quad (IIIa)
\end{eqnarray}
and
\begin{equation}
V(x):= \left( 1+\sqrt{a} x\right) x^2 \left( \frac{\sqrt{a} {\mathfrak W}(x) + (1-\sqrt{a} x) {\mathfrak W}^{'}(x)}{1-\sqrt{a} x}\right)
\end{equation}
where
\begin{eqnarray}
{\mathfrak W}(x):= x\left(
C_1 M_{\imath \sqrt{a} A,\sqrt{1-4 a A^2}/2}\left( \frac{2\imath A(1+\sqrt{a} x)}{x}\right)+
C_2 W_{\imath \sqrt{a} A,\sqrt{1-4 a A^2}/2}\left( \frac{2\imath A(1+\sqrt{a} x)}{x}\right)
\right)
\end{eqnarray}
where $M_{\cdot,\cdot}()$ and $W_{\cdot,\cdot}()$ are the Whittaker functions https://en.wikipedia.org/wiki/Whittaker_function . The code provides a "proof":
In[45]:= A =.; a =.; x =.; Clear[W];
W[x_] = (x) (C[1] WhittakerM[I Sqrt[a] A, 1/2 Sqrt[1 - 4 a A^2], (
2 I (A + Sqrt[a] A x))/x] +
C[2] WhittakerW[I Sqrt[a] A, 1/2 Sqrt[1 - 4 a A^2], (
2 I (A + Sqrt[a] A x))/x]);
P = {A^2, -2 Sqrt[a] A^2, -2 + a A^2, -8 Sqrt[a], 0,
4 a^(3/2), -2 a^2};
eX = (D[#, {x, 2}] +
Sum[P[[i + 1]] x^i, {i, 0,
6}]/( (x)^4 (a x^2 - 1)^2) #) & /@ {(1 +
Sqrt[a] x)^1 (x)^2 ((
Sqrt[a] W[x] + (1 - Sqrt[a] x) W'[x])/(1 - Sqrt[a] x))};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[50]= {(0.*10^-46 + 0.*10^-46 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}
Secondly:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) +
\left(\frac{A^2-\frac{4}{3} \sqrt{a} A^2 x+x^2 \left(-\frac{4 a A^2}{3}-6\right)+\frac{4}{3} \sqrt{a} x^3 \left(2 a A^2-9\right)+a x^4 \left(8-a A^2\right)+8 a^{3/2} x^5-6 a^2 x^6}
{x^4 (a x^2-1)^2}\right)\cdot V(x)=0 \quad (IIIb)
\end{eqnarray}
and
\begin{equation}
V(x):= {\mathfrak W}(x) + \frac{3 x^3 (1+\sqrt{a} x)}{A^2(-1+\sqrt{a} x)} {\mathfrak W}^{'}(x)
\end{equation}
where
\begin{eqnarray}
&&{\mathfrak W}(x):=\\
&&x\left( C_1 M_{-\frac{2}{3} i \sqrt{a} A,\frac{1}{2} \sqrt{\frac{8 a A^2}{3}+1}}\left(-\frac{2 i A \left(\sqrt{a} x+1\right)}{x}\right)+C_2 W_{-\frac{2}{3} i \sqrt{a} A,\frac{1}{2} \sqrt{\frac{8 a A^2}{3}+1}}\left(-\frac{2 i A
\left(\sqrt{a} x+1\right)}{x}\right) \right)
\end{eqnarray}
Again the code below provides the "proof":
In[3]:= A =.; a =.; x =.; Clear[W];
W[x_] = x (C[1] WhittakerM[-(2/3) I Sqrt[a] A,
1/2 Sqrt[1 + (8 a A^2)/3], -((2 I A (1 + Sqrt[a] x))/x)] +
C[2] WhittakerW[-(2/3) I Sqrt[a] A,
1/2 Sqrt[1 + (8 a A^2)/3], -((2 I A (1 + Sqrt[a] x))/x)]);
eX = (D[#, {x, 2}] + (
A^2 - 4/3 Sqrt[a] A^2 x + (-6 - (4 a A^2)/3) x^2 +
4/3 Sqrt[a] (-9 + 2 a A^2) x^3 + a (8 - a A^2) x^4 +
8 a^(3/2) x^5 - 6 a^2 x^6)/( x^4 (a x^2 - 1)^2) #) & /@ {W[
x] + (3 x^3 (1 + Sqrt[a] x))/(A^2 (-1 + Sqrt[a] x)) W'[x]};
{a, A, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[7]= {(0.*10^-45 + 0.*10^-44 I) C[1] + (0.*10^-45 + 0.*10^-45 I) C[
2]}
Thirdly:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) +
\left(\frac{A^2-2 \sqrt{a} A^2 x-2 x^2+2 \sqrt{a} x^3 \left(a A^2-2\right)-a x^4 \left(a A^2-3\right)+2 a^{3/2} x^5-2 a^2 x^6}
{x^4 (a x^2-1)^2}\right)\cdot V(x)=0 \quad (IIIc)
\end{eqnarray}
and
\begin{equation}
V(x):= w^{'}(x) x^2 \sqrt{\frac{1+\sqrt{a} x}{1-\sqrt{a} x}}
\end{equation}
and
\begin{eqnarray}
w(x):= x C_1 {\mathfrak W}_{\imath A \sqrt{a},-\frac{1}{2}} \left(\frac{2 \imath A (1+\sqrt{a} x)}{x} \right)
\end{eqnarray}
In here the 2nd solution needs to be derived separately by the Wronskian method.
In[127]:= A =.; x =.; a =.; Clear[V]; Clear[w];
w[x_] = x (C[1] WhittakerW[I A Sqrt[a], -1/2, (
2 I A (1 + Sqrt[a] x))/ x]);
V[x_] = (w'[x] x^2 Sqrt[1 + Sqrt[a] x]/Sqrt[1 - Sqrt[a] x]);
eX = (D[#, {x, 2}] + (
A^2 - 2 A^2 Sqrt[a] x - 2 x^2 + 2 Sqrt[a] (-2 + a A^2) x^3 -
a (-3 + a A^2) x^4 + 2 a^(3/2) x^5 - 2 a^2 x^6)/(
x^4 (-1 + a x^2)^2) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[132]= {(0.*10^-45 + 0.*10^-45 I) C[1]}
We obtained those results by starting from the Whittaker ODE then transforming the independent variable $x \rightarrow (A x+B)/(C x+D)$ and then gauge tranforming Gauge transformation of differential equations. the dependent variable $y(x) \rightarrow y(x) + r(x) \cdot y^{'}(x)$.
Now let us focus on the ODE $(I)$.
Firstly we have:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left(
\frac{-2 a^2-2 a^{3/2} x+a x^2 \left(3-a A^2\right)-2 \sqrt{a} x^3 \left(a A^2-2\right)-2 x^4+2 \sqrt{a} A^2 x^5+A^2 x^6}
{x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVa)
\end{eqnarray}
where
\begin{equation}
V(x):=\left(-\frac{w(x)}{x} + w^{'}(x)\right)\sqrt{\frac{x-\sqrt{a}}{x+\sqrt{a}}}
\end{equation}
and
\begin{equation}
w(x):=e^{i A \left(\sqrt{a}-x\right)} \left(c_1 U\left(i \sqrt{a} A,0,-2 i A \left(\sqrt{a}-x\right)\right)+c_2 L_{-i \sqrt{a} A}^{-1}\left(-2 i A \left(\sqrt{a}-x\right)\right)\right)
\end{equation}
where $U(\cdot,\cdot,x)$ and $L_n^{(a)}(x)$ is the confluent hypergeometric function and the Laguerre polynomials respectively.
Again the code below provides a "proof".
In[91]:= A =.; a =.; x =.; Clear[f]; Clear[V]; Clear[w];
w[x_] = E^(
I A (Sqrt[a] -
x)) (C[1] HypergeometricU[I Sqrt[a] A ,
0, -2 I A (Sqrt[a] - x)] +
C[2] LaguerreL[-I Sqrt[a] A , -1, -2 I A (Sqrt[a] - x)]);
V[x_] = (w[x]/(-x) + w'[x]) Sqrt[(x - Sqrt[a])/(x + Sqrt[a])];
P = {-8 a^2, -8 a^(3/2), 4 a (3 - a A^2 ), 8 Sqrt[a] (2 - a A^2 ), -8,
8 Sqrt[a] A^2 , 4 A^2 };
eX = (D[#, {x, 2}] + (-2 a^2 - 2 a^(3/2) x + a (3 - a A^2) x^2 -
2 Sqrt[a] (-2 + a A^2) x^3 - 2 x^4 + 2 Sqrt[a] A^2 x^5 +
A^2 x^6)/(-a x + x^3)^2 #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[97]= {(0.*10^-47 + 0.*10^-48 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}
Secondly we have:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left(
\frac{\frac{a^2}{4}+\frac{1}{2} a^{3/2} x \left(2 a A^2-1\right)+\frac{1}{4} a x^2 \left(4 a A^2-15\right)-2 \sqrt{a} x^3 \left(a A^2+1\right)-2 x^4 \left(a A^2+1\right)+\sqrt{a} A^2 x^5+A^2 x^6}
{x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVa)
\end{eqnarray}
where
\begin{equation}
V(x):=\left(\frac{w(x)}{\sqrt{a}-x} + w^{'}(x)\right)
\sqrt{\frac{x}{x+\sqrt{a}}}
\end{equation}
and
\begin{eqnarray}
w(x):=e^{-i A x} \left(C_1 U\left(\frac{1}{2} i \sqrt{a} A,0,2 i A x\right)+C_2 L_{-\frac{1}{2} i \sqrt{a} A}^{-1}(2 i A x)\right)
\end{eqnarray}
In[62]:= A =.; x =.; a =.; Clear[V]; Clear[w];
w[x_] = E^(-I A x) (C[1] HypergeometricU[1/2 I Sqrt[a] A, 0,
2 I A x] + C[2] LaguerreL[-(1/2) I Sqrt[a] A, -1, 2 I A x]);
V[x_] = (w[x]/(Sqrt[a] - x) + w'[x]) Sqrt[x/(Sqrt[a] + x)];
eX = (D[#, {x, 2}] + (
1/((-a x + x^3)^2))(a^2/4 + 1/2 a^(3/2) (-1 + 2 a A^2) x +
1/4 a (-15 + 4 a A^2) x^2 - 2 Sqrt[a] (1 + a A^2) x^3 -
2 (1 + a A^2) x^4 + Sqrt[a] A^2 x^5 + A^2 x^6) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[67]= {(0.*10^-47 + 0.*10^-47 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}
Thirdly we have:
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left(
\frac{-3/4 a^2 + 3/2 a^{3/2} x - 3/4 a x^2+\sqrt{a}(-6+a A^2) x^3+(-2-a A^2) x^4 - \sqrt{a} A^2 x^5+A^2 x^6}
{x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVc)
\end{eqnarray}
where
\begin{equation}
V(x):=\left(\frac{w(x)}{\sqrt{a}-x} + w^{'}(x)\right)
\sqrt{\frac{x+\sqrt{a}}{x}}
\end{equation}
and
\begin{eqnarray}
w(x):= e^{-i A \left(\sqrt{a}+x\right)} \left(C_1 U\left(-\frac{1}{2} i \sqrt{a} A,0,2 i A \left(x+\sqrt{a}\right)\right)+C_2 L_{\frac{1}{2} i \sqrt{a} A}^{-1}\left(2 i A \left(\sqrt{a}+x\right)\right)\right)
\end{eqnarray}
In[57]:= A =.; x =.; a =.; Clear[V]; Clear[w];
w[x_] = E^(-I A (Sqrt[a] +
x)) (C[1] HypergeometricU[-(1/2) I Sqrt[a] A, 0,
2 I A (Sqrt[a] + x)] +
C[2] LaguerreL[1/2 I Sqrt[a] A, -1, 2 I A (Sqrt[a] + x)]);
V[x_] = (w[x]/(Sqrt[a] - x) + w'[x]) Sqrt[(Sqrt[a] + x)/x];
eX = (D[#, {x, 2}] + (-((3 a^2)/4) + 3/2 a^(3/2) x - (3 a x^2)/4 +
Sqrt[a] (-6 + a A^2) x^3 + (-2 - a A^2) x^4 -
Sqrt[a] A^2 x^5 + A^2 x^6)/( x^2 (a - x^2)^2) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]
Out[62]= {(0.*10^-39 + 0.*10^-39 I) C[
1] + (0.*10^-45 + 0.*10^-45 I) C[2]}
Fourthly we have:
\begin{eqnarray}
V^{''}(x) + \left( \frac{-2 a^2-4 a^{3/2} x+8 \sqrt{a} x^3+x^4 \left(-a A^2-2\right)-2 \sqrt{a} A^2 x^5-A^2 x^6}{x^2 \left(x^2-a\right)^2} \right) V(x)=0\quad (IVd)
\end{eqnarray}
where
\begin{equation}
V(x):=(\sqrt{a}-x) \left( \frac{\sqrt{a}}{x(\sqrt{a}+x)} w(x) + w^{'}(x)\right)
\end{equation}
and
\begin{eqnarray}
w(x):=\frac{1}{x}\left( C_1 M_{-A \sqrt{a},1/2\sqrt{1+4 a A^2}}(2 A(x-\sqrt{a}))+
C_1 W_{-A \sqrt{a},1/2\sqrt{1+4 a A^2}}(2 A(x-\sqrt{a}))
\right)
\end{eqnarray}
In[760]:= A =.; B =.; CC =.; DD =.;
k =.; mu =.; x0 =.; x1 =.; a =.; Clear[f]; Clear[y]; Clear[w]; \
Clear[V]; Clear[m]; Clear[g]; x =.;
n = 2; CC = 0; DD =.;
k = -A Sqrt[a];
mu = Sqrt[a (A 2)^2 + 1]/2;
w[x_] = 1/
x (C[1] WhittakerM[k, mu, -Sqrt[a] A 2 + A 2 x] +
C[2] WhittakerW[k, mu, -Sqrt[a] A 2 + A 2 x]);
V[x_] = (Sqrt[a] - x) (Sqrt[a]/(x (Sqrt[a] + x)) w[x] + w'[x]);
eX = (D[#, {x, 2}] + (-2 a^2 - 4 a^(3/2) x +
8 Sqrt[a] x^3 + (-2 - a (A)^2) x^4 -
Sqrt[a] 2 (A)^2 x^5 - (A)^2 x^6)/( (
x^2) ((x^2 - a)^2) ) #) & /@ {V[x]};
{a, DD, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
Simplify[eX]
Out[769]= {(0.*10^-46 + 0.*10^-46 I) C[
1] + (0.*10^-45 + 0.*10^-45 I) C[2]}
Our results are only particular solutions and we believe there are more solutions that still can be found.
