Hypergeometric differential equation with fractional polynomial coefficient

Question

I have a differential equation of the form $$ y''(x)+P(x)y'(x)+Q(x)y(x) =0 $$ where $P$ and $Q$ are rational functions $$ P(x)=\frac{c-2ia-(c+2b-2ia)x}{cx(1-x)}\\ Q(x)=\frac{(-1+(1-x)^{2/c})a^{2}+x(-b^{2}+2iab+2a^{2})+x^{2}(b-ia)}{c^{2}x^{2}(1-x)^{2}} $$ for $a \in \mathbb{C}, b \in \mathbb{R},$ and $c$ some positive rational number (in the case of interest, $c = 14/5$). The equation has three regular singular points at $x = 0,1, \infty$ which I believe means it can be transformed into a hypergeometric differential equation. However, I am having trouble doing this due to the term with the fractional exponent in $Q$. Does anyone know whether this equation has hypergeometric solutions, and if so how to find them? Thanks!