Yes, it is (how simple, is a matter of opinion). You may always assume that $0,1,\infty$ on the boundary of the half-plane
are preimages of the vertices. Now suppose that the inner angles of your triangle
are $\pi\alpha_j,$ and let us assume that $\sum\alpha_j$ is not an odd integer (Euclidean triangles must be considered separately and this case is in fact simpler.

Then your mapping function is a solution of the Schwarz differential equation
$$\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2
=\frac{1-\alpha_1^2}{2z^2}+\frac{1-\alpha_2^2}{2(z-1)^2}+\frac{\alpha_1^2+\alpha_2^2-\alpha_3^2-1}{2z(z-1)}.$$
This is written in Hurwitz-Courant (with a misprint which I corrected), and in
Caratheodory, vol. II.   

This is not as scary as it may look, because in fact if $F(z)$ is the right hand-side, then $f=y_1/y_2$ where $y_1$ and $y_2$ are linearly independent solutions of the linear differential equation
$$y''+(F/2)w=0,$$
which in our case is a hypergeometric equation. Its solutions (hypergeometric functions with real parameters) are special functions, and "everything is known" about them. I mean explicit power series, integral representations, tables, and so on.
Function $f$ contains 3 arbitrary constants which can be determined from positions
of vertices of your triangle.

For the case of Euclidean triangle (bounded by straight lines, and with $\sum\alpha_j=1$) the function is more explicit: it is the Schwarz--Christoffel integral
$$f(z)=C\int_a^z\zeta^{\alpha_1-1}(\zeta-1)^{\alpha_2-1}d\zeta$$
where $a$ and $C$ are constants, and can be determined if you positions of two  vertices of your triangle.

Refs. A. Hurwitz and R. Courant, Vorlesungen uber allgemeine Funktionentheorie,
(available in German and Russian)

C. Caratheodory, Theory of functions of a complex variable, vol. II (available in German and English). Most of the vol. II is devoted to this subject.

F. Klein, Vorlesungen über die Hypergeometrische Funktion, Berlin 1033 (the whole book is devoted to the subject).