add mathastext compat (#663)

_data/tagging-status.yml

@@ -5943,13 +5943,14 @@
 
  - name: mathastext
    type: package
-   status: unknown
+   status: compatible
    included-in: [arxiv001]
    priority: 7
+   supported-through: [phase-III,math]
+   comments: "Use of math tagging currently requires support from external tools."
    issues:
-   tests: false
-   tasks: needs tests
-   updated: 2024-07-18
+   tests: true
+   updated: 2024-08-23
 
  - name: mathbbol
    type: package

tagging-status/testfiles/mathastext/mathastext-01.tex

@@ -0,0 +1,46 @@
+% sample taken from mathastext.dtx
+\DocumentMetadata
+  {
+    lang=en-US,
+    pdfversion=2.0,
+    pdfstandard=ua-2,
+    testphase={phase-III,math,title,table,firstaid}
+  }
+\documentclass{article}
+
+\usepackage[T1]{fontenc}
+\usepackage[default]{droidserif}
+\usepackage[LGRgreek]{mathastext}
+\let\varepsilon\epsilon
+
+\title{mathastext tagging test - pdftex}
+
+\begin{document}
+
+Let $(X,Y)$ be two functions of a variable $a$. If they obey  the differential
+system  $(VI_{\nu,n})$: 
+\begin{align*}
+a\frac{d}{da} X &= \nu
+X - (1 - X^2)\frac{2n a}{1 - a^2}\frac{aX+Y}{1+a XY} \\  
+a\frac{d}{da} Y &= -(\nu+1) Y
++ (1 - Y^2)\frac{2n a}{1 - a^2}\frac{X+aY}{1+a XY} 
+\end{align*}
+then the quantity $q = a \frac{aX+Y}{X+aY}$
+satisfies as function of $b= a^2$  the $P_{VI}$ differential equation:
+\begin{equation*}
+\begin{split}
+\frac{d^2 q}{db^2} = \frac12\left\{\frac1q+\frac1{q-1}
++\frac1{q-b}\right\}\left(\frac{dq}{db}\right)^2 - \left\{\frac1b+\frac1{b-1}
++\frac1{q-b}\right\}\frac{dq}{db}\\+\frac{q(q-1)(q-b)}{b^2(b-1)^2}\left\{\alpha+\frac{\beta
+b}{q^2} + \frac{\gamma (b-1)}{(q-1)^2}+\frac{\delta
+b(b-1)}{(q-b)^2}\right\}
+\end{split}
+\end{equation*}
+with
+parameters
+$(\alpha,\beta,\gamma,\delta) = (\frac{(\nu+n)^2}2,
+\frac{-(\nu+n+1)^2}2, \frac{n^2}2, \frac{1 - n^2}2)$.
+
+Test of uppercase Greek in math: $\Alpha\Beta\Gamma\Delta\Xi\Omega$.
+
+\end{document}

tagging-status/testfiles/mathastext/mathastext-02.tex

@@ -0,0 +1,47 @@
+% sample taken from mathastext.dtx
+\DocumentMetadata
+  {
+    lang=en-US,
+    pdfversion=2.0,
+    pdfstandard=ua-2,
+    testphase={phase-III,math,title,table,firstaid}
+  }
+\documentclass{article}
+
+\usepackage[no-math]{fontspec}
+\setmainfont[Ligatures=TeX]{Libertinus Serif}
+\usepackage[defaultmathsizes,LGRgreek]{mathastext}
+\MTgreekfont{LibertinusSerif-TLF}
+\Mathastext
+
+\title{mathastext tagging test - luatex/xetex}
+
+\begin{document}
+
+Let $(X,Y)$ be two functions of a variable $a$. If they obey  the differential
+system  $(VI_{\nu,n})$: 
+\begin{align*}
+a\frac{d}{da} X &= \nu
+X - (1 - X^2)\frac{2n a}{1 - a^2}\frac{aX+Y}{1+a XY} \\  
+a\frac{d}{da} Y &= -(\nu+1) Y
++ (1 - Y^2)\frac{2n a}{1 - a^2}\frac{X+aY}{1+a XY} 
+\end{align*}
+then the quantity $q = a \frac{aX+Y}{X+aY}$
+satisfies as function of $b= a^2$  the $P_{VI}$ differential equation:
+\begin{equation*}
+\begin{split}
+\frac{d^2 q}{db^2} = \frac12\left\{\frac1q+\frac1{q-1}
++\frac1{q-b}\right\}\left(\frac{dq}{db}\right)^2 - \left\{\frac1b+\frac1{b-1}
++\frac1{q-b}\right\}\frac{dq}{db}\\+\frac{q(q-1)(q-b)}{b^2(b-1)^2}\left\{\alpha+\frac{\beta
+b}{q^2} + \frac{\gamma (b-1)}{(q-1)^2}+\frac{\delta
+b(b-1)}{(q-b)^2}\right\}
+\end{split}
+\end{equation*}
+with
+parameters
+$(\alpha,\beta,\gamma,\delta) = (\frac{(\nu+n)^2}2,
+\frac{-(\nu+n+1)^2}2, \frac{n^2}2, \frac{1 - n^2}2)$.
+
+Test of uppercase Greek in math: $\Alpha\Beta\Gamma\Delta\Xi\Omega$.
+
+\end{document}