Suppose that $X$ and $Y$ are independent beta random variables (r.v.'s) with parameters $(a,b)$ and $(c,d)$, respectively. Let \begin{equation*} V:=\frac X{X+Y}. \tag{0} \end{equation*} The transformation $(x,v)\mapsto(x,x\frac{1-v}v)$ transforms $(X,V)$ to $(X,Y)$. The Jacobian determinant of this transformation is $-x/v^2$. Also, \begin{equation*} \Big(x,x\frac{1-v}v\Big)\in(0,1)\times(0,1) \iff (0 < v<1\ \&\ 0 < x < s), \end{equation*} where \begin{equation*} s:=1\wedge\frac1r=\min\Big(1,\frac1r\Big),\quad r:=r_v:=\frac{1-v}v. \end{equation*} The joint pdf of $(X,Y)$ is given by \begin{equation*} f_{X,Y}(x,y)=Cx^{a-1}(1-x)^{b-1}y^{c-1}(1-y)^{d-1}\,I\{0<x<1\ \&\ 0<y<1\}, \end{equation*} where $I$ denotes the indicator and \begin{equation*} C:=\frac1{B(a,b)B(c,d)}. \end{equation*} So, joint pdf of $(X,V)$ is given by \begin{align*} f_{X,V}(x,v)&=f_{X,Y}(x,xr)x/v^2 \\ &= Cx^{a+c-1}(1-x)^{b-1}v^{-c-1}(1-v)^{c-1}(1-rx)^{d-1} \\ &\times I\{0 < v<1\ \&\ 0 < x < s\}. \end{align*} So, the pdf of $V$ is given by \begin{align*} f_V(v)&=\int_{\mathbb R}f_{X,V}(x,v)\,dx \\ & =Cv^{-c-1}(1-v)^{c-1}J(v)I\{0 < v<1\}, \tag{1} \end{align*} where \begin{equation*} J(v):=\int_0^s x^{a+c-1}(1-x)^{b-1}(1-rx)^{d-1}\,dx. \end{equation*} To evaluate $J(v)$, use formula 3.197.3 of Table of Integrals, Series, and Products, Seventh Edition by Gradshteyn and Ryzhik: \begin{equation*} \int_0^1 x^{\lambda-1}(1-x)^{\mu-1}(1-\beta x)^{-\nu}\,dx =B(\lambda,\mu)\,_2F_1(\nu,\lambda;\lambda+\mu;\beta) \tag{*} \end{equation*} for $\lambda>0$, $\mu>0$, $|\beta|<1$, where $_2F_1$ is the hypergeometric function.
If $1/2<v<1$, then $0<r<1$, $s=1$, and (*) immediately yields \begin{equation*} J(v)=B(a+c,b)\,_2F_1(1-d,a+c;a+c+b;r). \tag{2} \end{equation*}
If $0<v<1/2$, then $r>1$, $s=1/r\in(0,1)$, and the substitution $t=rx$ together with (*) yields \begin{align*} J(v)&=r^{-a-c}\int_0^1 t^{a+c-1}(1-t/r)^{b-1}(1-t)^{d-1}\,dx \\ &=r^{-a-c}B(a+c,d)\,_2F_1(1-b,a+c;a+c+d;1/r). \tag{3} \end{align*}
Formulas (1), (2) (for $v\in(1/2,1)$), and (3) (for $v\in(0,1/2)$) give the pdf $f_V$ of $V$, defined by (0).