To begin, a partial/half answer based on the summation formula of Martin Rubey, given that the two-fold integration result in the Matt F. streamlined reformulation takes the form
\begin{equation}
v(b,\mu) + w(b,\mu) \log(\mu) = \int_{p=0}^1 \int_{q=0}^{1-p} (\mu p q (1-p-q))^b (\mu^2 q +p)^{-b-1} dq dp,
\end{equation}
is that (in line with the "hypergeometric" comment of Nemo above)
\begin{equation}
w(b,\mu)=\frac{\sqrt{\pi } 4^{-b} \mu^b \left(\mu^2-1\right)^{-2 b-1} \Gamma (b+1) \,
_2F_1\left(-b,-b;1;\mu^2\right)}{\Gamma \left(b+\frac{3}{2}\right)}.
\end{equation}
The first term in the commented Rubey summation formula, $\frac{b \left(\mu ^2-1\right)^{-2 b-1} \, _2F_1\left(-b,-b;1;\mu ^2\right)}{4 \left(4
b^2-1\right) \binom{2 (b-1)}{b-1}}$, for $w(b,\mu)$ above (with $m$ corrected to $\mu$) required an additional factor of $4 \mu^b$ (as Ruhey noted in a subsequent comment). (Mathematica insists, it seems, on converting his summation to hypergeometric form. It would be of interest to know how Ruhey arrived at his formula--and if it might be of potential use in obtaining $v(b,\mu)$, as well.)

We have, additionally, trying to address the other "half" of the problem, found that
\begin{equation}
v(b,1)=\frac{\pi 4^{-2 b-1} \Gamma (b+1)^2}{\Gamma \left(b+\frac{3}{2}\right)^2}.
\end{equation}

Further, for $v(b,\frac{1}{2})$, Mathematica has given

```
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {(1+\[FormalN])^2
\[FormalY][\[FormalN]] -
5 (3+2 \[FormalN])^2 \[FormalY][
1 + \[FormalN]] + (135 + 144 \[FormalN] +
36 \[FormalN]^2) \[FormalY][2 + \[FormalN]] ==
0, \[FormalY][1] == -(8/27), \[FormalY][2] == -(8/81)}]][b]
```

The TeX version of this that Mathematica gives is not accepted here because the "Argument to unicode must be a number". (This appears to be a problem with the TeXForm command of Mathematica https://mathematica.stackexchange.com/questions/191414/nonacceptance-by-stackexchange-site-of-mathematica-texform-employing-unicode .)

Along similar lines, for $v(b,1/3)$, we have

```
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {9 (1 + \[FormalN])^2
\[FormalY][\[FormalN]] -
60 (3 + 2 \[FormalN])^2 \[FormalY][
1 + \[FormalN]] + (3840 + 4096 \[FormalN] +
1024 \[FormalN]^2) \[FormalY][2 + \[FormalN]] ==
0, \[FormalY][1] == -(9/64), \[FormalY][2] == -(81/4096)}]][b]
```

So, "DifferenceRoot" formulas appear to be the case for other specific values of $v(b,\mu)$ with $\mu \neq 1$.

Perhaps, we will attempt ("manually") to present the DifferenceRoot formulas above in more standard, proper TeX form (per comment of Somos in https://mathematica.stackexchange.com/questions/191414/nonacceptance-by-stackexchange-site-of-mathematica-texform-employing-unicode ).

It may be helpful to note that the inner integral of the Matt F. reformulation can be performed, yielding
\begin{equation}
\int_{q=0}^{1-p} (\mu p q (1-p-q))^b (\mu^2 q +p)^{-b-1} dq =-\frac{(p-1) \Gamma (b+1)^2 \left((p-1)^2 \mu\right)^b \, _2\tilde{F}_1\left(b+1,b+1;2
(b+1);\frac{(p-1) \mu^2}{p}\right)}{p},
\end{equation}
where the regularized hypergeometric function is indicated.

So, we would like the counterpart for $v(b,\mu)$ of the Rubey formula for $w(b,\mu)$, that is,
\begin{equation}
4 u^b \left(u^2-1\right)^{-2 b-1} \frac{1}{4 \left(4 b^2-1\right)} \frac{b} {\binom{2 (b-1)}{b-1}} \Sigma_{k=0}^b u^{2 k} \binom{b}{k}^2,
\end{equation}
which he apparently obtained using the general purpose computer algebra system, FriCAS.

A new, interesting observation is that if we multiply $v(b,\mu)$ by
\begin{equation}
(-1+\mu^2)^{2 b+1} \mu^{-b} P,
\end{equation}
where, with $\gamma$ being Euler's constant, $\psi ^{(0)}(b+1)$ being the zero-th derivative of the digamma function, and $H_b$, the $b$-th harmonic number,
\begin{equation}
P= \frac{\sqrt{\pi } 2^{-2 b} b! (\psi ^{(0)}(b+1)+\gamma )}{\left(b+\frac{1}{2}\right)!} = \frac{\sqrt{\pi } 4^{-b} b! H_b}{\Gamma \left(b+\frac{3}{2}\right)},
\end{equation}
we obtain ("asymmetric"-type) polynomials in $\mu$ of degree $2 b$, all with constant terms equal to 1 and highest terms equal to $-u^{2 b}$. For example, for $b=1$, we have
\begin{equation}
1-\mu^2
\end{equation}
for $b=2$,
\begin{equation}
1-\mu^4
\end{equation}
for $b=3$,
\begin{equation}
-\mu ^6-\frac{27 \mu ^4}{11}+\frac{27 \mu ^2}{11}+1
\end{equation}
for $b=4$,
\begin{equation}
-\mu ^8-\frac{32 \mu ^6}{5}+\frac{32 \mu ^2}{5}+1,
\end{equation}
for $b=5$,
\begin{equation}
-\mu ^{10}-\frac{1625 \mu ^8}{137}-\frac{2000 \mu ^6}{137}+\frac{2000 \mu
^4}{137}+\frac{1625 \mu ^2}{137}+1, \ldots
\end{equation}
So, one needs to find the governing rule for these polynomials, which would complete the formula for $v(b,\mu)$.

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