Un lugar para comenzar:
Creo que la búsqueda de una más amigable la forma cerrada podrían beneficiarse de la ruptura de la serie en más componentes manejables. Empezar por la división de la suma en la suma de impares y pares de términos:
$$\begin{align}
S
&=\sum_{p=1}^{\infty}\frac{\psi{\left(\frac{p+1}{2}\right)}}{\binom{2p}{p}}\\
&=\sum_{n=1}^{\infty}\frac{\psi{\left(\frac{(2n-1)+1}{2}\right)}}{\binom{2(2n-1)}{(2n-1)}}+\sum_{n=1}^{\infty}\frac{\psi{\left(\frac{(2n)+1}{2}\right)}}{\binom{2(2n)}{(2n)}}\\
&=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n-2}{2n-1}}+\sum_{n=1}^{\infty}\frac{\psi{\left(n+\frac12\right)}}{\binom{4n}{2n}}\\
&=:S^{(-)}+S^{(+)}.\\
\end{align}$$
La proporción de los respectivos denominadores de los sumandos de $S^{(-)}$ $S^{(+)}$ resulta ser una simple función racional de $n$:
$$\frac{\binom{4n}{2n}}{\binom{4n-2}{2n-1}}=4-\frac{1}{n}.$$
Esto nos permite reescribir la suma de términos raros como,
$$\begin{align}
S^{(-)}
&=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n-2}{2n-1}}\\
&=\sum_{n=1}^{\infty}\frac{\left(4-\frac{1}{n}\right)\psi{\left(n\right)}}{\binom{4n}{2n}}\\
&=4\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{n\binom{4n}{2n}}\\
&=:4A-B.\\
\end{align}$$
La suma de condiciones puede ser dividido utilizando la función digamma identidad,
$$\psi{\left(n+\frac12\right)}=2\psi{\left(2n\right)}-\psi{\left(n\right)}-\ln{(4)}.$$
A continuación,
$$\begin{align}
S^{(+)}
&=\sum_{n=1}^{\infty}\frac{\psi{\left(n+\frac12\right)}}{\binom{4n}{2n}}\\
&=\sum_{n=1}^{\infty}\frac{2\psi{\left(2n\right)}-\psi{\left(n\right)}-\ln{(4)}}{\binom{4n}{2n}}\\
&=2\sum_{n=1}^{\infty}\frac{\psi{\left(2n\right)}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n}{2n}}-\ln{(4)}\sum_{n=1}^{\infty}\frac{1}{\binom{4n}{2n}}\\
&=:2C-A-\ln{(4)}D.\\
\end{align}$$
Al menos podemos encontrar una bonita forma cerrada para la suma de $D$ en la última línea de arriba:
$$D=\sum_{n=1}^{\infty}\frac{1}{\binom{4n}{2n}}=\frac{1}{15}+\frac{\sqrt{3}\,\pi}{27}-\frac{2\sqrt{5}}{25}\ln{\phi}\approx 0.182118141166556731177.$$
$$-\ln{(4)}\,D\approx -0.25246935215683381486272395984234.$$
Queda por evaluar las tres series $A,B,C$. Tenga en cuenta que las tres series caen bajo la forma general de
$$\sigma{\left(r,s\right)}:=\sum_{n=1}^{\infty}n^{s}\,\frac{\psi{\left(r\,n\right)}}{\binom{4n}{2n}}.$$
A continuación,
$$\begin{cases}
A=\sigma{\left(1,0\right)}\approx -0.08905771667254809;\\
B=\sigma{\left(1,-1\right)}\approx -0.0928236851612787689;\\
C=\sigma{\left(2,0\right)}\approx 0.0904248179545543815129359182;\\
3A-B+2C\approx 0.0065001710527432522437457174000.\\
\end{casos}$$
Integral de las representaciones:
Por conveniencia de notación, podemos definir tres parámetros auxiliar de la serie de $\tau{\left(r,s,t\right)}$,
$$\tau{\left(r,s,t\right)}:=\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}.$$
También vamos a seguir adelante y definir el parámetro de una serie de $\tau{\left(s\right)}$ en términos de $\tau{\left(r,s,t\right)}$ tomando $r=2$$t=1$:
$$\tau{\left(s\right)}:=\tau{\left(2,s,1\right)}=\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}.$$
Para $\Re{(z)}>0$, la digamma función tiene la siguiente representación integral:
$$\psi{\left(z\right)}=-\gamma+\int_{0}^{1}\frac{1-t^{z-1}}{1-t}\,\mathrm{d}t.$$
Los coeficientes binomiales puede ser escrito en términos de funciones gamma o beta funciones:
$$\begin{align}
\binom{z}{w}
&=\frac{z!}{w!(z-w)!};~z,w\in\mathbb{N}^{+};z\ge w\\
&=\frac{\Gamma{\left(z+1\right)}}{\Gamma{\left(w+1\right)}\Gamma{\left(z-w+1\right)}}\\
&=\frac{1}{w\operatorname{B}{\left(w,z-w+1\right)}}.\\
\end{align}$$
A continuación, podemos encontrar una sola representación integral para recíprocos de los coeficientes binomiales en términos de una que de la función beta:
$$\begin{align}
\frac{1}{\binom{z}{w}}
&=w\operatorname{B}{\left(w,z-w+1\right)}\\
&=w\int_{0}^{1}t^{w-1}\left(1-t\right)^{z-w}\,\mathrm{d}t;~\Re{(w)}>0\land\Re{(z)}>\Re{(w)}-1.\\
\end{align}$$
A partir de la representación integral de la función digamma, podemos llegar a una representación de la serie de $\sigma{\left(r,s\right)}$ (y por tanto, para la serie de $S$) en términos de la serie $\tau{\left(r,s,t\right)}$ a través de la sumatoria bajo el signo integral:
$$\begin{align}
\sigma{\left(r,s\right)}
&=\sum_{n=1}^{\infty}n^{s}\,\frac{\psi{\left(r\,n\right)}}{\binom{4n}{2n}}\\
&=\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}\left[-\gamma+\int_{0}^{1}\frac{1-t^{rn-1}}{1-t}\,\mathrm{d}t\right]\\
&=-\gamma\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}+\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}\int_{0}^{1}\frac{1-t^{rn-1}}{1-t}\,\mathrm{d}t\\
&=-\gamma\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}+\int_{0}^{1}\frac{\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}}{1-t}\,\mathrm{d}t\\
&=-\gamma\,\tau{\left(s\right)}+\int_{0}^{1}\frac{\tau{\left(s\right)}-\tau{\left(r,s,t\right)}}{1-t}\,\mathrm{d}t,\\
\end{align}$$
$$\begin{align}
\implies S
&=3A-B+2C-2\ln{(2)}D\\
&=3\sigma{\left(1,0\right)}-\sigma{\left(1,-1\right)}+2\sigma{\left(2,0\right)}-2\ln{(2)}D\\
&=3\left[-\gamma\,\tau{\left(0\right)}+\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t\right]+2\left[-\gamma\,\tau{\left(0\right)}+\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t\right]\\
&~~~~~ -\left[-\gamma\,\tau{\left(-1\right)}+\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t\right]-2\ln{(2)}\tau{\left(0\right)}\\
&=-3\gamma\,\tau{\left(0\right)}+3\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t-2\gamma\,\tau{\left(0\right)}+2\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t\\
&~~~~~ +\gamma\,\tau{\left(-1\right)}-\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t-2\ln{(2)}\tau{\left(0\right)}\\
&=-\left[5\gamma+2\ln{(2)}\right]\tau{\left(0\right)}+\gamma\,\tau{\left(-1\right)}+3\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t\\
&~~~~~ +2\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t-\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t.\\
\end{align}$$
Una representación integral para $\tau{\left(r,s,t\right)}$ también sería útil. Podemos partir de la representación integral de la reciprocidad coeficiente binomial $\frac{1}{\binom{4n}{2n}}$:
$$\begin{align}
\frac{1}{\binom{4n}{2n}}
&=2n\operatorname{B}{\left(2n,2n+1\right)}\\
&=\left(4n+1\right)\operatorname{B}{\left(2n+1,2n+1\right)}\\
&=\left(4n+1\right)\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u.\\
\end{align}$$
A continuación,
$$\begin{align}
\frac{n^{s}}{\binom{4n}{2n}}
&=\left(4n+1\right)n^{s}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=4n^{s+1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+n^{s}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u,\\
\end{align}$$
y,
$$\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}=4n^{s+1}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+n^{s}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u.$$
Suma más de $n$ y cambiar el orden de la suma y de la integración, llegamos a una representación integral para $\tau{\left(r,s,t\right)}$:
$$\begin{align}
\tau{\left(r,s,t\right)}
&=\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}\\
&=\sum_{n=1}^{\infty}4n^{s+1}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\sum_{n=1}^{\infty}n^{s}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=4\int_{0}^{1}\sum_{n=1}^{\infty}n^{s+1}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}r\sum_{n=1}^{\infty}n^{s}\cdot nt^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}t\sum_{n=1}^{\infty}n^{s}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\cdot rnt^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\frac{\partial}{\partial t}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\sum_{n=1}^{\infty}n^{s}\left[t^{r}u^{2}\left(1-u\right)^{2}\right]^n\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\left[t^{r}u^{2}\left(1-u\right)^{2}\right]^n\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u\\
&=\frac{4}{r}\int_{0}^{1}\frac{r}{t}\operatorname{Li}_{-s-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u\\
&=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{-s-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u,\\
\end{align}$$
donde$r>0$$t>0$.
Cálculo de los principales integral:
Por el bien de este problema, sólo tenemos que evaluar $\tau{\left(r,s,t\right)}$ en los dos casos particulares donde $s=0$ o $s=-1$:
$$\begin{cases}
\tau{\left(r,0,t\right)}=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{0}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u,\\
\tau{\left(r,-1,t\right)}=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{0}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u.\\
\end{casos}$$
Definir la función de $F{\left(z\right)}$ por la integral,
$$F{\left(z\right)}:=\int_{0}^{1}\operatorname{Li}_{1}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x.$$
A continuación,
$$F^{\prime}{\left(z\right)}=\frac{1}{z}\int_{0}^{1}\operatorname{Li}_{0}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x,$$
y
$$F^{\prime\prime}{\left(z\right)}=\frac{1}{z^2}\int_{0}^{1}\operatorname{Li}_{-1}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x-\frac{1}{z}F^{\prime}{\left(z\right)}.$$
Por lo tanto, podemos representar el $\tau$ funciones necesitamos enteramente en términos de la función de $F$ y sus derivados:
$$\begin{cases}
\tau{\left(r,0,t\right)}=4t^{2r-1}F^{\prime\prime}{\left(t^r\right)}+5t^{r-1}F^{\prime}{\left(t^r\right)},\\
\tau{\left(r,-1,t\right)}=4t^{r-1}F^{\prime}{\left(t^r\right)}+\frac{1}{t}F{\left(t^r\right)}.\\
\end{casos}$$
Así que nos concentramos en encontrar una forma cerrada integral para la definición de $F{(z)}$.
Dejando $\sqrt{z}=:a$$\sqrt{\frac{a}{4+a}}=:b$, y usando el hecho de que $\operatorname{Li}_{1}{\left(x\right)}=-\ln{\left(1-x\right)}$, tenemos:
$$\begin{align}
F{\left(z\right)}
&=F{\left(a^2\right)}\\
&=\int_{0}^{1}\operatorname{Li}_{1}{\left(a^2x^2\left(1-x\right)^2\right)}\,\mathrm{d}x\\
&=-\int_{0}^{1}\ln{\left(1-a^2x^2\left(1-x\right)^2\right)}\,\mathrm{d}x\\
&=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+ax\left(1-x\right)\right)}\,\mathrm{d}x\\
&=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+\frac{4b^2}{1-b^2}x\left(1-x\right)\right)}\,\mathrm{d}x\\
&=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(\left(1+\frac{2b}{1-b}x\right)\left(1-\frac{2b}{1+b}x\right)\right)}\,\mathrm{d}x\\
&=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+\frac{2b}{1-b}x\right)}\,\mathrm{d}x\\
&~~~~~ -\int_{0}^{1}\ln{\left(1-\frac{2b}{1+b}x\right)}\,\mathrm{d}x\\
&=\left[2-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}\right]+\left[1+\frac{1+b}{2b}\ln{\left(\frac{1-b}{1+b}\right)}\right]\\
&~~~~~ +\left[1+\frac{1-b}{2b}\ln{\left(\frac{1-b}{1+b}\right)}\right]\\
&=4-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}+\frac{1}{b}\ln{\left(\frac{1-b}{1+b}\right)}\\
&=4-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}+\sqrt{\frac{4+a}{a}}\ln{\left(1+\frac{a}{2}-\frac12\sqrt{a\left(a+4\right)}\right)}\\
&=4-2\frac{\sqrt{4-\sqrt{z}}}{z^{1/4}}\tan^{-1}{\left(\frac{z^{1/4}}{\sqrt{4-\sqrt{z}}}\right)}+\frac{\sqrt{4+\sqrt{z}}}{z^{1/4}}\ln{\left(1+\frac{\sqrt{z}}{2}-\frac{z^{1/4}}{2}\sqrt{\sqrt{z}+4}\right)}.\\
\end{align}$$
