New trigamma identity for $\Psi_1(\frac3{20})+6\,\Psi_1(\frac15)+10\,\Psi_1(\frac25)-\Psi_1(\frac1{20})$ I play with Maple, and I find this relation for the trigamma function:

$$\begin{align}
\Psi_1\left({\frac{3}{20}}\right)+6\,\Psi_1\left(\frac15\right)+
10\,\Psi_1\left(\frac25 \right)-\Psi_1\left(\frac1{20}\right)
&=-96\,{G}-{\frac{24\,
\pi^2\sqrt{5}}{5}}+16\,\pi^2\\[0.5em]
&\quad-2\pi^2{\frac{15+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}}
\end{align}$$
where $G$ is the Catalan's constant.

But I don't know if this relation is well-known or not.
Please suggest how to prove it.
Thanks.

Identities involving linear combinations of trigamma functions at rational arguments can be proved semi-automatically. The arguments here are of the form $\frac k{20}$ for $1\le k\le 19$, so denote $a_k=\psi_1(k/20)$ and write down some identities:
$$a_k+a_{20-k}=\frac{\pi^2}{\sin^2k\pi/20},1\le k\le 10\tag{reflection}$$
$$a_5=\pi^2+8G\tag{special value}$$
$$a_k+a_{k+10}=4a_{2k},1\le k\le9\tag{duplication}$$
$$a_k+a_{k+5}+a_{k+10}+a_{k+15}=16a_{4k},1\le k\le4\tag{quadruplication}$$
$$a_k+a_{k+4}+a_{k+8}+a_{k+12}+a_{k+16}=25a_{5k},1\le k\le3\tag{quintuplication}$$
Treat the $a_k$ as variables and convert the identities into rows of a matrix equation $(a_1,a_2,\dots,a_{19},b)$, so e.g. reflection at $k=1$ becomes
$$\left[\begin{array}{ccccccccccccccccccc|c}1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&\frac{\pi^2}{\sin^2\pi/20}\end{array}\right]$$
Now drop the last ($\mathbf b$) column and see if the desired linear combination $\mathbf c$ – in this case $(-1,0,1,6,0,0,0,10,0,0,0,0,0,0,0,0,0,0,0)$ – is in the row space of the remaining matrix $\mathbf A$, which can be done by trying to solve $\mathbf A^T\mathbf x=\mathbf c^T$. For the question's $\mathbf c$ there is a solution:
$$\mathbf x=\left(\color{blue}{0, 0,\frac12, 0, 0, 2, 0, 2,\frac12, 0}, -12, \color{blue}{-\frac12, -2,\frac12, -2, 0, 0, 0, 0, 0}, 0, 0, 0, 0,\color{blue}{-\frac12, 0, 0}\right)^T$$
Then $\mathbf x\cdot\mathbf b$ gives an explicit expression for the trigamma linear combination:
$$\frac12\frac{\pi^2}{\sin^23\pi/20}+2\frac{\pi^2}{\sin^26\pi/20}+2\frac{\pi^2}{\sin^28\pi/20}+\frac12\frac{\pi^2}{\sin^29\pi/20}-12(\pi^2+8G)$$
Simplifying shows this is equal to $-96G-\frac{24\pi^2\sqrt5}5+16\pi^2-2\pi^2\frac{15+\sqrt5}{\sqrt{10+2\sqrt5}}$ as suspected. (The simplification I get from Mathematica is $-96G+\pi^2(16-24/\sqrt5-2\sqrt{25-2\sqrt5})$).
