I am interested to know if the following sum has a closed form
$$s_4 = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^4}\tag{1}$$
I stumbled on this question while studying a very useful book about harmonic series and logarithmic integrals whichh has appeared recently [1]. Checking it for possible missing entries I was led to consider this family alternating Euler sums
$$s(a) = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^a}$$
where $H_{n} =\sum_{k=1}^{n}\frac{1}{k}$ is the harmonic number, as well as the corresponding integrals
$$i(a) = \int_0^1 \frac{\log ^{a-1}\left(\frac{1}{x}\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
These are related by
$$i(a) = - \Gamma(a) s(a)$$
As listed in detail in the appendix, closed forms exist for all odd $a=2q+1$. For even $a=2q$ only the case $a=2, q=1$ is known.
Hence, the natural question is to aks for a closed form for the smallest case open up to now, $a = 4$.
What did I do so far
The integral $i(a)$ can be found by differentiation with respect to a parameter $u$ from the generating integral
$$g_i(u) = \int_0^1 \frac{t^u \log \left(t^2+1\right)}{t^2+1} \, dt$$
which is evaluated by Mathematica in terms of hypergeometric functions as follows
$$g_i(u) = \frac{1}{4} \left(-\frac{2^{\frac{u+5}{2}} \, _3F_2\left(\frac{1}{2}-\frac{u}{2},\frac{1}{2}-\frac{u}{2},\frac{1}{2}-\frac{u}{2};\frac{3}{2}-\frac{u}{2},\frac{3}{2}-\frac{u}{2};\frac{1}{2}\right)}{(u-1)^2}\\-2 \pi H_{-\frac{u}{2}-\frac{1}{2}} \sec \left(\frac{\pi u}{2}\right)-\log (4) B_{\frac{1}{2}}\left(\frac{1-u}{2},\frac{u+1}{2}\right)\right)$$
As the parameter $u$ appears in 7 places each derivative generates a factor 7 in the length of the result.
Unless someone comes up with a very clever idea to simplify the hypergeometric expressions this path seems to be hopeless.
Appendix: known closed forms
For positive integer values of $a$ the following results have been obtained:
a) for odd $a=2q+1$
the closed form was calculated in [1], 4.1.15 (4.91) as:
$$s(2q+1) = (2q+1)\beta(2q+2) + \frac{\pi}{(2q)! 4^{q+1}}\lim_{m\to \frac12 }\frac{\mathrm{d}^{2q}}{\mathrm{d} m^{2q}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}$$
Here
$$\beta(z)=\sum_{k=1}^{\infty}\frac{(-1)^k}{2k+1)^z}$$
is Dirichelt's beta function.
As for a simplification of r.h.s. see https://math.stackexchange.com/a/4139359/198592
b) for even $a=2q$
there is a closed form just for $a=2, i.e. q=1$, found in [1] 4.5.5 (4.187)
$$s(2) = 2 \;\Im \text{Li}_3(1-i)+ \frac{3\pi^3}{32}+\frac{\pi}{8}\log^2(2) -\log(2) G$$
where $G = \beta(2)$ is Catalan's constant.
References
[1] Ali Shadhar Olaikhan, "An introduction to harmonic series and logarithmic integrals", April 2021, ISBN 978-1-7367360-0-5