The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was proposed by @Tolaso (The original question had been closed and deleted).
My friend Cornel and I managed to compute it in different ways and would like to see other solutions if possible.
Cornel's solution:
We start with the following key identity presented in the book (Almost) Impossible Integrals, Sums, and Series) , Sect. $4.19$, pages $290-291$, $$\varphi(n)=\sum_{k=1}^{\infty} \frac{H_k H_k^{(2)}}{(k+1)(k+n+1)}$$ $$\small =2\zeta(3)\frac{H_n}{n}+\frac{\zeta(2)}{2}\frac{H_n^2}{n}-\frac{\zeta(2)}{2}\frac{H_n^{(2)}}{n}-\frac{H_n^{(4)}}{4n}-\frac{(H_n^{(2)})^2}{4n}-\frac{H_n}{n}\sum_{i=1}^n \frac{H_i}{i^2}+\frac{1}{2n}\sum_{i=1}^{n} \frac{H_i^2}{i^2}.\tag1$$ Then, based on $(1)$ we obtain that $$\sum_{k=1}^{\infty} \frac{H_k H_k^{(2)}}{(k+1)(k+2)(k+3)}=\varphi(1)-\varphi(2)=\frac{1}{2}\zeta(3)-\frac{1}{4}\zeta(2)-\frac{1}{32}.$$
End of story
A note: the whole process is completed by using series manipulations only, with no use of integrals.
My solution:
Replace $x$ by $xyz$ in the generating function: $$\sum_{n=1}^\infty H_nH_n^{(2)}x^n= \frac{\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\frac12\ln x\ln^2(1-x)-\zeta(2)\ln(1-x)-\zeta(3)}{1-x}=f(x),$$
we get
$$\sum_{n=1}^\infty H_nH_n^{(2)}(xyz)^n=f(xyz).$$
Multiply both sides by $yz^2$ then integrate w.r.t $x$, $y$ and $z$ using the fact$$\int_0^1\int_0^1\int_0^1 x^n y^{n+1}z^{n+2}dxdydz=\frac{1}{(n+1)(n+2)(n+3)}$$
we obtain
$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}=\int_0^1\int_0^1\int_0^1 yz^2f(xyz)dxdydz$$
$$\overset{x=t/y}{=}\int_0^1\left[\int_0^1\int_0^y z^2f(zt)dtdy\right]dz=\int_0^1\left[\int_0^1\int_t^1 z^2f(zt)dydt\right]dz$$
$$=\int_0^1\left[\int_0^1 z^2f(zt)(1-t)dt\right]dz\overset{t=u/z}{=}\int_0^1\int_0^z zf(u)\left(1-\frac{u}{z}\right)dudz$$
$$=\int_0^1\int_u^1 f(u)(z-u)dzdu=\frac12\int_0^1(1-u)^2f(u)du=\frac12\int_0^1 u^2f(1-u)du$$
$$=\frac12\underbrace{\int_0^1 u\operatorname{Li}_3(1-u)du}_{1-u\to u}+\frac12\int_0^1 u\operatorname{Li}_3(u)du$$ $$+\frac12\int_0^1 u\left[\frac12\ln(1-u)\ln^2(u)-\zeta(2)\ln(u)-\zeta(3)\right]du$$
$$=\frac12\int_0^1 \operatorname{Li}_3(u)du+\frac12\int_0^1 u\left[\frac12\ln(1-u)\ln^2(u)-\zeta(2)\ln(u)-\zeta(3)\right]du$$
$$=\frac12\left(\zeta(3)-\zeta(2)+1\right)+\frac12\left(\frac12\zeta(2)-\frac{17}{16}\right)$$
$$=\frac12\zeta(3)-\frac14\zeta(2)-\frac1{32}.$$