Proving an equality involving harmonic numbers

I am working through the book (Almost) Impossible Integrals, Sums, and Series, and I am on the following problem:

Let $$n$$ be a positive integer. Prove that the following equality holds. $$K_n = \int_0^1 x^{n-1} \log^3 (1-x) \, dx = -\frac{3}{n} \sum_{k=1}^n \frac{H_k^2 + H_k^{(2)}}{k} = - \frac{H_n^3 + 3H_n H_n^{(2)} + 2H_n^{(3)}}{n},$$ where $$H_n^{(m)} = 1 + \frac{1}{2^m} + \cdots + \frac{1}{n^m}, \, m \geq 1,$$ is the $$n$$th generalized harmonic number of order $$m$$.

I was able to evaluate the integral, but I was unable to evaluate the resulting series. Here is what I have so far:

Using Abel’s summation, we get that

\begin{align} \sum_{k=1}^{n} \frac{H_k^2 + H_k^{(2)}}{k} &= H_n(H_{n+1}^2 + H_{n+1}^{(2)}) + \sum_{k=1}^{n}H_k \left(H_k^2 + H_k^{(2)} - H_{k+1}^2 - H_{k+1}^{(2)} \right) \tag{1} \\ &= H_n(H_{n+1}^2 + H_{n+1}^{(2)}) + \sum_{k=1}^{n}H_k \left(H_k^2 - H_{k+1}^2 - \frac{1}{(k+1)^2} \right) \tag{2} \end{align}

I got stuck here so I went and looked at the author’s solution. In the author’s solution, he skips right over (1) and (2), and writes

$$\sum_{k=1}^{n} \frac{H_k^2 + H_k^{(2)}}{k} = H_n(H_{n+1}^2 + H_{n+1}^{(2)}) - 2 \sum_{k=1}^n \left( \frac{H_k^2}{k+1} + \frac{H_k}{(k+1)^2} \right) \tag{3},$$

but I do not see how does this follows. How does one get from (2) to (3)?

Their simplification amounts to $$H_k^2-H_{k+1}^2-\frac{1}{(k+1)^2}=-\frac{2H_k}{k+1}-\frac{2}{(k+1)^2}.$$ Now recall that $$H_{k+1}=H_k+\dfrac{1}{k+1}$$.