Initially in the solution the following equilibria exist
$$\ce{HA <=> H+ +A-}$$
and
$$\ce{H2O <=> H+ +OH-}$$
Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ and the ionic product of water, $K_w$. By adding acid, you have disturbed the equilibrium, and from Le Chatelier's principle, both the reactions go backwards.
$$\ce{HA <<=> H+ + A- }\tag{1} \label{eqn:1}$$
$K_{eq}=K_a$
$$\ce{H2O <<=> H+ +OH-} \tag{2} \label{eqn:2}$$
$K_{eq}=K_w$
Let the new equilibrium concentrations of $\ce{HA},~\ce{A-},~\ce{H+},~\ce{OH-}$ be $a_1+\mathrm dx,~a_2-\mathrm dx,~\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy,~\frac{K_w\cdot a_2}{K_a.a_1} - \mathrm dy$ respectively.
At equilibrium, for $\ref{eqn:1}$,
$$K_a = \frac{(\frac{K_a\cdot a_1}{a_2} +\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a_2 - \mathrm dx)}{a_1+\mathrm dx} \tag{3} \label{eqn:3}$$
and for $\ref{eqn:2}$,
$$K_w = (\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (\frac{K_w\cdot a_2}{K_a.a_1}-\mathrm dy) \tag{4} \label{eqn:4}$$
Dividing $\ref{eqn:3}$ by $\ref{eqn:4}$, we get
$$(K_a)\cdot (\frac{K_w\cdot a_2}{K_a\cdot a_1}-\mathrm dy)=(K_w)\cdot (\frac{a_2-\mathrm dx}{a_1+\mathrm dx})\\
K_w\cdot a_2+\frac{K_w\cdot a_2}{a_1}\cdot \mathrm dx-K_aa_1\mathrm dy-K_a\mathrm dx\mathrm dy=K_wa_2-K_w\mathrm dx$$
Neglecting the term $K_a\mathrm dx\mathrm dy$, we get
$$dx= \frac{K_a\cdot {a_1}^2}{K_w(a_1+a_2)}\cdot \mathrm dy$$
and also from $\ref{eqn:3}$
$$K_a\cdot (a_1+\mathrm dx)=(\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a_2-\mathrm dx)$$
Which on simplifying and neglecting terms of the type $\mathrm da\cdot \mathrm db$ will give,
$$\mathrm dn= (\frac{{K_a}^2\cdot {a_1}^2}{K_w\cdot {a_2}^2}+\frac{K_a\cdot {a_1}^2}{K_w\cdot (a_1+a_2)}+1)\cdot \mathrm dy \tag{5} \label{eqn:5}$$
By definition,
$\beta = \frac{\mathrm dn}{\mathrm{dpH}}$ and $\mathrm{pH}=-\log_{10} {[\ce{H+}]}$.
On differentiating this, we get
$$\mathrm{dpH} = \frac{-1}{[\ce{H+}]}\cdot \frac{1}{\log_e10}\cdot \mathrm d[\ce{H+}]$$
Substituting this in the definition of buffer capacity,
$$\beta = \frac{-\log_e10\cdot [\ce{H+}]\cdot \mathrm dn}{\mathrm d[\ce{H+}]} \tag{6} \label{eqn:6}$$
Here,
$$\mathrm d[\ce{H+}] = \mathrm dx+\mathrm dy-\mathrm dn = -\frac{{K_a}^2\cdot {a_1}^2}{{a_2^2}\cdot K_w}\cdot \mathrm dy$$
Substituting this expression of $\mathrm d[\ce{H+}]$ and $\mathrm dn$ from $\ref{eqn:5}$ in $\ref{eqn:6}$, we get,
$$\beta = \log_e10\cdot [\ce{H+}]\cdot \frac{(\frac{{K_a}^2\cdot {a_1}^2}{{a_2}^2\cdot K_w}+\frac{K_a\cdot {a_1^2}}{K_w\cdot (a_1 + a_2)} +1)}{\frac{{K_a}^2\cdot {a_1}^2}{{a_2}^2\cdot K_w}}\\
\beta = \log_e(10)[\ce{H+}](1 + \frac{K_w}{{[\ce{H+}]}^2} + \frac{{a_2}^2}{(a_1+a_2)\cdot K_a})\\
\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{{a_2}^2\cdot [\ce{H+}]}{(a_1+a_2)\cdot K_a})$$
Substituting $[\ce{H+}]= \frac{K_a\cdot a_1}{a_2}$ in the third term, we get
$$\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{a_1\cdot a_2}{(a_1+a_2)})$$
Multiplying numerator and denominator of the third term with $K_a$,
$$\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{K_a\cdot a_1\cdot a_2}{K_a\cdot (a_1+a_2)})$$
Taking $a_2$ common from the denominator and cancelling the $a_2$ in the numerator and substitute $\frac{k_a\cdot a_1}{a_2}=[\ce{H+}]$ and multiplying numerator and denominator with $[\ce{H+}]+K_a$, we get
$$\beta = \log_e10([\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{(\frac{K_a\cdot a_1}{a_2})\cdot (K_aa_1+K_aa_2)}{{([\ce{H+}]+{K_a})}^2}$$
Taking $K_a$ out from the numerator and substituting $[\ce{H+}]=\frac{K_a\cdot a_1}{a_2}$, we get
$$\beta= \log_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K_a(a_1+a_2)}{([\ce{H+}]+K_a)^2})$$
So, in your expression, of course, $C_{\text{buf}}=a_1+a_2=[\ce{HA}]+[\ce{A-}]$ and not $[\ce{HA}]+[\ce{A-}]+[\ce{H+}]$
$$\beta= \log_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K_aC_{\text{buf}}}{([\ce{H+}]+K_a)^2})$$
You will get the same result if a small amount of base is added.
