If $$\int \left(\frac{1}{x} + \frac{1}{x^3}\right)\left(\sqrt[23]{3x^{-24} + x^{-26}}\right) dx = -\frac{\alpha}{3(\alpha+1)}(3x^\beta + x^\gamma)^{\frac{\alpha+1}{\alpha}} + C$$, $$x > 0$$, $$(\alpha, \beta, \gamma \in \mathbb{Z})$$, where C is the constant of integration, then $$\alpha + \beta + \gamma$$ is equal to _____.

We begin with the given integrand

$$I=\left(\frac1x+\frac1{x^{3}}\right)\left(3x^{-24}+x^{-26}\right)^{\frac1{23}}, \qquad x\gt 0.$$

The powers inside the radical differ by $$2$$, so take out the larger common power $$x^{-23}$$ from the second bracket:

$$3x^{-24}+x^{-26}=x^{-23}\!\left(3x^{-1}+x^{-3}\right).$$

Using $$\bigl(x^{-23}\bigr)^{\!\frac1{23}}=x^{-1},$$ the radical becomes

$$\left(3x^{-24}+x^{-26}\right)^{\frac1{23}}=x^{-1}\!\left(3x^{-1}+x^{-3}\right)^{\frac1{23}}.$$

Substituting this back in $$I$$ gives

$$I=\left(\frac1x+\frac1{x^{3}}\right)x^{-1}\left(3x^{-1}+x^{-3}\right)^{\frac1{23}}=\left(\frac1{x^{2}}+\frac1{x^{4}}\right)\left(3x^{-1}+x^{-3}\right)^{\frac1{23}}.$$

Now observe the derivative of the inner expression:

$$\frac{d}{dx}\!\left(3x^{-1}+x^{-3}\right)=-3x^{-2}-3x^{-4}=-3\!\left(\frac1{x^{2}}+\frac1{x^{4}}\right).$$

Hence, apart from the constant factor $$-3$$, the term $$\dfrac1{x^{2}}+\dfrac1{x^{4}}$$ in $$I$$ is the derivative of $$3x^{-1}+x^{-3}$$. Rewrite $$I$$ accordingly:

$$I=-\frac13\;\frac{d}{dx}\!\left(3x^{-1}+x^{-3}\right)\left(3x^{-1}+x^{-3}\right)^{\frac1{23}}.$$

This is of the standard form $$f'(x)\,f(x)^{\lambda}$$ with $$f(x)=3x^{-1}+x^{-3}$$ and $$\lambda=\dfrac1{23}$$. Integrating directly,

$$\int I\,dx=-\frac13\int f'(x)\,f(x)^{\frac1{23}}dx =-\frac13\;\frac{f(x)^{\frac{1}{23}+1}}{\frac{1}{23}+1}+C =-\frac{23}{3\!\left(23+1\right)}\!\left(3x^{-1}+x^{-3}\right)^{\frac{24}{23}}+C.$$

Comparing with the form provided in the question,

$$-\frac{\alpha}{3(\alpha+1)}\!\left(3x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C,$$

we identify

$$\alpha=23,\qquad \beta=-1,\qquad \gamma=-3.$$

Finally,

$$\alpha+\beta+\gamma=23+(-1)+(-3)=19.$$

Therefore, the required value is $$\boxed{19}$$.