If $$\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = A\left(\frac{\alpha x - 1}{\beta x + 3}\right)^B + C$$, where C is the constant of integration, then the value of $$\alpha + \beta + 20AB$$ is _____

We need to evaluate $$\int \frac{dx}{\sqrt[5]{(x-1)^4(x+3)^6}}$$.

First rewrite it as $$\int \frac{dx}{(x-1)^{4/5}(x+3)^{6/5}}$$.

Equivalently, the integral takes the form $$= \int \frac{dx}{(x+3)^2 \cdot \left(\frac{x-1}{x+3}\right)^{4/5}}$$.

Let $$t = \frac{x-1}{x+3}$$. Then $$dt = \frac{(x+3) - (x-1)}{(x+3)^2}dx = \frac{4}{(x+3)^2}dx$$, so $$\frac{dx}{(x+3)^2} = \frac{dt}{4}$$.

Substituting these gives $$\int \frac{1}{t^{4/5}} \cdot \frac{dt}{4} = \frac{1}{4}\int t^{-4/5}\,dt = \frac{1}{4} \cdot 5t^{1/5} + C = \frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1/5} + C$$.

Comparing this with $$A\left(\frac{\alpha x - 1}{\beta x + 3}\right)^B + C$$, we have $$A = \frac{5}{4}$$, $$\alpha = 1$$, $$\beta = 1$$, and $$B = \frac{1}{5}$$.

Hence $$\alpha + \beta + 20AB = 1 + 1 + 20 \cdot \frac{5}{4} \cdot \frac{1}{5} = 2 + 20 \cdot \frac{1}{4} = 2 + 5 = 7$$.

The answer is $$\boxed{7}$$.