Summary
Comparing parameterized families of distributions is usually a matter of algebraic manipulation of the most convenient form of expression (usually a density (PDF) or the distribution function (CDF) but sometimes a characteristic function or cumulant generating function).
The procedure can be simplified to the point where the comparison can almost always be performed by inspecting the formulas, using these basic techniques:
Ignore all normalizing constants.
Recognize common location and scale parameters and set them to values that simplify the formulas (almost always $0$ for the location and $1$ for the scale).
Compare their supports, if these might vary within either family.
Isolate algebraic factors common to both expressions and equate corresponding elements (often powers).
These techniques are illustrated through their application to the present question. The conclusions section at the end summarizes the result of this comparison.
The Feller-Pareto Distribution
For analytical purposes, the alternative construction
$$W = \mu + \sigma\left(\frac{X}{Y}\right)^\gamma$$
may be more congenial, where independently
$$X\sim\Gamma(\alpha);\quad Y \sim\Gamma(\beta).$$
Since the support of both $X$ and $Y$ is the set of positive real numbers, it is immediate that the support of $W$ is $(\mu, \infty)$ for $\sigma\gt 0$ and $(-\infty, \mu)$ otherwise. In what follows, all expressions for density functions are understood as applying only for values within the support: the densities otherwise are zero.
Clearly $\mu$ is a location parameter and $\sigma$ is a scale parameter (with negative values of $\sigma$ reversing the distribution). To understand $W$ we may therefore begin with $\mu=0$ and $\sigma=1$. As such, it is recognizable as being a multiple of the $\gamma$ power of an F-Ratio distribution with parameters $(2\alpha,2\beta).$
To understand probability distributions we may ignore constant factors, because in the end we know the distribution will integrate to unity. Consider, then, the density of the F-ratio distribution (without its scaling constants),
$$f(z;2\alpha, 2\beta)\frac{dz}{z} \ \propto\ z^{\alpha}(1+z)^{-(\alpha+\beta)}\frac{dz}{z}.$$
Writing $z=w^{1/\gamma}$ produces
$$\frac{dz}{z} = d\log(z) = \frac{1}{\gamma} d\log(w) = \frac{1}{\gamma} \frac{dw}{w}\ \propto\ \frac{dw}{dw},$$
whence the density of $w$ is found simply by substituting $z=w^{1/\gamma}$ into $f$:
$$f(w^{1/\gamma};\alpha,\beta)\frac{dw}{w} \ \propto\ w^{\alpha/\gamma}(1+w^{1/\gamma})^{-(\alpha+\beta)}\frac{dw}{w}.$$
The Generalized Beta (GB) Distribution
Again ignoring constants, the density of a distribution in this family is
$$g(w; a,b,c,p,q)\frac{dw}{w}\ \propto\ w^{ap}\left(1 - (1-c)(w/b)^a\right)^{q-1}\left(1 + c(w/b)^a\right)^{-(p+q)}\frac{dw}{w}$$
where $0 \lt w^a \lt b^a/(1-c),$ $0\le c\le 1,$ and $b, p, q$ are positive. Obviously $b$ is a (positive) scale parameter and plays the role of $\sigma$ (up to a multiple). Let us therefore consider $b=1.$
Comparing the Distributions
A distribution will occur in both families provided its density has an expression of the form $f(w;2\alpha,2\beta)dw/w$ (for some $\mu$) and $g(w;a,1,c,p,q)dw/w.$
By comparing the supports of these two families, we must have $c=1$ (so that $W$ is unbounded) and $\sigma\gt 0$ (so that $W$ is not negative).
Equate these two forms of density function (neglecting the common differential element $dw/w$) and solve for the parameters:
$$w^{\alpha/\gamma}(1+w^{1/\gamma})^{-(\alpha+\beta)} = w^{ap}\left(1 - (1-c)w^a\right)^{q-1}\left(1 + cw^a\right)^{-(p+q)}.$$
Comparing powers of $w$ gives
$$\alpha/\gamma = ap.$$
Since the left hand side has no power of $(1-\lambda w)$ (for any nonzero constant $\lambda$), neither can the right hand side, whence
$$1-c=0,$$
a conclusion we arrived at previously by considering the support of $W.$
Comparing the remaining terms with $c=1$ shows
$$(1+w^{1/\gamma})^{-(\alpha+\beta)} = \left(1 + w^a\right)^{-(p+q)}$$
implying
$$a = \frac{1}{\gamma};\ \alpha+\beta=p+q.$$
Collecting these results we find
$$\alpha=p;\ \beta = q;\ \gamma=1/a;\ c=1.$$
There is no analog of the location parameter $\mu$ in the GB distribution.
Conclusions
For $\sigma\gt 0,$ members of the Feller-Pareto distribution with parameters $(\mu=0,\sigma=1,\gamma,\alpha,\beta)$ are also members of the GB distribution with parameters
$$(a,b,c,p,q) = (1/\gamma, 1, 1, \alpha,\beta).$$
Moreover, $\sigma$ and $b$ are both scale parameters.
The two five-parameter families are not the same, but they share a four-parameter family of distributions. The Feller-Pareto includes a location parameter $\mu$ not present in the GB family and the GB includes another shape parameter $c$ not present in the Feller-Pareto.
