Closed form for $\int_0^1 \frac{x^a dx}{(1+x^b)^c}$. By means of a substitution, the integral can be reduced to

$$\int_0^1 \frac{x^p dx}{(1+x)^q}$$
but no method that I've thusfar tried seems tenable.

This Beta-like integral cannot be reduced into the Beta function (i.e. $\int_0^\infty \frac{x^p dx}{(1+x)^q}$) by any immediately obvious method (the denominator ranges from $1$ to $2$, not $0$ to $1$, which no subsitution can resolve). Numerically the integral tends to be a fraction (suggesting that $\Gamma(x)$ may be involved), unless $c=1$, in which case the result is easy to work out.
As @OL points out the integral is a hypergeometric function:

$$\frac{\, _2F_1(p+1,q;p+2;-1)}{p+1}$$

You can get this by expanding the denominator in a series (by the binomial theorem), integrating term by term, and observing that you get a hypergeometric series. If there is a relationship of some sort between $p$ and $q,$ this might be simplifiable.
You can use the variable $t=1+x$ and use the binomial theorem (for extended binomials coefficient to real numbers)