Is there a useful closed form for $\int \frac{x^{p/q}}{1-x}dx$?

Question

Consider the function $$f=f(a,b,x)=\sum_{n=0}^\infty\frac{x^n}{an+b}=\frac 1 a\sum_{n=0}^\infty\frac{x^n}{n+\frac b a}$$ Where $a,b$ are coprime integers. This function is interesting because of the BBP formula: $$\pi=4f\left(8,1,\frac 1{16}\right)-\frac12f\left(2,1,\frac 1{16}\right)-f\left(8,5,\frac 1{16}\right)-\frac12f\left(4,3,\frac 1{16}\right)$$ $$\left(fx^\frac b a\right)' =\frac 1 a \sum_{n=0}^\infty x^{n+\frac b a -1}=\frac{x^{\frac b a -1}}{a(1-x)}$$ $$f=\frac 1 a x^{-\frac b a}\int\frac{x^{\frac {b-a} a}}{1-x}dx$$

Let $x=y^a$

$$\int\frac{x^{\frac {b-a} a}}{1-x}dx=a\int \frac{y^{b-1}}{1-y^a}dy$$

Let $\omega$ be the primitive $a$-th root of unity. Then the partial fraction decomposition of the integrand is
$$\frac{y^{b-1}}{1-y^a}=\frac{y^{b-1}}{a}\sum_{h=0}^{a-1}\frac{\omega^h}{\omega^h-y}$$ And the rest is boring computation to find that

$$\int\frac{x^{\frac {b-a} a}}{1-x}dx=\sum_{h=0}^{a-1}\left(\omega^{bh}\ln\left(1-x^\frac 1 a \omega ^{-h}\right)+\sum_{n=1}^{b-1}\frac{x^\frac n a \omega^{(b-n)h}} n \right)$$ Where the principal branch of the logarithm is taken. Which means $$f=\frac 1 a x^{-\frac b a}\sum_{h=0}^{a-1}\left(\omega^{bh}\ln\left(1-x^\frac 1 a \omega ^{-h}\right)+\sum_{n=1}^{b-1}\frac{x^\frac n a \omega^{(b-n)h}} n \right)$$ This is pretty inelegant. Is there a way to simplify this expression, preferably without using complex logarithms?

Answer 1

In the BBP formula and similar formulas it is more practical to exploit $$ \frac{1}{n+\frac{a}{b}}=\int_{0}^{1}z^{(n-1)+\frac{a}{b}}\,dz $$ to convert the series into the integral of a rational function, rather than exploiting $$ \int_{0}^{1} z^{\alpha-1}(1-z)^{\beta-1}\,dz = B(\alpha,\beta) = \frac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)} $$ or, even worse, removing the integration bounds and dealing with incomplete Beta functions/Lerch trascendents. With the former approach the BBP formula turns into $$ \pi \stackrel{?}{=}\int_{0}^{1}\sum_{k\geq 0} (4-2x^3-x^4-x^5)\frac{x^{8k}}{16^k}\,dx =16\int_{0}^{1}\frac{(1-x)(2+x^2)(2+2x+x^2)}{16-x^8}\,dx$$ where the last integral is straightforward to compute through partial fraction decomposition.