Power series expansion of function involving logarithm and power function

Question

It is well-known that the power series $$\log\left(\frac{2}{1+\sqrt{1-x}}\right)=\sum_{n=1}^\infty \frac{\binom{2n}{n}}{n4^n}x^n\quad (x\in[-1,1]).$$ However, for the following function $$\log\left(1+(1-x)^{1/p}\right)\quad (p>2)$$ Does it have a simple power series expansion? More general, $$\log\left(1+c_1(a_1-x)^{1/{p_1}}+c_2(a_2-x)^{1/{p_2}}+\cdots+c_r(a_r-x)^{1/{p_r}}\right)$$ Does it have a simple power series expansion? Here $p_1,\ldots,p_r\ge 2$ are positive integers, $a_1,\ldots,a_r\in [-1,1]$ and $c_1,\ldots,c_r$ are real numbers.

Answer 1

I think that, because of the required composition of Taylor series, it could be difficult to find the general term $$(1-x)^{\frac{1}{p}}=1-\frac{1}{p}x+\frac{(1-p)}{2 p^2} x^2+\frac{ (1 - p) ( 2 p-1)}{6 p^3}x^3+$$ $$\frac {(1-p) (2 p-1) (3 p-1) }{24p^4}x^4+\frac {(1-p) (2 p-1) (3 p-1)(4p-1) }{120p^5}x^5+O\left(x^6\right)$$ $$\log \left(1+(1-x)^{\frac{1}{p}}\right)=\log (2)-\frac{1}{2 p}x+\frac{(1-2 p) }{8 p^2}x^2+\frac{(3-4 p) }{24 p^2}x^3+\frac{(1-4 p) \left(6 p^2-4 p-1\right)}{192 p^4} x^4-\frac{\left(48 p^3-50 p^2+5\right) }{480 p^4}x^5+O\left(x^6\right)$$

Edit

Let $k=\frac 1p$ and we have $$\log \left(1+(1-x)^k\right)=\log (2)+\sum_{n=1}^\infty \frac {a_n}{b_n}x^n$$ where the $b_n$'s make the sequence $$\{2,8,24,192,480,5760,13440\}$$ and the $a_n$'s are polynomials of degree $n$ in $k$; the first are given below $$\left( \begin{array}{cc} n & a_n \\ 1 & -k \\ 2 & (k-2) k \\ 3 & k (3 k-4) \\ 4 & -(k-4) k \left(k^2+4 k-6\right) \\ 5 & -k \left(5 k^3-50 k+48\right) \\ 6 & (k-6) (k-2) k \left(2 k^3+16 k^2+19 k-40\right) \\ 7 & k \left(14 k^5-245 k^3+1176 k-960\right) \end{array} \right)$$