Let be the functions
$$ f(x)=\sqrt{2}\frac{\cos(\frac{\pi}{4}-\frac{1}{2}\arctan(x))}{(1+x^2)^{1/4}} \qquad g(x)=\frac{1}{\sqrt{1-x}} $$
Obviously, they are holomorphic functions in a neighborhood of $0$. The series of $g(z)$ is well known and using Mathematica we have
$$ f(x)= 1+\frac{x}{2}- \frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}-\frac{231x^6}{1024}-\frac{429x^7}{2048}+\dotsc $$ $$ g(x)= 1+\frac{x}{2}+ \frac{3x^2}{8}+\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}+\frac{231x^6}{1024}+\frac{429x^7}{2048}+\dotsc $$
Exercise. Proof the absolute value of the coefficients of the two series are the same.
Attemp. With trigonometric manipulation I get $$ f(x)=\frac{1+x+\sqrt{1+x^2}}{\sqrt{2}(1+x^2)^{3/4}\sqrt{1+\frac{1}{\sqrt{1+x^2}}}} $$ My idea is use the Cauchy Integral Theorem to calculate the coefficients $$ a_n = \frac{1}{2\pi i}\int_C \frac{f(z)}{z^{n+1}}dz $$ and with changes of variables get something that seems to $$ \frac{1}{2\pi i}\int_C \frac{g(z)}{z^{n+1}}dz $$ But if I work in $\mathbb{C}$ I need to define $\log(1+z^2)$ and work with it in changes of variables doesn't look so friendly.
Can you give me a hint to prove it?
Hint
Let $\phi(z)=\sum_k a_k z^k$.
The first step is to see how to express $a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots$ using $\phi$.
The four roots of $X^4=1$ being $\lbrace \pm 1, \pm i \rbrace$. You have to solve the system: $$\begin{pmatrix}1&1&1&1\\1&i&-i&-1\\1^2&i^2&(-i)^2&(-1)^2\\1^3&i^3&(-i)^3&(-1)^3 \end{pmatrix} \begin{pmatrix} a\\ b \\c \\d \end{pmatrix}=\begin{pmatrix} 1\\ 1 \\-1 \\-1 \end{pmatrix}$$
As the solutions are $a=d=0$, $b=\frac{1-i}{2}, c=\frac{1+i}{2}$ you obtain: $$a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots = \frac{1}{2} \left((1-i)\phi(ix)+(1+i)\phi(-ix) \right)$$
Here you then have to prove that near $0$: $$f(x)=\frac{1}{2} \left(\frac{1-i}{\sqrt{1-ix}}+\frac{1+i}{\sqrt{1+ix}} \right)$$
Using your expression of $f$ this is the same to prove that near $0$: $$(1+i) \sqrt{1-ix}+(1-i) \sqrt{1+ix}=\sqrt{2}\frac{1+x+\sqrt{1+x^2}}{\sqrt{1+\sqrt{1+x^2}}}=2\sqrt{x+\sqrt{1+x^2}}$$ which seems easier.