Series of inverse function $A(s) = \sum_{k>0}a_ks^k$ and $A(s)+A(s)^3=s$.
I want calculate $a_5$. What ways to do it most efficiently?

We can elaborate on the Lagrange inversion concept.
Suppose we have $$A(s) = \sum_{n\ge 0} a_n s^n$$
and $A(s)+A(s)^3=s$ and we seek $a_n.$

Using the Cauchy Residue Theorem to prepare for Lagrange inversion we have that
$$a_n = 
\frac{1}{2\pi i}
\int_{|s|=\epsilon}
\frac{1}{s^{n+1}} A(s) \; ds.$$
Now put $A(s)=w$ so that $w+w^3 = s$ and
$$ds = 1 + 3w^2 \;dw.$$
This yields
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{(w+w^3)^{n+1}} w \times (1+3w^2) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n+1}} \times (3w^2+3-2) \; dw.$$
The first component here is
$$3 \times\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n}} \; dw
= 3\times[w^{n-1}] \frac{1}{(1+w^2)^{n}}.$$
This is zero when $n$ is even
and when $n$ is odd it yields
$$3\times (-1)^{(n-1)/2}\times {(n-1)/2+n-1\choose n-1}
= 3 (-1)^{(n-1)/2} {3/2n-3/2\choose n-1}.$$
The second component is
$$-2\times \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n+1}} \; dw
= -2\times [w^{n-1}] \frac{1}{(1+w^2)^{n+1}}.$$
This is again zero when $n$ is even 
and when $n$ is odd it yields
$$-2\times (-1)^{(n-1)/2} {(n-1)/2+n\choose n}
= -2 (-1)^{(n-1)/2} {3/2n-1/2\choose n}.$$
Combining these two yields
$$(-1)^{(n-1)/2}
\left(3 {3/2n-3/2\choose n-1}
-2 {3/2n-1/2\choose n}\right)$$
when $n$ is odd and zero otherwise.

If desired this can be simplified to
$$(-1)^{(n-1)/2}
\left(3-2 \frac{3/2n-1/2}{n}\right)
{3/2n-3/2\choose n-1}
= (-1)^{(n-1)/2}
\frac{1}{n}{3/2n-3/2\choose n-1}.$$
