Given the series expansion $$1+b_1\arctan x+b_2 \arctan^2 x=1+\sum_{k=1}^\infty a_k x^k$$ one has the relationships $$a_{2k+1}=(-1)^k\frac{b_1}{2k+1},$$ $$a_{2k}=(-1)^k b_2\sum_{i=0}^k \frac{1}{(2i+1)(2k-2i+1)}$$ $$\qquad=(-1)^k \frac{b_2}{4k+4}\left(H_{-k-\frac{3}{2}}+H_{k+\frac{1}{2}}+4\ln 2\right),$$ with $H_n$ the harmonic number.

[This answers the question as originally stated. It has now been changed.]

Given the series expansion $$1+b_1\arctan x+b_2 \arctan^2 x=1+\sum_{k=1}^\infty a_k x^k$$ one has the relationships $$a_{2k+1}=(-1)^k\frac{b_1}{2k+1},$$ $$a_{2k}=(-1)^k b_2\sum_{i=0}^k \frac{1}{(2i+1)(2k-2i+1)}$$ $$\qquad=(-1)^k \frac{b_2}{4k+4}\left(H_{-k-\frac{3}{2}}+H_{k+\frac{1}{2}}+4\ln 2\right),$$ with $H_n$ the harmonic number.

Given the series expansion $$1+b_1\arctan x+b_2 \arctan^2 x=1+\sum_{k=1}^\infty a_k x^k$$ one has the relationships $$a_{2k+1}=(-1)^k\frac{b_1}{2k+1},$$ $$a_{2k}=(-1)^k b_2\sum_{i=0}^k \frac{1}{(2i+1)(2k-2i+1)}.$$

Given the series expansion $$1+b_1\arctan x+b_2 \arctan^2 x=1+\sum_{k=1}^\infty a_k x^k$$ one has the relationships $$a_{2k+1}=(-1)^k\frac{b_1}{2k+1},$$ $$a_{2k}=(-1)^k b_2\sum_{i=0}^k \frac{1}{(2i+1)(2k-2i+1)}$$ $$\qquad=(-1)^k \frac{b_2}{4k+4}\left(H_{-k-\frac{3}{2}}+H_{k+\frac{1}{2}}+4\ln 2\right),$$ with $H_n$ the harmonic number.