I have no references for this particular series, but here's some hints to get a closed formula for the coefficients listed above by Michael Renardy.

If we let $u:=\frac {1-x} 2$, an expansion $$\arctan(1-2u)=\arctan(1) + \sum_{k=1}^\infty \frac {c_k} k \arctan(u^k)$$ can be obtained by term-wise integration over on $[0,u]$ of a (somehow more common) expansion into rational fractions $$\frac 2 {1 + (1-2u)^2}= \sum_{k=1}^\infty \\ c_k \frac {u^{k-1}} {1+u^{2k} }\, ,$$ (such expansions have a role in number theory, and are related to Dirichlet series). Here the coefficients may be identified expanding formally the geometric series $ (1+ u^{2k} )^{-1}$ and rearranging into a series of powers of $u$, to be compared with the power series of the LHS. One finds an equality with an arithmetic convolution, that inverted gives the $c_k$'s. The exponential growth of the $c_k$ give a positive radius of convergence (I guess $1/\sqrt 2$), that in particular allows the term-wise integration. Note that $\frac 2 {1 + (1-2u)^2}= \mathrm{Im} { \frac 2 {1-2u+i} }$, that simplifies things a bit.

I have no references for this particular series, but here's some hints to get a closed formula for the coefficients listed above by Michael Renardy.

If we let $u:=\frac {1-x} 2$, an expansion $$\arctan(1-2u)=\arctan(1) + \sum_{k=1}^\infty \frac {c_k} k \arctan(u^k)$$ can be obtained by term-wise integration over on $[0,u]$ of a (somehow more common) expansion into rational fractions $$\frac 2 {1 + (1-2u)^2}= \sum_{k=1}^\infty \, c_k \frac {u^{k-1}} {1+u^{2k} }\, ,$$ (such expansions have a role in number theory, and are related to Dirichlet series). Here the coefficients may be identified expanding formally the geometric series $ (1+ u^{2k} )^{-1}$ and rearranging into a series of powers of $u$, to be compared with the power series of the LHS. One finds an equality with an arithmetic convolution, that inverted gives the $c_k$'s. The exponential growth of the $c_k$ give a positive radius of convergence (I guess $1/\sqrt 2$), that in particular allows the term-wise integration. Note that $\frac 2 {1 + (1-2u)^2}= \mathrm{Im} { \frac 2 {1-2u+i} }$, that simplifies things a bit.

I have no references for this particular series, but here's some hints to get a closed formula for the coefficients listed above by Michael Renardy.

If we let $u:=\frac {1-x} 2$, an expansion $$\arctan(1-2u)=\arctan(1) + \sum_{k=1}^\infty \frac {c_k} k \arctan(u^k)$$ can be obtained by term-wise integration over on $[0,u]$ of a (somehow more common) expansion into rational fractions $$\frac 2 {1 + (1-2u)^2}= \sum_{k=1}^\infty \\ c_k \frac {u^{k-1}} {1+u^{2k} }\\ ,$$ (such expansions have a role in number theory, and are related to Dirichlet series). Here the coefficients may be identified expanding formally the geometric series $ (1+ u^{2k} )^{-1}$ and rearranging into a series of powers of $u$, to be compared with the power series of the LHS. One finds an equality with an arithmetic convolution, that inverted gives the $c_k$'s. The exponential growth of the $c_k$ give a positive radius of convergence (I guess $1/\sqrt 2$), that in particular allows the term-wise integration. Note that $\frac 2 {1 + (1-2u)^2}= \mathrm{Im} { \frac 2 {1-2u+i} }$, that simplifies things a bit.

I have no references for this particular series, but here's some hints to get a closed formula for the coefficients listed above by Michael Renardy.

If we let $u:=\frac {1-x} 2$, an expansion $$\arctan(1-2u)=\arctan(1) + \sum_{k=1}^\infty \frac {c_k} k \arctan(u^k)$$ can be obtained by term-wise integration over on $[0,u]$ of a (somehow more common) expansion into rational fractions $$\frac 2 {1 + (1-2u)^2}= \sum_{k=1}^\infty \\ c_k \frac {u^{k-1}} {1+u^{2k} }\, ,$$ (such expansions have a role in number theory, and are related to Dirichlet series). Here the coefficients may be identified expanding formally the geometric series $ (1+ u^{2k} )^{-1}$ and rearranging into a series of powers of $u$, to be compared with the power series of the LHS. One finds an equality with an arithmetic convolution, that inverted gives the $c_k$'s. The exponential growth of the $c_k$ give a positive radius of convergence (I guess $1/\sqrt 2$), that in particular allows the term-wise integration. Note that $\frac 2 {1 + (1-2u)^2}= \mathrm{Im} { \frac 2 {1-2u+i} }$, that simplifies things a bit.