How to solve $\int\frac{x\arctan x}{x^4+1}dx$ in a practical way I need to evaluate the following indefinite integral for some other definite integral
$$\int\frac{x\arctan x}{x^4+1}dx$$
I found that
$$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{4}-2I(1)$$
where
$$I(t)=\int_0^1\frac{x\operatorname{Ti}_2(tx)}{x^4+1}dx$$
taking derivatives
$$tI'(t)=\int_0^1\frac{x\arctan(tx)}{x^4+1}dx$$
$$(tI')'=\int_0^1\frac{x^2}{(t^2x^2+1)(x^4+1)}dx$$
The answer for the integral is the integrand in my indefinite integral plus some stuff I can integrate on my own, which is where my problem comes from. In reality, I need
$$\int\frac{1}{t}\left(\int\frac{t\arctan{t}}{t^4+1}dt\right)dt$$
But it's fine if I just get an answer for the inside integral.
Back to the indefinite integral. I wanted to use partial fractions
$$\int\frac{x\arctan x}{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}dx$$
We would get four integrals of the form $\int\frac{\arctan{x}}{x+c}dx$. Using the logarithmic definition of arctangent, we get 8 integrals of the form $\int\frac{\ln{(a+bx)}}{x+c}$. That's 8 dilogarithms. I need this for another calculation, and having 8 dilogarithms is too numerous to be practical to work with. I want another way to evaluate this integral.
WAs answer
TO CLARIFY
I am fine with dilogarithms

I am not sure that you could avoid dilogarithms  except if you accept an infinite summation of Gaussian hypergeometric functions.
$$I=\int \frac{x }{x^4+1}\tan ^{-1}(x)\,dx$$ Using integration by parts
$$I=\frac{1}{2} \tan ^{-1}(x) \tan ^{-1}\left(x^2\right)-\frac 12\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ looks a bit more pleasant.
Using
$$\tan ^{-1}\left(x^2\right)=\sum_{n=0}^\infty\frac{(-1)^n}{2 n+1}x^{4n+2}$$
$$J=\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ $$J=\sum_{n=0}^\infty (-1)^n\,\frac{x^{4 n+3}}{(2n+1)(4 n+3)}\,\,\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)$$
Notice that
$$x^{4 n+3}\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)=-(4n+3)\tan ^{-1}(x)+P_{4n+1}(x)$$ where the first polynomials are
$$\left(
\begin{array}{cc}
 0 & 3 x \\
 1 & \frac{7 x^5}{5}-\frac{7 x^3}{3}+7 x \\
 2 & \frac{11 x^9}{9}-\frac{11 x^7}{7}+\frac{11 x^5}{5}-\frac{11
   x^3}{3}+11 x \\
 3 & \frac{15 x^{13}}{13}-\frac{15 x^{11}}{11}+\frac{5
   x^9}{3}-\frac{15 x^7}{7}+3 x^5-5 x^3+15 x \\
 4 & \frac{19 x^{17}}{17}-\frac{19 x^{15}}{15}+\frac{19
   x^{13}}{13}-\frac{19 x^{11}}{11}+\frac{19 x^9}{9}-\frac{19
   x^7}{7}+\frac{19 x^5}{5}-\frac{19 x^3}{3}+19 x \\
\end{array}
\right)$$
