<s>[INST] Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way of doing it.
 The integral is $$\int\frac{4x^4}{{x^8+1}}\;dx$$
Thanks in advance.
 [/INST] Not an answer.  Maybe this result inspires a by-hand method.  The integral is "elementary".
\begin{align*}
&\int \frac{4x^4}{x^8+1} \,\mathrm{d}x  \\
&= 4 \left(-\frac{1}{8} \cos \left(\frac{\pi }{8}\right) \log \left(x^2-2 x \sin
   \left(\frac{\pi }{8}\right)+1\right)  
\right.  \\  &\quad \left. {}
+\frac{1}{8} \cos \left(\frac{\pi
   }{8}\right) \log \left(x^2+2 x \sin \left(\frac{\pi
   }{8}\right)+1\right)
\right.  \\  &\quad \left. {}
+\frac{1}{8} \sin \left(\frac{\pi }{8}\right) \log
   \left(x^2-2 x \cos \left(\frac{\pi }{8}\right)+1\right)
\right.  \\  &\quad \left. {}
-\frac{1}{8} \sin
   \left(\frac{\pi }{8}\right) \log \left(x^2+2 x \cos \left(\frac{\pi
   }{8}\right)+1\right)
\right.  \\  &\quad \left. {}
+\frac{1}{4} \cos \left(\frac{\pi }{8}\right) \tan
   ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(x-\cos \left(\frac{\pi
   }{8}\right)\right)\right)
\right.  \\  &\quad \left. {}
+\frac{1}{4} \cos \left(\frac{\pi }{8}\right) \tan
   ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(x+\cos \left(\frac{\pi
   }{8}\right)\right)\right)
\right.  \\  &\quad \left. {}
-\frac{1}{4} \sin \left(\frac{\pi }{8}\right) \tan
   ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(x-\sin \left(\frac{\pi
   }{8}\right)\right)\right)
\right.  \\  &\quad \left. {}
-\frac{1}{4} \sin \left(\frac{\pi }{8}\right) \tan
   ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(x+\sin \left(\frac{\pi
   }{8}\right)\right)\right)\right)  \text{.}
\end{align*}
Produced by a CAS.  There are some "simplifications", but they don't shorten the expression much.
</s>