Proving the given language isn't context free using Pumping Lemma
The pumping segments $v$ and $x$ can be anywhere in the string assuming they are at most $n$ apart, i.e., $|vwx|\le n$. Like you I get five possibilities, but I think you wrote some special cases rather than the general picture.

Inside one of the $a$-segments, for instance :


the first $u=a^i$, $vwx=a^j$, $y=a^{n-i-j}ba^nba^n$ where $i+j\le n$ and $j\ge 1$. Or  
the second $u=a^nba^i$, $vwx=a^j$, $y=a^{n-i-j}ba^n$ where $i+j\le n$ and $j\ge 1$. Or 
the third $\dots$ .


Or overlapping with one of the $b$'s. Of course because of the lengths we cannot overlap with both $b$'s.


First: $u=a^{n-i}$, $vwx=a^iba^j$, $y=a^{n-j}ba^n$ where $0\le i,j\le n$, $i+j&lt; n$ .
Second $\dots$ .


However this does not solve the pumping problem. Note we repeat the strings $v,x$ when pumping. In these last cases we do not know how $v,x$ look. Do they contain the $b$? More cases.

A shortcut to the proof would be to make the following observation.


The pumping segments cannot contain $b$.


That leaves five cases for the position of $v$, $x$ in the three $a$-segments. Either deal with them or argue


Now that $v$, $x$ can only contain $a$'s we can pump either one or two of the three $a$-segments in $z$. That means after pumping they cannot all contain the same number of $a$'s