Try substitute x=sin(t); The integral becomes
∫dt/√[1+a*sin^{6}(t)*(1+sin^{2}(t)+sin^{4}(t))];
This integral can be evaluated using the following expansion:
(1x)^{½}=∑_{0}^{∞}(2k1)!!/(2k)!!*x^{k}, summed over k. This series converges for x<1.
Expression f(t)=a*sin^{6}(t)(1+sin^{2}(t)+sin^{4}(t)) needs to be less than 1. This puts restrictions on the a parameter. [f(t)]^{k} is integrable, however, integrals will be very ugly looking. For example,
∫f(t)dt=a*
213 x 337 Sin[2 x] 41 Sin[4 x] 31 Sin[6 x]
{   +    +
256 12 256 1024
7 Sin[8 x] Sin[10 x]
   }
2048 5120
10/31/2013

Kirill Z.
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