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:

(1-x)^{-½}=∑_{0}^{∞}(2k-1)!!/(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

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