here is another...

multiply and divide by x
it becomes
∫1/x(x^3m+x^2m+x^m)(2x^3m+3x^2m+6x^m)^1/mdx

now

take 6 common in the 2nd bracket

∫1/x 6^1/m(x^3m+x^2m+x^m)(1/3x^3m+1/2x^2m+x^m)^1/mdx

now
∫6^1/m(x^3m-1+x^2m-1+x^m-1)(1/3x^3m+1/2x^2m+x^m)dx

6^1/m∫...............................

now u can proceed
too lazy to proceed
yime 4 lunch :P

thank u :)

put\ x^m=t\\ \\ x^mdx=\frac{t^{\frac{1}{m}}dt}{m}\\ \\ I=\int (x^{2m}+x^m+1)(2x^{2m}+3x^m+6)^{\frac{1}{m}}x^m\ dx\\ \\ I=\frac{1}{m}\int (t^2+t+1)(2t^3+3t^2+6t)^{\frac{1}{m}}dt\\ \\ put\ 2t^3+3t^2+6t = u\\ \\ 6(t^2+t+1)dt = du\\ \\ I=\frac{1}{6m}\int u^{\frac{1}{m}}du=\frac{u^(\frac{m+1}{m})}{6(m+1)}+C

put back u in terms of x and you will get it...