What is the integral of sec^6(x)tan^10(x)dx

∫sec^6(x)tan^10(x)dx=????
I have tried so many different ways to do this question but I haven't figured out the right answer!
Can anyone help?

The solution is (26cos(2x)+cos(4x)+168)tan^11(x)sec^4(x) / 2145 + Constant.

Some intermediate steps:

For the integrand tan^10(x)sec^6(x), use the identity sec^2(x)=tan^2(x)+1:
int(tan^10(x)(tan^2(x)+1)^2sec^2(x))dx

For the integrand tan^10(x)(tan^2(x)+1)2sec^2(x), substitute u=tan(x) and du=sec^2(x)dx:
int(u^10(u^2+1)^2du

Expanding the integrand u^10(u^2+1)^2 gives u^14+2u^12+u^10:
int(u^14+2u^12+u^10)du

Integrate the sum term and factor out constants:
int(u^14)du+2int(u^12)du+int(u^10)du

The integral of u^12 is u^13/13:
int(u^14)du+2u^13/13+int(u^10)du

The integral of u^10 is u^11/11:
int(u^14du+2u^13/13+u^11/11

The integral of u^14 is u^15/15:
u^15/15+2u^13/13+u^11/11+constant

Substitute back for u = tan(x):
tan^15(x)/15+2tan^13(x)/13+tan^11(x) / 11+constant

Which is equal to:
(26cos(2x)+cos(4x)+168)tan^11(x)sec^4(x… / 2145 + Constant

∫sec^6(x)tan^10(x) dx
Use the fact that 1 + tan^2(x) = sec^2(x)
= ∫sec^2(x)tan^10(x)(1 + tan^2(x))^2 dx
Let u = tan(x), du/dx = sec^2(x)
= ∫u^10(u^2 + 1)^2 du
= ∫u^10(u^4 + 2u^2 + 1) du
= ∫u^14 + 2u^12 + u^10 du
= (1/15)u^15 + (2/13)u^13 + (1/11)u^11 + C
= (1/15)tan^15(x) + (2/13)tan^13(x) + (1/11)tan^11(x) + C