As per Table 6.2.8, the given integral yields to the substitution $u\=\tan \left(x\right)$ because the exponent on the factor ${\sec }^6\left(x\right)$ is the even number 6. The complete solution appears in Table 6.2.3(a).

The annotated stepwise solution in Table 6.2.15 is obtained with the  tutor after both the Sum and Constant Multiple rules were selected as Understood Rules. There are two major ideas in the solution, the rewrite and the change of variables $u\=\tan \left(x\right)$. Left to its own devices, the tutor would make the first rewrite, but then expand ${\left(1\+{\tan }^2\left(x\right)\right)}^2$and make another rewrite in which the integrand itself is completely expanded. The guided solution in Table 6.2.3(b) is a bit more compact, but follows the same strategy, namely, that of Table 6.2.8.

Table 6.2.3(c) contains a command-based solution from first principles.

Maple itself seems to take a different approach to the evaluation of the integral:

$$ $\int {\tan }^4\left(x\right){\sec }^6\left(x\right) \mathit{\ⅆ}x$ = $\frac{1}{9}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^9}\+\frac{4}{63}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^7}\+\frac{8}{315}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^5}$$$

That these two forms of the solution are equivalent can be established by comparing the difference:

$\mathrm{simplify}\left(\frac{1}{9}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^9}\+\frac{4}{63}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^7}\+\frac{8}{315}\frac{{\sin \left(x\right)}^5}{{\cos \left(x\right)}^5}-\left(\frac{1}{9}{\tan \left(x\right)}^9\+\frac{2}{7}{\tan \left(x\right)}^7\+\frac{1}{5}{\tan \left(x\right)}^5\right)\right)$ = $0$$$

The reader who is so inclined can massage one form to the other by applying the identities $\sin \left(x\right)\=\cos \left(x\right)\cdot \tan \left(x\right)\,\cos \left(x\right)\=1\/\sec \left(x\right)$, and ${\sec }^2\left(x\right)\=1\+{\tan }^2\left(x\right)$.