$u\={\sec }^{2 k-1}\left(x\right)$

$\mathrm{dv}\={\sec }^2\left(x\right)\mathrm{dx}$

$\mathrm{du}\=\left(2 k-1\right){\sec }^{2 k-2}\left(x\right)\cdot \left(\sec \left(x\right)\tan \left(x\right)\mathrm{dx}\right)$

$v\=\tan \left(x\right)$

$\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)-\left(2 k-1\right)\int \tan \left(x\right)\cdot \left({\sec }^{2 k-2}\left(x\right)\cdot \sec \left(x\right)\cdot \tan \left(x\right)\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)-\left(2 k-1\right)\int {\tan }^2\left(x\right){\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)-\left(2 k-1\right)\int \left({\sec }^2\left(x\right)-1\right)\cdot {\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)-\left(2 k-1\right)\left(\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}-\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}\right)$

$\left(1\+\left(2 k-1\right)\right)\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$2 k\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\=\frac{1}{2 k}\left({\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}\right)$

$\left(1\+\left(2 k-1\right)\right)\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$2 k\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\={\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}$

$\int {\sec }^{2 k\+1}\left(x\right) \mathrm{dx}$

$\=\frac{1}{2 k}\left({\sec }^{2 k-1}\left(x\right)\tan \left(x\right)\+\left(2 k-1\right)\int {\sec }^{2 k-1}\left(x\right) \mathrm{dx}\right)$

Initialize

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$

$Q≔\int {\sec }^{2 k\+1}\left(x\right) \mathit{\ⅆ}x\:$

Integrate by parts, setting $u\={\sec }^{2 k-1}\left(x\right)$ and $\mathrm{dv}\={\sec }^2\left(x\right) \mathrm{dx}$ 

$q_1≔\mathrm{Parts}\left(Q\,{\sec }^{2 k-1}\left(x\right)\right)$

$\frac{\sin \left(x\right){\sec \left(x\right)}^{2k-1}}{\cos \left(x\right)}-\left(\∫\frac{\sin \left(x\right){\sec \left(x\right)}^{2k-1}\left(2k-1\right)\tan \left(x\right)}{\cos \left(x\right)}\ⅆx\right)$

Restore $\tan \left(x\right)$, replace $2 k-1$ with $K$, and apply the identity ${\tan }^2\left(x\right)\={\sec }^2\left(x\right)-1$ 

$$\begin{gathered} q_2≔\mathrm{eval}\left(q_1\,\left[2 k-1\=K\, \sin \left(x\right)\=\tan \left(x\right)\cdot \cos \left(x\right)\right]\right)\: \\ {q}_3≔\mathrm{eval}\left(q_2\,{\tan }^2\left(x\right)\={\sec }^2\left(x\right)-1\right) \end{gathered}$$

$\tan \left(x\right){\sec \left(x\right)}^K-\left(\∫\left({\sec \left(x\right)}^2-1\right){\sec \left(x\right)}^{K}K\ⅆx\right)$

Expand the product in the integrand and combine the resulting secant terms
Split the integral and restore $2 k-1$

$$\begin{gathered} q_4≔\mathrm{combine}\left(\mathrm{expand}\left(q_3\right)\right) assuming K\colon\colon \mathrm{posint}\: \\ {q}_5≔\mathrm{eval}\left(\mathrm{expand}\left(q_4\,K\+2\right)\,K\=2 k-1\right) \end{gathered}$$

$\left(2k-1\right)\left(\∫{\sec \left(x\right)}^{2k-1}\ⅆx\right)-\left(2k-1\right)\left(\∫{\sec \left(x\right)}^{2k\+1}\ⅆx\right)\+\tan \left(x\right){\sec \left(x\right)}^{2k-1}$

Solve for the unknown integral

$\mathrm{isolate}\left(Q-q_5\,Q\right)$

$\∫{\sec \left(x\right)}^{2k\+1}\ⅆx\=\frac{1}{2}\frac{\left(2k-1\right)\left(\∫{\sec \left(x\right)}^{2k-1}\ⅆx\right)\+\tan \left(x\right){\sec \left(x\right)}^{2k-1}}{k}$