\[ \int (e \sec (c+d x))^{-1-n} (a+i a \tan (c+d x))^n \, dx \]
Optimal antiderivative \[ \frac {i \left (e \sec \left (d x +c \right )\right )^{-1-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}}{d \left (1-n \right )}-\frac {i \left (e \sec \left (d x +c \right )\right )^{-1-n} \left (a +i a \tan \left (d x +c \right )\right )^{1+n}}{a d \left (-n^{2}+1\right )} \]
command
integrate((e*sec(d*x+c))**(-1-n)*(a+I*a*tan(d*x+c))**n,x)
Sympy 1.10.1 under Python 3.10.4 output
\[ \frac {\begin {cases} \frac {x \left (e \sec {\left (c \right )}\right )^{- n} \left (i a \tan {\left (c \right )} + a\right )^{n}}{\sec {\left (c \right )}} & \text {for}\: d = 0 \\\frac {d e x \tan {\left (c + d x \right )}}{2 a d \tan {\left (c + d x \right )} - 2 i a d} - \frac {i d e x}{2 a d \tan {\left (c + d x \right )} - 2 i a d} + \frac {e}{2 a d \tan {\left (c + d x \right )} - 2 i a d} & \text {for}\: n = -1 \\\frac {a x \tan ^{2}{\left (c + d x \right )}}{2 e \sec ^{2}{\left (c + d x \right )}} + \frac {a x}{2 e \sec ^{2}{\left (c + d x \right )}} + \frac {a \tan {\left (c + d x \right )}}{2 d e \sec ^{2}{\left (c + d x \right )}} - \frac {i a}{2 d e \sec ^{2}{\left (c + d x \right )}} & \text {for}\: n = 1 \\- \frac {i n \left (i a \tan {\left (c + d x \right )} + a\right )^{n}}{d n^{2} \left (e \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )} - d \left (e \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )}} - \frac {\left (i a \tan {\left (c + d x \right )} + a\right )^{n} \tan {\left (c + d x \right )}}{d n^{2} \left (e \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )} - d \left (e \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )}} & \text {otherwise} \end {cases}}{e} \]
Sympy 1.8 under Python 3.8.8 output
\[ \int \left (e \sec {\left (c + d x \right )}\right )^{- n - 1} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]________________________________________________________________________________________