\(\int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx\) [110]

Optimal result

Integrand size = 22, antiderivative size = 28 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \sec (c+d x)}{d (a+i a \tan (c+d x))} \]

[Out]

I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3569} \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \sec (c+d x)}{d (a+i a \tan (c+d x))} \]

[In]

Int[Sec[c + d*x]/(a + I*a*Tan[c + d*x]),x]

[Out]

(I*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x]))

Rule 3569

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {i \sec (c+d x)}{d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\sec (c+d x)}{a d (-i+\tan (c+d x))} \]

[In]

Integrate[Sec[c + d*x]/(a + I*a*Tan[c + d*x]),x]

[Out]

Sec[c + d*x]/(a*d*(-I + Tan[c + d*x]))

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{a d}\)	\(19\)
derivativedivides	\(\frac {2}{d a \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(23\)
default	\(\frac {2}{d a \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(23\)

[In]

int(sec(d*x+c)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

I/a/d*exp(-I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \, e^{\left (-i \, d x - i \, c\right )}}{a d} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

I*e^(-I*d*x - I*c)/(a*d)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\begin {cases} \frac {\sec {\left (c + d x \right )}}{a d \tan {\left (c + d x \right )} - i a d} & \text {for}\: d \neq 0 \\\frac {x \sec {\left (c \right )}}{i a \tan {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise((sec(c + d*x)/(a*d*tan(c + d*x) - I*a*d), Ne(d, 0)), (x*sec(c)/(I*a*tan(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2}{{\left (-i \, a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )} d} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

2/((-I*a + a*sin(d*x + c)/(cos(d*x + c) + 1))*d)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2}{a d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

2/(a*d*(tan(1/2*d*x + 1/2*c) - I))

Mupad [B] (verification not implemented)

Time = 4.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2{}\mathrm {i}}{a\,d\,\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )} \]

[In]

int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)),x)

[Out]

2i/(a*d*(tan(c/2 + (d*x)/2)*1i + 1))