\(\int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\) [14]

Optimal result

Integrand size = 20, antiderivative size = 27 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

[Out]

a*arctanh(sin(d*x+c))/d+I*a*sec(d*x+c)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3567, 3855} \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 3567

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {i a \sec (c+d x)}{d}+a \int \sec (c+d x) \, dx \\ & = \frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {\frac {i a}{\cos \left (d x +c \right )}+a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(34\)
default	\(\frac {\frac {i a}{\cos \left (d x +c \right )}+a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(34\)
risch	\(\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(68\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.04 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {2 i \, a e^{\left (i \, d x + i \, c\right )} + {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

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

[Out]

(2*I*a*e^(I*d*x + I*c) + (a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c) + I) - (a*e^(2*I*d*x + 2*I*c) + a)*lo
g(e^(I*d*x + I*c) - I))/(d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 2.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\begin {cases} \frac {a \log {\left (\tan {\left (c + d x \right )} + \sec {\left (c + d x \right )} \right )} + i a \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\left (c \right )} + a\right ) \sec {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((a*log(tan(c + d*x) + sec(c + d*x)) + I*a*sec(c + d*x))/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*sec(c), T
rue))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {i \, a}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (25) = 50\).

Time = 0.37 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 i \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

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

[Out]

(a*log(tan(1/2*d*x + 1/2*c) + 1) - a*log(tan(1/2*d*x + 1/2*c) - 1) - 2*I*a/(tan(1/2*d*x + 1/2*c)^2 - 1))/d

Mupad [B] (verification not implemented)

Time = 4.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,2{}\mathrm {i}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (a*2i)/(d*(tan(c/2 + (d*x)/2)^2 - 1))