\(\int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [289]

Optimal result

Integrand size = 24, antiderivative size = 31 \[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 i a \sec (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3574} \[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 i a \sec (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 (i \cos (c+d x)+\sin (c+d x)) \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

(2*(I*Cos[c + d*x] + Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [A] (verified)

Time = 5.63 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}{d}\)	\(37\)

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{d} \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/d

Sympy [F]

\[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec {\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x), x)

Maxima [F]

\[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2\,\left (\sin \left (c+d\,x\right )+\cos \left (c+d\,x\right )\,1{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{d} \]

[In]

int((a + a*tan(c + d*x)*1i)^(1/2)/cos(c + d*x),x)

[Out]

(2*(cos(c + d*x)*1i + sin(c + d*x))*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^
(1/2))/d