\(\int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [302]

Optimal result

Integrand size = 24, antiderivative size = 69 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

8/3*I*a^2*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+2/3*I*a*sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3575, 3574} \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[In]

Int[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((8*I)/3)*a^2*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((2*I)/3)*a*Sec[c + d*x]*Sqrt[a + I*a*Tan[c + d
*x]])/d

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]

Rule 3575

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 - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{3} (4 a) \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 a (\cos (c)-i \sin (c)) (\cos (d x)-i \sin (d x)) (-5 i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[In]

Integrate[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(-2*a*(Cos[c] - I*Sin[c])*(Cos[d*x] - I*Sin[d*x])*(-5*I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

Maple [A] (verified)

Time = 6.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {2 i a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (8 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )-4 i \sin \left (d x +c \right )+8 \left (\cos ^{3}\left (d x +c \right )\right )+\sin \left (d x +c \right ) \tan \left (d x +c \right )-3 \cos \left (d x +c \right )\right )}{3 d}\)	\(80\)

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*I/d*a*(a*(1+I*tan(d*x+c)))^(1/2)*(8*sin(d*x+c)^2*cos(d*x+c)-4*I*sin(d*x+c)+8*cos(d*x+c)^3+sin(d*x+c)*tan(d
*x+c)-3*cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {4 \, \sqrt {2} {\left (-3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-4/3*sqrt(2)*(-3*I*a*e^(2*I*d*x + 2*I*c) - 2*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(2*I*d*x + 2*I*c) + d
)

Sympy [F]

\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*sec(c + d*x), x)

Maxima [F]

\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c), x)

Giac [F]

\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c), x)

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.42 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2\,a\,\sqrt {\frac {a\,\left (2\,{\cos \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (c+d\,x\right )}^2}}\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,8{}\mathrm {i}+{\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2\,2{}\mathrm {i}+\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )-5{}\mathrm {i}\right )}{3\,d\,{\cos \left (c+d\,x\right )}^2} \]

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/cos(c + d*x),x)

[Out]

(2*a*((a*(sin(2*c + 2*d*x)*1i + 2*cos(c + d*x)^2))/(2*cos(c + d*x)^2))^(1/2)*(sin(c + d*x) + sin(3*c + 3*d*x)
+ cos(c/2 + (d*x)/2)^2*8i + cos((3*c)/2 + (3*d*x)/2)^2*2i - 5i))/(3*d*cos(c + d*x)^2)