\(\int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [343]

Optimal result

Integrand size = 26, antiderivative size = 35 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i a \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {3574} \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i a \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

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

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 ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sec (c+d x) (i+\tan (c+d x))}{3 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.64 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {\frac {2 i \sec \left (d x +c \right )}{3}+\frac {2 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{3}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(44\)

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {4 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (27) = 54\).

Time = 0.33 (sec) , antiderivative size = 206, normalized size of antiderivative = 5.89 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \, {\left (-i \, \sqrt {a} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {i \, \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1}}{3 \, {\left (a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.80 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,1{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{3\,a\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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