\(\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [357]

Optimal result

Integrand size = 26, antiderivative size = 35 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i a \sec ^5(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

2/5*I*a*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.08 (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} \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i a \sec ^5(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((2*I)/5)*a*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2))

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 ^5(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^3(c+d x) (1-i \tan (c+d x))}{5 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^3*(1 - I*Tan[c + d*x]))/(5*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F(-1)]

Timed out.

\[\int \frac {\sec ^{5}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x)

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. 59 vs. \(2 (27) = 54\).

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{5 \, {\left (a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

8/5*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(a^2*d*e^(4*I*d*x + 4*I*c) + 2*a^2*d*e^(2*I*d*x + 2*I*c) + a^2
*d)

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**5/(I*a*(tan(c + d*x) - I))**(3/2), 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. 350 vs. \(2 (27) = 54\).

Time = 0.34 (sec) , antiderivative size = 350, normalized size of antiderivative = 10.00 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-i \, \sqrt {a} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 i \, \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {6 \, \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 i \, \sqrt {a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {i \, \sqrt {a} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {3}{2}}}{5 \, {\left (a^{2} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/5*(-I*sqrt(a) - 2*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 6*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 6*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*I*sqrt(a)
*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + I*sqrt(a)*sin(d*x + c)^
8/(cos(d*x + c) + 1)^8)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(3/2
)/((a^2 - 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^2*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) +
1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(3/2))

Giac [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.97 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )-\sin \left (c\right )\,1{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+1-\sin \left (2\,c+2\,d\,x\right )\,2{}\mathrm {i}-\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,a^2\,d\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )} \]

[In]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

((cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*
d*x) + 1))^(1/2)*(2*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) - sin(2*c + 2*d*x)*2i - sin(4*c + 4*d*x)*1i + 1)*4i)/(
5*a^2*d*(4*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) + 3))