\(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [364]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d} \]

[Out]

-8*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+8/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^4/d-2/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^5/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d}-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^3 d} \]

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-8*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d) + (((8*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^4*d) - (((2*I)/5)*(a
+ I*a*Tan[c + d*x])^(5/2))/(a^5*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {(a-x)^2}{\sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {4 a^2}{\sqrt {a+x}}-4 a \sqrt {a+x}+(a+x)^{3/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \sqrt {a+i a \tan (c+d x)} \left (-43+14 i \tan (c+d x)+3 \tan ^2(c+d x)\right )}{15 a^3 d} \]

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((2*I)/15)*Sqrt[a + I*a*Tan[c + d*x]]*(-43 + (14*I)*Tan[c + d*x] + 3*Tan[c + d*x]^2))/(a^3*d)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{5}}\)	\(63\)
default	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{5}}\)	\(63\)

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^5*(-1/5*(a+I*a*tan(d*x+c))^(5/2)+4/3*a*(a+I*a*tan(d*x+c))^(3/2)-4*a^2*(a+I*a*tan(d*x+c))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (8 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 20 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 15 i \, e^{\left (i \, d x + i \, c\right )}\right )}}{15 \, {\left (a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-8/15*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(8*I*e^(5*I*d*x + 5*I*c) + 20*I*e^(3*I*d*x + 3*I*c) + 15*I*e^(
I*d*x + I*c))/(a^3*d*e^(4*I*d*x + 4*I*c) + 2*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{6}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**6/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 i \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 60 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{15 \, a^{5} d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/15*I*(3*(I*a*tan(d*x + c) + a)^(5/2) - 20*(I*a*tan(d*x + c) + a)^(3/2)*a + 60*sqrt(I*a*tan(d*x + c) + a)*a^
2)/(a^5*d)

Giac [F]

\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{6}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.80 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {4\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,321{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,132{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,23{}\mathrm {i}+35\,\sin \left (2\,c+2\,d\,x\right )+28\,\sin \left (4\,c+4\,d\,x\right )+7\,\sin \left (6\,c+6\,d\,x\right )+212{}\mathrm {i}\right )}{15\,a^3\,d\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

-(4*((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(2*c + 2*d*x)*321i + c
os(4*c + 4*d*x)*132i + cos(6*c + 6*d*x)*23i + 35*sin(2*c + 2*d*x) + 28*sin(4*c + 4*d*x) + 7*sin(6*c + 6*d*x) +
212i))/(15*a^3*d*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))