\(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [377]

Optimal result

Integrand size = 26, antiderivative size = 270 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {1155 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d} \] Output:

1155/8192*I*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2 
))*2^(1/2)/a^(5/2)/d+1/8*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(5/2)+11/96*I 
*cos(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)+385/2048*I*cos(d*x+c)/a^2/d/(a+ 
I*a*tan(d*x+c))^(1/2)+33/256*I*cos(d*x+c)^3/a^2/d/(a+I*a*tan(d*x+c))^(1/2) 
-1155/4096*I*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d-77/512*I*cos(d*x+c) 
^3*(a+I*a*tan(d*x+c))^(1/2)/a^3/d
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^3(c+d x) \left (-3325-3465 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-1605 \cos (2 (c+d x))+1800 \cos (4 (c+d x))+80 \cos (6 (c+d x))+1111 i \sin (2 (c+d x))+2552 i \sin (4 (c+d x))+176 i \sin (6 (c+d x))\right )}{24576 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((I/24576)*Sec[c + d*x]^3*(-3325 - 3465*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2 
*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - 1605*Cos[2*(c + d 
*x)] + 1800*Cos[4*(c + d*x)] + 80*Cos[6*(c + d*x)] + (1111*I)*Sin[2*(c + d 
*x)] + (2552*I)*Sin[4*(c + d*x)] + (176*I)*Sin[6*(c + d*x)]))/(a^2*d*(-I + 
 Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3042, 3983, 3042, 3983, 3042, 3983, 3042, 3978, 3042, 3983, 3042, 3971, 3042, 3970, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^3 (a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {11 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {11 \left (\frac {3 \int \frac {\cos ^3(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \left (\frac {3 \int \frac {1}{\sec (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3971

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

((I/8)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (11*(((I/6)*Cos[ 
c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (3*(((I/4)*Cos[c + d*x]^3)/ 
(d*Sqrt[a + I*a*Tan[c + d*x]]) + (7*(((-1/3*I)*Cos[c + d*x]^3*Sqrt[a + I*a 
*Tan[c + d*x]])/d + (5*a*(((I/2)*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x 
]]) + (3*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a* 
Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) 
/d))/(4*a)))/6))/(8*a)))/(4*a)))/(16*a)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3970

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S 
ymbol] :> Simp[-2*(a/(b*f))   Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ 
Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 
]
 

rule 3971

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

rule 3978

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/( 
a*f*m)), x] + Simp[a*((m + n)/(m*d^2))   Int[(d*Sec[e + f*x])^(m + 2)*(a + 
b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b 
^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 3983

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

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (221 ) = 442\).

Time = 9.09 (sec) , antiderivative size = 480, normalized size of antiderivative = 1.78

method	result	size
default	\(-\frac {i \left (13860 \cos \left (d x +c \right )^{2}+6930 \cos \left (d x +c \right )-3465\right ) \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {i-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\left (13860 \cos \left (d x +c \right )^{3}+6930 \cos \left (d x +c \right )^{2}-10395 \cos \left (d x +c \right )-3465\right ) \operatorname {arctanh}\left (\frac {i-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \left (-13860 \cos \left (d x +c \right )^{2}+3465\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \cos \left (d x +c \right ) \left (13860 \cos \left (d x +c \right )^{2}-10395\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-5632 \cos \left (d x +c \right )^{4}-5632 \cos \left (d x +c \right )^{3}-14784 \cos \left (d x +c \right )^{2}+12936 \cos \left (d x +c \right )+6930\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \cos \left (d x +c \right ) \left (2560 \cos \left (d x +c \right )^{5}+2560 \cos \left (d x +c \right )^{4}+10560 \cos \left (d x +c \right )^{3}-17160 \cos \left (d x +c \right )^{2}-16170 \cos \left (d x +c \right )+4620\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{24576 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+2\right )+2 \cos \left (d x +c \right )^{3}+2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\)	\(480\)

Input:

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24576/d*(I*(13860*cos(d*x+c)^2+6930*cos(d*x+c)-3465)*sin(d*x+c)*arctanh 
(1/2/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(I-cot(d*x+c)+csc(d*x+c)))+(13860* 
cos(d*x+c)^3+6930*cos(d*x+c)^2-10395*cos(d*x+c)-3465)*arctanh(1/2/(-cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*(I-cot(d*x+c)+csc(d*x+c)))+sin(d*x+c)*(-13860*c 
os(d*x+c)^2+3465)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*cos(d*x+c 
)*(13860*cos(d*x+c)^2-10395)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+ 
sin(d*x+c)*cos(d*x+c)*(-5632*cos(d*x+c)^4-5632*cos(d*x+c)^3-14784*cos(d*x+ 
c)^2+12936*cos(d*x+c)+6930)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*cos(d*x+c 
)*(2560*cos(d*x+c)^5+2560*cos(d*x+c)^4+10560*cos(d*x+c)^3-17160*cos(d*x+c) 
^2-16170*cos(d*x+c)+4620)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(-cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)/(I*sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)+2)+2*cos(d*x 
+c)^3+2*cos(d*x+c)^2-cos(d*x+c)-1)/(a*(1+I*tan(d*x+c)))^(1/2)/a^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-128 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 2176 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 247 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3325 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1358 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 376 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{24576 \, a^{3} d} \] Input:

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

Output:

1/24576*(-3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log 
(-1155/2048*(sqrt(2)*sqrt(1/2)*(I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqr 
t(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^ 
2*d)) + 3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log(- 
1155/2048*(sqrt(2)*sqrt(1/2)*(-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt 
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2 
*d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-128*I*e^(12*I*d*x + 12* 
I*c) - 2176*I*e^(10*I*d*x + 10*I*c) + 247*I*e^(8*I*d*x + 8*I*c) + 3325*I*e 
^(6*I*d*x + 6*I*c) + 1358*I*e^(4*I*d*x + 4*I*c) + 376*I*e^(2*I*d*x + 2*I*c 
) + 48*I))*e^(-8*I*d*x - 8*I*c)/(a^3*d)
 

Sympy [F]

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

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

Output:

Integral(cos(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), 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. 3789 vs. \(2 (207) = 414\).

Time = 0.43 (sec) , antiderivative size = 3789, normalized size of antiderivative = 14.03 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/98304*(4*(cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + sin( 
1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*cos(1/4*arctan2(sin 
(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)^(3/4)*(15*((-I*sqrt(2)*cos(8*d*x + 
8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d 
*x + 8*c)))^2 + (-I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*s 
in(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(-I*sqrt(2)*cos( 
8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), 
 cos(8*d*x + 8*c))) - I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c 
))*cos(7/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), c 
os(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + (55*I*sqrt(2)* 
cos(8*d*x + 8*c) + 960*I*sqrt(2)*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d 
*x + 8*c))) - 1296*I*sqrt(2)*cos(1/2*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 
 8*c))) + 55*sqrt(2)*sin(8*d*x + 8*c) + 960*sqrt(2)*sin(3/4*arctan2(sin(8* 
d*x + 8*c), cos(8*d*x + 8*c))) - 1296*sqrt(2)*sin(1/2*arctan2(sin(8*d*x + 
8*c), cos(8*d*x + 8*c))) + 128*I*sqrt(2))*cos(3/2*arctan2(sin(1/4*arctan2( 
sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), co 
s(8*d*x + 8*c))) + 1)) + 15*((sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d 
*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + (sqrt( 
2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(1/4*arctan2(sin(8*d* 
x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\cos \left (d x +c \right )^{3}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

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

Output:

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