\(\int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [392]

Optimal result

Integrand size = 24, antiderivative size = 227 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {315 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2048 \sqrt {2} a^{7/2} d}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}+\frac {3 i \cos (c+d x)}{32 a d (a+i a \tan (c+d x))^{5/2}}+\frac {21 i \cos (c+d x)}{256 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {105 i \cos (c+d x)}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {315 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{2048 a^4 d} \] Output:

315/4096*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^(7/2)/d+1/8*I*cos(d*x+c)/d/(a+I*a*tan(d*x+c))^(7/2)+3/32*I*cos 
(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(5/2)+21/256*I*cos(d*x+c)/a^2/d/(a+I*a*tan( 
d*x+c))^(3/2)+105/1024*I*cos(d*x+c)/a^3/d/(a+I*a*tan(d*x+c))^(1/2)-315/204 
8*I*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^4/d
 

Mathematica [A] (verified)

Time = 2.02 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.62 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {\sec ^3(c+d x) \left (420+\frac {630 e^{4 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+826 \cos (2 (c+d x))-224 \cos (4 (c+d x))+474 i \sin (2 (c+d x))-288 i \sin (4 (c+d x))\right )}{4096 a^3 d (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

-1/4096*(Sec[c + d*x]^3*(420 + (630*E^((4*I)*(c + d*x))*ArcTanh[Sqrt[1 + E 
^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)*(c + d*x))] + 826*Cos[2*(c + d*x)] 
 - 224*Cos[4*(c + d*x)] + (474*I)*Sin[2*(c + d*x)] - (288*I)*Sin[4*(c + d* 
x)]))/(a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {3042, 3983, 3042, 3983, 3042, 3983, 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.

\(\displaystyle \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x) (a+i a \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9 \int \frac {\cos (c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 \int \frac {1}{\sec (c+d x) (i \tan (c+d x) a+a)^{5/2}}dx}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9 \left (\frac {7 \int \frac {\cos (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 \left (\frac {7 \int \frac {1}{\sec (c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3971

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\right )}{12 a}+\frac {i \cos (c+d x)}{6 d (a+i a \tan (c+d x))^{5/2}}\right )}{16 a}+\frac {i \cos (c+d x)}{8 d (a+i a \tan (c+d x))^{7/2}}\)

Input:

Int[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((I/8)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (9*(((I/6)*Cos[c + 
 d*x])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (7*(((I/4)*Cos[c + d*x])/(d*(a + 
 I*a*Tan[c + d*x])^(3/2)) + (5*(((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)))/(8*a)))/(12*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 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. 491 vs. \(2 (184 ) = 368\).

Time = 10.00 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.17

method	result	size
default	\(-\frac {\tan \left (d x +c \right ) \sec \left (d x +c \right )^{2} \left (2520 \cos \left (d x +c \right )^{3}+1260 \cos \left (d x +c \right )^{2}-1260 \cos \left (d x +c \right )-315\right ) \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (-2520 \cos \left (d x +c \right )-1260+2520 \sec \left (d x +c \right )+945 \sec \left (d x +c \right )^{2}-315 \sec \left (d x +c \right )^{3}\right )+i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2520 \sin \left (d x +c \right )-1260 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2520 \cos \left (d x +c \right )-2520 \sec \left (d x +c \right )+315 \sec \left (d x +c \right )^{3}\right )+i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \left (2304 \cos \left (d x +c \right )^{3}-2736 \cos \left (d x +c \right )^{2}-2100 \cos \left (d x +c \right )+420\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1792 \cos \left (d x +c \right )^{2}-3248 \cos \left (d x +c \right )-3444+1596 \sec \left (d x +c \right )+630 \sec \left (d x +c \right )^{2}\right )}{4096 d \left (\cos \left (d x +c \right )+1\right ) \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(492\)

Input:

int(cos(d*x+c)/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4096/d/(cos(d*x+c)+1)/(-tan(d*x+c)+I)^3/(-cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)/a^3/(a*(1+I*tan(d*x+c)))^(1/2)*(tan(d*x+c)*sec(d*x+c)^2*(2520*cos(d*x+ 
c)^3+1260*cos(d*x+c)^2-1260*cos(d*x+c)-315)*arctanh(1/2/(cot(d*x+c)^2-2*co 
t(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2 
))+I*arctanh(1/2/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/ 
2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2))*(-2520*cos(d*x+c)-1260+2520*sec(d*x+ 
c)+945*sec(d*x+c)^2-315*sec(d*x+c)^3)+I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*(2520*sin(d*x+c)-1260*sec(d*x+c)*tan(d*x+c))+2^(1/2)*(-2*cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*(2520*cos(d*x+c)-2520*sec(d*x+c)+315*sec(d*x+c) 
^3)+I*tan(d*x+c)*sec(d*x+c)*(2304*cos(d*x+c)^3-2736*cos(d*x+c)^2-2100*cos( 
d*x+c)+420)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(-cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*(1792*cos(d*x+c)^2-3248*cos(d*x+c)-3444+1596*sec(d*x+c)+630*sec(d* 
x+c)^2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.32 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {315 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{1024 \, a^{3} d}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {315 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{1024 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-128 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 197 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 535 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 298 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 104 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{4096 \, a^{4} d} \] Input:

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/4096*(-315*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(8*I*d*x + 8*I*c)*log(- 
315/1024*(sqrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x - I*c)/(a^3*d 
)) + 315*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(8*I*d*x + 8*I*c)*log(-315/ 
1024*(sqrt(2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e 
^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x - I*c)/(a^3*d)) 
+ sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-128*I*e^(10*I*d*x + 10*I*c) 
+ 197*I*e^(8*I*d*x + 8*I*c) + 535*I*e^(6*I*d*x + 6*I*c) + 298*I*e^(4*I*d*x 
 + 4*I*c) + 104*I*e^(2*I*d*x + 2*I*c) + 16*I))*e^(-8*I*d*x - 8*I*c)/(a^4*d 
)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))**(7/2),x)
 

Output:

Timed out
 

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. 2779 vs. \(2 (172) = 344\).

Time = 0.36 (sec) , antiderivative size = 2779, normalized size of antiderivative = 12.24 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

-1/16384*(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)*(325*((-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))* 
sin(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))), 
cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + 643*(-I*sqrt( 
2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(3/2*arctan2(sin(1/4*ar 
ctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8* 
c), cos(8*d*x + 8*c))) + 1)) + 325*((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(s 
in(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(sqrt(2)*cos(8*d*x + 8*c) - I*sq 
rt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c) 
)) + sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(7/2*arctan 
2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(si 
n(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + 643*(sqrt(2)*cos(8*d*x + 8*c...
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(7/2),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. \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \] Input:

int(cos(c + d*x)/(a + a*tan(c + d*x)*1i)^(7/2),x)
 

Output:

int(cos(c + d*x)/(a + a*tan(c + d*x)*1i)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right ) \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) i -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right )\right )}{a^{4}} \] Input:

int(cos(d*x+c)/(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

(sqrt(a)*(int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)*tan(c + d*x))/(tan(c 
+ d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 
- 3*tan(c + d*x)*i - 1),x)*i - int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)) 
/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + 
d*x)**2 - 3*tan(c + d*x)*i - 1),x)))/a**4