\(\int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [147]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 139 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {56 \sin (c+d x)}{99 a^3 d}-\frac {56 \sin ^3(c+d x)}{99 a^3 d}+\frac {56 \sin ^5(c+d x)}{165 a^3 d}-\frac {8 \sin ^7(c+d x)}{99 a^3 d}+\frac {i \cos ^5(c+d x)}{11 d (a+i a \tan (c+d x))^3}+\frac {16 i \cos ^7(c+d x)}{99 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

56/99*sin(d*x+c)/a^3/d-56/99*sin(d*x+c)^3/a^3/d+56/165*sin(d*x+c)^5/a^3/d- 
8/99*sin(d*x+c)^7/a^3/d+1/11*I*cos(d*x+c)^5/d/(a+I*a*tan(d*x+c))^3+16/99*I 
*cos(d*x+c)^7/d/(a^3+I*a^3*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (-5775-16632 \cos (2 (c+d x))+5940 \cos (4 (c+d x))+440 \cos (6 (c+d x))+27 \cos (8 (c+d x))-11088 i \sin (2 (c+d x))+7920 i \sin (4 (c+d x))+880 i \sin (6 (c+d x))+72 i \sin (8 (c+d x)))}{63360 a^3 d (-i+\tan (c+d x))^3} \] Input:

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

Output:

(Sec[c + d*x]^3*(-5775 - 16632*Cos[2*(c + d*x)] + 5940*Cos[4*(c + d*x)] + 
440*Cos[6*(c + d*x)] + 27*Cos[8*(c + d*x)] - (11088*I)*Sin[2*(c + d*x)] + 
(7920*I)*Sin[4*(c + d*x)] + (880*I)*Sin[6*(c + d*x)] + (72*I)*Sin[8*(c + d 
*x)]))/(63360*a^3*d*(-I + Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3983, 3042, 3981, 3042, 3113, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

\(\displaystyle \frac {8 \left (\frac {7 \int \cos ^7(c+d x)dx}{9 a^2}+\frac {2 i \cos ^7(c+d x)}{9 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{11 a}+\frac {i \cos ^5(c+d x)}{11 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \left (\frac {7 \int \sin \left (c+d x+\frac {\pi }{2}\right )^7dx}{9 a^2}+\frac {2 i \cos ^7(c+d x)}{9 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{11 a}+\frac {i \cos ^5(c+d x)}{11 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {8 \left (-\frac {7 \int \left (-\sin ^6(c+d x)+3 \sin ^4(c+d x)-3 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{9 a^2 d}+\frac {2 i \cos ^7(c+d x)}{9 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{11 a}+\frac {i \cos ^5(c+d x)}{11 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8 \left (-\frac {7 \left (\frac {1}{7} \sin ^7(c+d x)-\frac {3}{5} \sin ^5(c+d x)+\sin ^3(c+d x)-\sin (c+d x)\right )}{9 a^2 d}+\frac {2 i \cos ^7(c+d x)}{9 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{11 a}+\frac {i \cos ^5(c+d x)}{11 d (a+i a \tan (c+d x))^3}\)

Input:

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

Output:

((I/11)*Cos[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^3) + (8*((-7*(-Sin[c + d 
*x] + Sin[c + d*x]^3 - (3*Sin[c + d*x]^5)/5 + Sin[c + d*x]^7/7))/(9*a^2*d) 
 + (((2*I)/9)*Cos[c + d*x]^7)/(d*(a^2 + I*a^2*Tan[c + d*x]))))/(11*a)
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3113
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3981
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
 f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) 
Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] 
 && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

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 [A] (verified)

Time = 1.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12

method result size
risch \(\frac {i {\mathrm e}^{-7 i \left (d x +c \right )}}{64 a^{3} d}+\frac {i {\mathrm e}^{-9 i \left (d x +c \right )}}{288 a^{3} d}+\frac {i {\mathrm e}^{-11 i \left (d x +c \right )}}{2816 a^{3} d}+\frac {7 i \cos \left (d x +c \right )}{64 a^{3} d}+\frac {21 \sin \left (d x +c \right )}{64 a^{3} d}+\frac {11 i \cos \left (5 d x +5 c \right )}{256 a^{3} d}+\frac {57 \sin \left (5 d x +5 c \right )}{1280 a^{3} d}+\frac {31 i \cos \left (3 d x +3 c \right )}{384 a^{3} d}+\frac {13 \sin \left (3 d x +3 c \right )}{128 a^{3} d}\) \(155\)
derivativedivides \(\frac {\frac {217 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {303 i}{64 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {169 i}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {i}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{4}}-\frac {5 i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {8}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {106}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {33}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {623}{20 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {365}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {219}{128 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {23 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}+\frac {1}{40 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{5}}-\frac {7}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {37}{128 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{a^{3} d}\) \(273\)
default \(\frac {\frac {217 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {303 i}{64 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {169 i}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {i}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{4}}-\frac {5 i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {8}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {106}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {33}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {623}{20 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {365}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {219}{128 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {23 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}+\frac {1}{40 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{5}}-\frac {7}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {37}{128 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{a^{3} d}\) \(273\)

Input:

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

Output:

1/64*I/a^3/d*exp(-7*I*(d*x+c))+1/288*I/a^3/d*exp(-9*I*(d*x+c))+1/2816*I/a^ 
3/d*exp(-11*I*(d*x+c))+7/64*I/a^3/d*cos(d*x+c)+21/64*sin(d*x+c)/a^3/d+11/2 
56*I/a^3/d*cos(5*d*x+5*c)+57/1280/a^3/d*sin(5*d*x+5*c)+31/384*I/a^3/d*cos( 
3*d*x+3*c)+13/128/a^3/d*sin(3*d*x+3*c)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (-99 i \, e^{\left (16 i \, d x + 16 i \, c\right )} - 1320 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 13860 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 27720 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 11550 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 5544 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1980 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 440 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 45 i\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{126720 \, a^{3} d} \] Input:

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

Output:

1/126720*(-99*I*e^(16*I*d*x + 16*I*c) - 1320*I*e^(14*I*d*x + 14*I*c) - 138 
60*I*e^(12*I*d*x + 12*I*c) + 27720*I*e^(10*I*d*x + 10*I*c) + 11550*I*e^(8* 
I*d*x + 8*I*c) + 5544*I*e^(6*I*d*x + 6*I*c) + 1980*I*e^(4*I*d*x + 4*I*c) + 
 440*I*e^(2*I*d*x + 2*I*c) + 45*I)*e^(-11*I*d*x - 11*I*c)/(a^3*d)
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (122) = 244\).

Time = 0.46 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.40 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 626985510622986240 i a^{24} d^{8} e^{41 i c} e^{5 i d x} - 8359806808306483200 i a^{24} d^{8} e^{39 i c} e^{3 i d x} - 87777971487218073600 i a^{24} d^{8} e^{37 i c} e^{i d x} + 175555942974436147200 i a^{24} d^{8} e^{35 i c} e^{- i d x} + 73148309572681728000 i a^{24} d^{8} e^{33 i c} e^{- 3 i d x} + 35111188594887229440 i a^{24} d^{8} e^{31 i c} e^{- 5 i d x} + 12539710212459724800 i a^{24} d^{8} e^{29 i c} e^{- 7 i d x} + 2786602269435494400 i a^{24} d^{8} e^{27 i c} e^{- 9 i d x} + 284993413919539200 i a^{24} d^{8} e^{25 i c} e^{- 11 i d x}\right ) e^{- 36 i c}}{802541453597422387200 a^{27} d^{9}} & \text {for}\: a^{27} d^{9} e^{36 i c} \neq 0 \\\frac {x \left (e^{16 i c} + 8 e^{14 i c} + 28 e^{12 i c} + 56 e^{10 i c} + 70 e^{8 i c} + 56 e^{6 i c} + 28 e^{4 i c} + 8 e^{2 i c} + 1\right ) e^{- 11 i c}}{256 a^{3}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((-626985510622986240*I*a**24*d**8*exp(41*I*c)*exp(5*I*d*x) - 83 
59806808306483200*I*a**24*d**8*exp(39*I*c)*exp(3*I*d*x) - 8777797148721807 
3600*I*a**24*d**8*exp(37*I*c)*exp(I*d*x) + 175555942974436147200*I*a**24*d 
**8*exp(35*I*c)*exp(-I*d*x) + 73148309572681728000*I*a**24*d**8*exp(33*I*c 
)*exp(-3*I*d*x) + 35111188594887229440*I*a**24*d**8*exp(31*I*c)*exp(-5*I*d 
*x) + 12539710212459724800*I*a**24*d**8*exp(29*I*c)*exp(-7*I*d*x) + 278660 
2269435494400*I*a**24*d**8*exp(27*I*c)*exp(-9*I*d*x) + 284993413919539200* 
I*a**24*d**8*exp(25*I*c)*exp(-11*I*d*x))*exp(-36*I*c)/(8025414535974223872 
00*a**27*d**9), Ne(a**27*d**9*exp(36*I*c), 0)), (x*(exp(16*I*c) + 8*exp(14 
*I*c) + 28*exp(12*I*c) + 56*exp(10*I*c) + 70*exp(8*I*c) + 56*exp(6*I*c) + 
28*exp(4*I*c) + 8*exp(2*I*c) + 1)*exp(-11*I*c)/(256*a**3), True))
 

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {33 \, {\left (555 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1920 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2710 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1760 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 463\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{5}} + \frac {108405 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 784080 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2901195 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 6652800 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 10407474 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 11435424 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8949270 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4899840 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1816265 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 411664 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 47279}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{11}}}{63360 \, d} \] Input:

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

Output:

1/63360*(33*(555*tan(1/2*d*x + 1/2*c)^4 + 1920*I*tan(1/2*d*x + 1/2*c)^3 - 
2710*tan(1/2*d*x + 1/2*c)^2 - 1760*I*tan(1/2*d*x + 1/2*c) + 463)/(a^3*(tan 
(1/2*d*x + 1/2*c) + I)^5) + (108405*tan(1/2*d*x + 1/2*c)^10 - 784080*I*tan 
(1/2*d*x + 1/2*c)^9 - 2901195*tan(1/2*d*x + 1/2*c)^8 + 6652800*I*tan(1/2*d 
*x + 1/2*c)^7 + 10407474*tan(1/2*d*x + 1/2*c)^6 - 11435424*I*tan(1/2*d*x + 
 1/2*c)^5 - 8949270*tan(1/2*d*x + 1/2*c)^4 + 4899840*I*tan(1/2*d*x + 1/2*c 
)^3 + 1816265*tan(1/2*d*x + 1/2*c)^2 - 411664*I*tan(1/2*d*x + 1/2*c) - 472 
79)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^11))/d
 

Mupad [B] (verification not implemented)

Time = 2.68 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {\cos \left (7\,c+7\,d\,x\right )}{64}+\frac {\cos \left (9\,c+9\,d\,x\right )}{288}+\frac {\cos \left (11\,c+11\,d\,x\right )}{2816}-\frac {\sin \left (7\,c+7\,d\,x\right )\,1{}\mathrm {i}}{64}-\frac {\sin \left (9\,c+9\,d\,x\right )\,1{}\mathrm {i}}{288}-\frac {\sin \left (11\,c+11\,d\,x\right )\,1{}\mathrm {i}}{2816}+\frac {\sqrt {224}\,\cos \left (5\,c+5\,d\,x+\mathrm {atanh}\left (\frac {57}{55}\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{1280}+\frac {\sqrt {560}\,\cos \left (3\,c+3\,d\,x+\mathrm {atanh}\left (\frac {39}{31}\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{384}+\frac {\sqrt {2}\,\cos \left (c+d\,x+\mathrm {atanh}\left (3\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{32}\right )\,1{}\mathrm {i}}{a^3\,d} \] Input:

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

Output:

((cos(7*c + 7*d*x)/64 + cos(9*c + 9*d*x)/288 + cos(11*c + 11*d*x)/2816 - ( 
sin(7*c + 7*d*x)*1i)/64 - (sin(9*c + 9*d*x)*1i)/288 - (sin(11*c + 11*d*x)* 
1i)/2816 + (224^(1/2)*cos(5*c + atanh(57/55)*1i + 5*d*x)*1i)/1280 + (560^( 
1/2)*cos(3*c + atanh(39/31)*1i + 3*d*x)*1i)/384 + (2^(1/2)*cos(c + atanh(3 
)*1i + d*x)*7i)/32)*1i)/(a^3*d)
 

Reduce [F]

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

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

Output:

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