\(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [164]

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 = 156 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {10 \sin (c+d x)}{33 a^4 d}-\frac {20 \sin ^3(c+d x)}{99 a^4 d}+\frac {2 \sin ^5(c+d x)}{33 a^4 d}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

10/33*sin(d*x+c)/a^4/d-20/99*sin(d*x+c)^3/a^4/d+2/33*sin(d*x+c)^5/a^4/d+1/ 
11*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^4+7/99*I*cos(d*x+c)^3/a/d/(a+I*a*ta 
n(d*x+c))^3+4/33*I*cos(d*x+c)^5/d/(a^4+I*a^4*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i \sec ^4(c+d x) (-924 \cos (c+d x)-1188 \cos (3 (c+d x))+308 \cos (5 (c+d x))+12 \cos (7 (c+d x))-231 i \sin (c+d x)-891 i \sin (3 (c+d x))+385 i \sin (5 (c+d x))+21 i \sin (7 (c+d x)))}{6336 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

((-1/6336*I)*Sec[c + d*x]^4*(-924*Cos[c + d*x] - 1188*Cos[3*(c + d*x)] + 3 
08*Cos[5*(c + d*x)] + 12*Cos[7*(c + d*x)] - (231*I)*Sin[c + d*x] - (891*I) 
*Sin[3*(c + d*x)] + (385*I)*Sin[5*(c + d*x)] + (21*I)*Sin[7*(c + d*x)]))/( 
a^4*d*(-I + Tan[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3983, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^3 (a+i a \tan (c+d x))^4}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {7 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^3}dx}{11 a}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^3}dx}{11 a}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {7 \left (\frac {2 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^2}dx}{3 a}+\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}\right )}{11 a}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \left (\frac {2 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^2}dx}{3 a}+\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}\right )}{11 a}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3113

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

((I/11)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + (7*(((I/9)*Cos[c + 
d*x]^3)/(d*(a + I*a*Tan[c + d*x])^3) + (2*((-5*(-Sin[c + d*x] + (2*Sin[c + 
 d*x]^3)/3 - Sin[c + d*x]^5/5))/(7*a^2*d) + (((2*I)/7)*Cos[c + d*x]^5)/(d* 
(a^2 + I*a^2*Tan[c + d*x]))))/(3*a)))/(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 = 1.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.88

method result size
risch \(\frac {7 i {\mathrm e}^{-5 i \left (d x +c \right )}}{128 a^{4} d}+\frac {3 i {\mathrm e}^{-7 i \left (d x +c \right )}}{128 a^{4} d}+\frac {7 i {\mathrm e}^{-9 i \left (d x +c \right )}}{1152 a^{4} d}+\frac {i {\mathrm e}^{-11 i \left (d x +c \right )}}{1408 a^{4} d}+\frac {7 i \cos \left (d x +c \right )}{64 a^{4} d}+\frac {7 \sin \left (d x +c \right )}{32 a^{4} d}+\frac {17 i \cos \left (3 d x +3 c \right )}{192 a^{4} d}+\frac {3 \sin \left (3 d x +3 c \right )}{32 a^{4} d}\) \(138\)
derivativedivides \(\frac {-\frac {i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}-\frac {67 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {44 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {385 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {201 i}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {208}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {61}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {105}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {267}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{4} d}\) \(240\)
default \(\frac {-\frac {i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}-\frac {67 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {44 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {385 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {201 i}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {208}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {61}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {105}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {267}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{4} d}\) \(240\)

Input:

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

7/128*I/a^4/d*exp(-5*I*(d*x+c))+3/128*I/a^4/d*exp(-7*I*(d*x+c))+7/1152*I/a 
^4/d*exp(-9*I*(d*x+c))+1/1408*I/a^4/d*exp(-11*I*(d*x+c))+7/64*I/a^4/d*cos( 
d*x+c)+7/32*sin(d*x+c)/a^4/d+17/192*I/a^4/d*cos(3*d*x+3*c)+3/32/a^4/d*sin( 
3*d*x+3*c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (-33 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 693 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 2079 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 1155 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 693 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 297 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 77 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{12672 \, a^{4} d} \] Input:

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/12672*(-33*I*e^(14*I*d*x + 14*I*c) - 693*I*e^(12*I*d*x + 12*I*c) + 2079* 
I*e^(10*I*d*x + 10*I*c) + 1155*I*e^(8*I*d*x + 8*I*c) + 693*I*e^(6*I*d*x + 
6*I*c) + 297*I*e^(4*I*d*x + 4*I*c) + 77*I*e^(2*I*d*x + 2*I*c) + 9*I)*e^(-1 
1*I*d*x - 11*I*c)/(a^4*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. 299 vs. \(2 (136) = 272\).

Time = 0.48 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (- 167196136166129664 i a^{28} d^{7} e^{39 i c} e^{3 i d x} - 3511118859488722944 i a^{28} d^{7} e^{37 i c} e^{i d x} + 10533356578466168832 i a^{28} d^{7} e^{35 i c} e^{- i d x} + 5851864765814538240 i a^{28} d^{7} e^{33 i c} e^{- 3 i d x} + 3511118859488722944 i a^{28} d^{7} e^{31 i c} e^{- 5 i d x} + 1504765225495166976 i a^{28} d^{7} e^{29 i c} e^{- 7 i d x} + 390124317720969216 i a^{28} d^{7} e^{27 i c} e^{- 9 i d x} + 45598946227126272 i a^{28} d^{7} e^{25 i c} e^{- 11 i d x}\right ) e^{- 36 i c}}{64203316287793790976 a^{32} d^{8}} & \text {for}\: a^{32} d^{8} e^{36 i c} \neq 0 \\\frac {x \left (e^{14 i c} + 7 e^{12 i c} + 21 e^{10 i c} + 35 e^{8 i c} + 35 e^{6 i c} + 21 e^{4 i c} + 7 e^{2 i c} + 1\right ) e^{- 11 i c}}{128 a^{4}} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**3/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Piecewise(((-167196136166129664*I*a**28*d**7*exp(39*I*c)*exp(3*I*d*x) - 35 
11118859488722944*I*a**28*d**7*exp(37*I*c)*exp(I*d*x) + 105333565784661688 
32*I*a**28*d**7*exp(35*I*c)*exp(-I*d*x) + 5851864765814538240*I*a**28*d**7 
*exp(33*I*c)*exp(-3*I*d*x) + 3511118859488722944*I*a**28*d**7*exp(31*I*c)* 
exp(-5*I*d*x) + 1504765225495166976*I*a**28*d**7*exp(29*I*c)*exp(-7*I*d*x) 
 + 390124317720969216*I*a**28*d**7*exp(27*I*c)*exp(-9*I*d*x) + 45598946227 
126272*I*a**28*d**7*exp(25*I*c)*exp(-11*I*d*x))*exp(-36*I*c)/(642033162877 
93790976*a**32*d**8), Ne(a**32*d**8*exp(36*I*c), 0)), (x*(exp(14*I*c) + 7* 
exp(12*I*c) + 21*exp(10*I*c) + 35*exp(8*I*c) + 35*exp(6*I*c) + 21*exp(4*I* 
c) + 7*exp(2*I*c) + 1)*exp(-11*I*c)/(128*a**4), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {33 \, {\left (12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{3}} + \frac {5940 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 39501 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 141075 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 313236 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 479556 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 516054 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 397914 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 214500 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 79024 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 17765 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2155}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{11}}}{3168 \, d} \] Input:

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

1/3168*(33*(12*tan(1/2*d*x + 1/2*c)^2 + 21*I*tan(1/2*d*x + 1/2*c) - 11)/(a 
^4*(tan(1/2*d*x + 1/2*c) + I)^3) + (5940*tan(1/2*d*x + 1/2*c)^10 - 39501*I 
*tan(1/2*d*x + 1/2*c)^9 - 141075*tan(1/2*d*x + 1/2*c)^8 + 313236*I*tan(1/2 
*d*x + 1/2*c)^7 + 479556*tan(1/2*d*x + 1/2*c)^6 - 516054*I*tan(1/2*d*x + 1 
/2*c)^5 - 397914*tan(1/2*d*x + 1/2*c)^4 + 214500*I*tan(1/2*d*x + 1/2*c)^3 
+ 79024*tan(1/2*d*x + 1/2*c)^2 - 17765*I*tan(1/2*d*x + 1/2*c) - 2155)/(a^4 
*(tan(1/2*d*x + 1/2*c) - I)^11))/d
 

Mupad [B] (verification not implemented)

Time = 2.83 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {269\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-\frac {1307\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{64}+\frac {1307\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{64}-\frac {1099\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{32}+\frac {203\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{32}-\frac {21\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{64}+\frac {21\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{64}+\frac {\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,231{}\mathrm {i}}{16}-\frac {\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,231{}\mathrm {i}}{16}+\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,33{}\mathrm {i}-\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,5{}\mathrm {i}+\frac {\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,3{}\mathrm {i}}{16}-\frac {\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,3{}\mathrm {i}}{16}\right )\,2{}\mathrm {i}}{99\,a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^{11}\,{\left (\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^3} \] Input:

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

-(cos(c/2 + (d*x)/2)*((cos((3*c)/2 + (3*d*x)/2)*231i)/16 - (cos((5*c)/2 + 
(5*d*x)/2)*231i)/16 + cos((7*c)/2 + (7*d*x)/2)*33i - cos((9*c)/2 + (9*d*x) 
/2)*5i + (cos((11*c)/2 + (11*d*x)/2)*3i)/16 - (cos((13*c)/2 + (13*d*x)/2)* 
3i)/16 + (269*sin(c/2 + (d*x)/2))/16 - (1307*sin((3*c)/2 + (3*d*x)/2))/64 
+ (1307*sin((5*c)/2 + (5*d*x)/2))/64 - (1099*sin((7*c)/2 + (7*d*x)/2))/32 
+ (203*sin((9*c)/2 + (9*d*x)/2))/32 - (21*sin((11*c)/2 + (11*d*x)/2))/64 + 
 (21*sin((13*c)/2 + (13*d*x)/2))/64)*2i)/(99*a^4*d*(cos(c/2 + (d*x)/2) + s 
in(c/2 + (d*x)/2)*1i)^11*(cos(c/2 + (d*x)/2)*1i + sin(c/2 + (d*x)/2))^3)
 

Reduce [F]

\[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x}{a^{4}} \] Input:

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x)
 

Output:

int(cos(c + d*x)**3/(tan(c + d*x)**4 - 4*tan(c + d*x)**3*i - 6*tan(c + d*x 
)**2 + 4*tan(c + d*x)*i + 1),x)/a**4