\(\int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [175]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (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 = 333 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {33 x}{2048 a^8}+\frac {i a^2}{80 d (a+i a \tan (c+d x))^{10}}+\frac {i a}{48 d (a+i a \tan (c+d x))^9}+\frac {3 i}{128 d (a+i a \tan (c+d x))^8}+\frac {5 i}{224 a d (a+i a \tan (c+d x))^7}+\frac {5 i}{256 a^2 d (a+i a \tan (c+d x))^6}+\frac {21 i}{1280 a^3 d (a+i a \tan (c+d x))^5}+\frac {3 i}{256 a^5 d (a+i a \tan (c+d x))^3}+\frac {7 i}{512 d \left (a^2+i a^2 \tan (c+d x)\right )^4}-\frac {i}{4096 d \left (a^4-i a^4 \tan (c+d x)\right )^2}+\frac {45 i}{4096 d \left (a^4+i a^4 \tan (c+d x)\right )^2}-\frac {11 i}{4096 d \left (a^8-i a^8 \tan (c+d x)\right )}+\frac {55 i}{4096 d \left (a^8+i a^8 \tan (c+d x)\right )} \] Output:

33/2048*x/a^8+1/80*I*a^2/d/(a+I*a*tan(d*x+c))^10+1/48*I*a/d/(a+I*a*tan(d*x 
+c))^9+3/128*I/d/(a+I*a*tan(d*x+c))^8+5/224*I/a/d/(a+I*a*tan(d*x+c))^7+5/2 
56*I/a^2/d/(a+I*a*tan(d*x+c))^6+21/1280*I/a^3/d/(a+I*a*tan(d*x+c))^5+3/256 
*I/a^5/d/(a+I*a*tan(d*x+c))^3+7/512*I/d/(a^2+I*a^2*tan(d*x+c))^4-1/4096*I/ 
d/(a^4-I*a^4*tan(d*x+c))^2+45/4096*I/d/(a^4+I*a^4*tan(d*x+c))^2-11/4096*I/ 
d/(a^8-I*a^8*tan(d*x+c))+55/4096*I/d/(a^8+I*a^8*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\sec ^{12}(c+d x) (48510 i+88704 i \cos (2 (c+d x))+69300 i \cos (4 (c+d x))+52800 i \cos (6 (c+d x))+21538 i \cos (8 (c+d x))-2240 i \cos (10 (c+d x))-84 i \cos (12 (c+d x))-22176 \sin (2 (c+d x))-34650 \sin (4 (c+d x))-39600 \sin (6 (c+d x))+27720 \arctan (\tan (c+d x)) (\cos (8 (c+d x))+i \sin (8 (c+d x)))-18073 \sin (8 (c+d x))+2800 \sin (10 (c+d x))+126 \sin (12 (c+d x)))}{1720320 a^8 d (-i+\tan (c+d x))^{10} (i+\tan (c+d x))^2} \] Input:

Integrate[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

(Sec[c + d*x]^12*(48510*I + (88704*I)*Cos[2*(c + d*x)] + (69300*I)*Cos[4*( 
c + d*x)] + (52800*I)*Cos[6*(c + d*x)] + (21538*I)*Cos[8*(c + d*x)] - (224 
0*I)*Cos[10*(c + d*x)] - (84*I)*Cos[12*(c + d*x)] - 22176*Sin[2*(c + d*x)] 
 - 34650*Sin[4*(c + d*x)] - 39600*Sin[6*(c + d*x)] + 27720*ArcTan[Tan[c + 
d*x]]*(Cos[8*(c + d*x)] + I*Sin[8*(c + d*x)]) - 18073*Sin[8*(c + d*x)] + 2 
800*Sin[10*(c + d*x)] + 126*Sin[12*(c + d*x)]))/(1720320*a^8*d*(-I + Tan[c 
 + d*x])^10*(I + Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 54, 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.

\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^4 (a+i a \tan (c+d x))^8}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i a^5 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{11}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 54

\(\displaystyle -\frac {i a^5 \int \left (\frac {11}{4096 a^{12} (a-i a \tan (c+d x))^2}+\frac {55}{4096 a^{12} (i \tan (c+d x) a+a)^2}+\frac {1}{2048 a^{11} (a-i a \tan (c+d x))^3}+\frac {45}{2048 a^{11} (i \tan (c+d x) a+a)^3}+\frac {9}{256 a^{10} (i \tan (c+d x) a+a)^4}+\frac {7}{128 a^9 (i \tan (c+d x) a+a)^5}+\frac {21}{256 a^8 (i \tan (c+d x) a+a)^6}+\frac {15}{128 a^7 (i \tan (c+d x) a+a)^7}+\frac {5}{32 a^6 (i \tan (c+d x) a+a)^8}+\frac {3}{16 a^5 (i \tan (c+d x) a+a)^9}+\frac {3}{16 a^4 (i \tan (c+d x) a+a)^{10}}+\frac {1}{8 a^3 (i \tan (c+d x) a+a)^{11}}+\frac {33}{2048 a^{12} \left (\tan ^2(c+d x) a^2+a^2\right )}\right )d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i a^5 \left (\frac {33 i \arctan (\tan (c+d x))}{2048 a^{13}}+\frac {11}{4096 a^{12} (a-i a \tan (c+d x))}-\frac {55}{4096 a^{12} (a+i a \tan (c+d x))}+\frac {1}{4096 a^{11} (a-i a \tan (c+d x))^2}-\frac {45}{4096 a^{11} (a+i a \tan (c+d x))^2}-\frac {3}{256 a^{10} (a+i a \tan (c+d x))^3}-\frac {7}{512 a^9 (a+i a \tan (c+d x))^4}-\frac {21}{1280 a^8 (a+i a \tan (c+d x))^5}-\frac {5}{256 a^7 (a+i a \tan (c+d x))^6}-\frac {5}{224 a^6 (a+i a \tan (c+d x))^7}-\frac {3}{128 a^5 (a+i a \tan (c+d x))^8}-\frac {1}{48 a^4 (a+i a \tan (c+d x))^9}-\frac {1}{80 a^3 (a+i a \tan (c+d x))^{10}}\right )}{d}\)

Input:

Int[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

((-I)*a^5*((((33*I)/2048)*ArcTan[Tan[c + d*x]])/a^13 + 1/(4096*a^11*(a - I 
*a*Tan[c + d*x])^2) + 11/(4096*a^12*(a - I*a*Tan[c + d*x])) - 1/(80*a^3*(a 
 + I*a*Tan[c + d*x])^10) - 1/(48*a^4*(a + I*a*Tan[c + d*x])^9) - 3/(128*a^ 
5*(a + I*a*Tan[c + d*x])^8) - 5/(224*a^6*(a + I*a*Tan[c + d*x])^7) - 5/(25 
6*a^7*(a + I*a*Tan[c + d*x])^6) - 21/(1280*a^8*(a + I*a*Tan[c + d*x])^5) - 
 7/(512*a^9*(a + I*a*Tan[c + d*x])^4) - 3/(256*a^10*(a + I*a*Tan[c + d*x]) 
^3) - 45/(4096*a^11*(a + I*a*Tan[c + d*x])^2) - 55/(4096*a^12*(a + I*a*Tan 
[c + d*x]))))/d
 

Defintions of rubi rules used

rule 54
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

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 3968
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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]
 
Maple [A] (verified)

Time = 1.34 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.59

method result size
derivativedivides \(\frac {-\frac {33 i \ln \left (-i+\tan \left (d x +c \right )\right )}{4096}+\frac {7 i}{512 \left (-i+\tan \left (d x +c \right )\right )^{4}}+\frac {3 i}{128 \left (-i+\tan \left (d x +c \right )\right )^{8}}-\frac {i}{80 \left (-i+\tan \left (d x +c \right )\right )^{10}}-\frac {5 i}{256 \left (-i+\tan \left (d x +c \right )\right )^{6}}-\frac {45 i}{4096 \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {1}{48 \left (-i+\tan \left (d x +c \right )\right )^{9}}-\frac {5}{224 \left (-i+\tan \left (d x +c \right )\right )^{7}}+\frac {21}{1280 \left (-i+\tan \left (d x +c \right )\right )^{5}}-\frac {3}{256 \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {55}{4096 \left (-i+\tan \left (d x +c \right )\right )}+\frac {i}{4096 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {33 i \ln \left (\tan \left (d x +c \right )+i\right )}{4096}+\frac {11}{4096 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{8}}\) \(197\)
default \(\frac {-\frac {33 i \ln \left (-i+\tan \left (d x +c \right )\right )}{4096}+\frac {7 i}{512 \left (-i+\tan \left (d x +c \right )\right )^{4}}+\frac {3 i}{128 \left (-i+\tan \left (d x +c \right )\right )^{8}}-\frac {i}{80 \left (-i+\tan \left (d x +c \right )\right )^{10}}-\frac {5 i}{256 \left (-i+\tan \left (d x +c \right )\right )^{6}}-\frac {45 i}{4096 \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {1}{48 \left (-i+\tan \left (d x +c \right )\right )^{9}}-\frac {5}{224 \left (-i+\tan \left (d x +c \right )\right )^{7}}+\frac {21}{1280 \left (-i+\tan \left (d x +c \right )\right )^{5}}-\frac {3}{256 \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {55}{4096 \left (-i+\tan \left (d x +c \right )\right )}+\frac {i}{4096 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {33 i \ln \left (\tan \left (d x +c \right )+i\right )}{4096}+\frac {11}{4096 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{8}}\) \(197\)
risch \(\frac {33 x}{2048 a^{8}}+\frac {33 i {\mathrm e}^{-6 i \left (d x +c \right )}}{1024 a^{8} d}+\frac {231 i {\mathrm e}^{-8 i \left (d x +c \right )}}{8192 a^{8} d}+\frac {99 i {\mathrm e}^{-10 i \left (d x +c \right )}}{5120 a^{8} d}+\frac {165 i {\mathrm e}^{-12 i \left (d x +c \right )}}{16384 a^{8} d}+\frac {55 i {\mathrm e}^{-14 i \left (d x +c \right )}}{14336 a^{8} d}+\frac {33 i {\mathrm e}^{-16 i \left (d x +c \right )}}{32768 a^{8} d}+\frac {i {\mathrm e}^{-18 i \left (d x +c \right )}}{6144 a^{8} d}+\frac {i {\mathrm e}^{-20 i \left (d x +c \right )}}{81920 a^{8} d}+\frac {247 i \cos \left (4 d x +4 c \right )}{8192 a^{8} d}+\frac {31 \sin \left (4 d x +4 c \right )}{1024 a^{8} d}+\frac {13 i \cos \left (2 d x +2 c \right )}{512 a^{8} d}+\frac {29 \sin \left (2 d x +2 c \right )}{1024 a^{8} d}\) \(222\)

Input:

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^8*(-33/4096*I*ln(-I+tan(d*x+c))+7/512*I/(-I+tan(d*x+c))^4+3/128*I/(- 
I+tan(d*x+c))^8-1/80*I/(-I+tan(d*x+c))^10-5/256*I/(-I+tan(d*x+c))^6-45/409 
6*I/(-I+tan(d*x+c))^2+1/48/(-I+tan(d*x+c))^9-5/224/(-I+tan(d*x+c))^7+21/12 
80/(-I+tan(d*x+c))^5-3/256/(-I+tan(d*x+c))^3+55/4096/(-I+tan(d*x+c))+1/409 
6*I/(tan(d*x+c)+I)^2+33/4096*I*ln(tan(d*x+c)+I)+11/4096/(tan(d*x+c)+I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {{\left (55440 \, d x e^{\left (20 i \, d x + 20 i \, c\right )} - 210 i \, e^{\left (24 i \, d x + 24 i \, c\right )} - 5040 i \, e^{\left (22 i \, d x + 22 i \, c\right )} + 92400 i \, e^{\left (18 i \, d x + 18 i \, c\right )} + 103950 i \, e^{\left (16 i \, d x + 16 i \, c\right )} + 110880 i \, e^{\left (14 i \, d x + 14 i \, c\right )} + 97020 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 66528 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 34650 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 13200 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3465 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 560 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 42 i\right )} e^{\left (-20 i \, d x - 20 i \, c\right )}}{3440640 \, a^{8} d} \] Input:

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
 

Output:

1/3440640*(55440*d*x*e^(20*I*d*x + 20*I*c) - 210*I*e^(24*I*d*x + 24*I*c) - 
 5040*I*e^(22*I*d*x + 22*I*c) + 92400*I*e^(18*I*d*x + 18*I*c) + 103950*I*e 
^(16*I*d*x + 16*I*c) + 110880*I*e^(14*I*d*x + 14*I*c) + 97020*I*e^(12*I*d* 
x + 12*I*c) + 66528*I*e^(10*I*d*x + 10*I*c) + 34650*I*e^(8*I*d*x + 8*I*c) 
+ 13200*I*e^(6*I*d*x + 6*I*c) + 3465*I*e^(4*I*d*x + 4*I*c) + 560*I*e^(2*I* 
d*x + 2*I*c) + 42*I)*e^(-20*I*d*x - 20*I*c)/(a^8*d)
 

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\begin {cases} \frac {\left (- 11433487528543532372369386809707411904921600 i a^{88} d^{11} e^{114 i c} e^{4 i d x} - 274403700685044776936865283432977885718118400 i a^{88} d^{11} e^{112 i c} e^{2 i d x} + 5030734512559154243842530196271261238165504000 i a^{88} d^{11} e^{108 i c} e^{- 2 i d x} + 5659576326629048524322846470805168892936192000 i a^{88} d^{11} e^{106 i c} e^{- 4 i d x} + 6036881415070985092611036235525513485798604800 i a^{88} d^{11} e^{104 i c} e^{- 6 i d x} + 5282271238187111956034656706084824300073779200 i a^{88} d^{11} e^{102 i c} e^{- 8 i d x} + 3622128849042591055566621741315308091479162880 i a^{88} d^{11} e^{100 i c} e^{- 10 i d x} + 1886525442209682841440948823601722964312064000 i a^{88} d^{11} e^{98 i c} e^{- 12 i d x} + 718676358937022034834647170895894462595072000 i a^{88} d^{11} e^{96 i c} e^{- 14 i d x} + 188652544220968284144094882360172296431206400 i a^{88} d^{11} e^{94 i c} e^{- 16 i d x} + 30489300076116086326318364825886431746457600 i a^{88} d^{11} e^{92 i c} e^{- 18 i d x} + 2286697505708706474473877361941482380984320 i a^{88} d^{11} e^{90 i c} e^{- 20 i d x}\right ) e^{- 110 i c}}{187326259667657234388900033490246236650235494400 a^{96} d^{12}} & \text {for}\: a^{96} d^{12} e^{110 i c} \neq 0 \\x \left (\frac {\left (e^{24 i c} + 12 e^{22 i c} + 66 e^{20 i c} + 220 e^{18 i c} + 495 e^{16 i c} + 792 e^{14 i c} + 924 e^{12 i c} + 792 e^{10 i c} + 495 e^{8 i c} + 220 e^{6 i c} + 66 e^{4 i c} + 12 e^{2 i c} + 1\right ) e^{- 20 i c}}{4096 a^{8}} - \frac {33}{2048 a^{8}}\right ) & \text {otherwise} \end {cases} + \frac {33 x}{2048 a^{8}} \] Input:

integrate(cos(d*x+c)**4/(a+I*a*tan(d*x+c))**8,x)
 

Output:

Piecewise(((-11433487528543532372369386809707411904921600*I*a**88*d**11*ex 
p(114*I*c)*exp(4*I*d*x) - 274403700685044776936865283432977885718118400*I* 
a**88*d**11*exp(112*I*c)*exp(2*I*d*x) + 5030734512559154243842530196271261 
238165504000*I*a**88*d**11*exp(108*I*c)*exp(-2*I*d*x) + 565957632662904852 
4322846470805168892936192000*I*a**88*d**11*exp(106*I*c)*exp(-4*I*d*x) + 60 
36881415070985092611036235525513485798604800*I*a**88*d**11*exp(104*I*c)*ex 
p(-6*I*d*x) + 5282271238187111956034656706084824300073779200*I*a**88*d**11 
*exp(102*I*c)*exp(-8*I*d*x) + 36221288490425910555666217413153080914791628 
80*I*a**88*d**11*exp(100*I*c)*exp(-10*I*d*x) + 188652544220968284144094882 
3601722964312064000*I*a**88*d**11*exp(98*I*c)*exp(-12*I*d*x) + 71867635893 
7022034834647170895894462595072000*I*a**88*d**11*exp(96*I*c)*exp(-14*I*d*x 
) + 188652544220968284144094882360172296431206400*I*a**88*d**11*exp(94*I*c 
)*exp(-16*I*d*x) + 30489300076116086326318364825886431746457600*I*a**88*d* 
*11*exp(92*I*c)*exp(-18*I*d*x) + 22866975057087064744738773619414823809843 
20*I*a**88*d**11*exp(90*I*c)*exp(-20*I*d*x))*exp(-110*I*c)/(18732625966765 
7234388900033490246236650235494400*a**96*d**12), Ne(a**96*d**12*exp(110*I* 
c), 0)), (x*((exp(24*I*c) + 12*exp(22*I*c) + 66*exp(20*I*c) + 220*exp(18*I 
*c) + 495*exp(16*I*c) + 792*exp(14*I*c) + 924*exp(12*I*c) + 792*exp(10*I*c 
) + 495*exp(8*I*c) + 220*exp(6*I*c) + 66*exp(4*I*c) + 12*exp(2*I*c) + 1)*e 
xp(-20*I*c)/(4096*a**8) - 33/(2048*a**8)), True)) + 33*x/(2048*a**8)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {33 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{4096 \, a^{8} d} - \frac {33 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{4096 \, a^{8} d} + \frac {3465 \, \tan \left (d x + c\right )^{11} - 27720 i \, \tan \left (d x + c\right )^{10} - 91245 \, \tan \left (d x + c\right )^{9} + 147840 i \, \tan \left (d x + c\right )^{8} + 82698 \, \tan \left (d x + c\right )^{7} + 114576 i \, \tan \left (d x + c\right )^{6} + 255222 \, \tan \left (d x + c\right )^{5} - 190080 i \, \tan \left (d x + c\right )^{4} - 21395 \, \tan \left (d x + c\right )^{3} - 72776 i \, \tan \left (d x + c\right )^{2} - 66953 \, \tan \left (d x + c\right ) + 34816 i}{215040 \, a^{8} d {\left (\tan \left (d x + c\right ) + i\right )}^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{10}} \] Input:

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
 

Output:

33/4096*I*log(tan(d*x + c) + I)/(a^8*d) - 33/4096*I*log(tan(d*x + c) - I)/ 
(a^8*d) + 1/215040*(3465*tan(d*x + c)^11 - 27720*I*tan(d*x + c)^10 - 91245 
*tan(d*x + c)^9 + 147840*I*tan(d*x + c)^8 + 82698*tan(d*x + c)^7 + 114576* 
I*tan(d*x + c)^6 + 255222*tan(d*x + c)^5 - 190080*I*tan(d*x + c)^4 - 21395 
*tan(d*x + c)^3 - 72776*I*tan(d*x + c)^2 - 66953*tan(d*x + c) + 34816*I)/( 
a^8*d*(tan(d*x + c) + I)^2*(tan(d*x + c) - I)^10)
 

Mupad [B] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {33\,x}{2048\,a^8}-\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,66953{}\mathrm {i}}{215040\,a^8}+\frac {17}{105\,a^8}-\frac {9097\,{\mathrm {tan}\left (c+d\,x\right )}^2}{26880\,a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,4279{}\mathrm {i}}{43008\,a^8}-\frac {99\,{\mathrm {tan}\left (c+d\,x\right )}^4}{112\,a^8}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,42537{}\mathrm {i}}{35840\,a^8}+\frac {341\,{\mathrm {tan}\left (c+d\,x\right )}^6}{640\,a^8}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^7\,1969{}\mathrm {i}}{5120\,a^8}+\frac {11\,{\mathrm {tan}\left (c+d\,x\right )}^8}{16\,a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^9\,869{}\mathrm {i}}{2048\,a^8}-\frac {33\,{\mathrm {tan}\left (c+d\,x\right )}^{10}}{256\,a^8}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{11}\,33{}\mathrm {i}}{2048\,a^8}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^{12}\,1{}\mathrm {i}+8\,{\mathrm {tan}\left (c+d\,x\right )}^{11}-{\mathrm {tan}\left (c+d\,x\right )}^{10}\,26{}\mathrm {i}-40\,{\mathrm {tan}\left (c+d\,x\right )}^9+{\mathrm {tan}\left (c+d\,x\right )}^8\,15{}\mathrm {i}-48\,{\mathrm {tan}\left (c+d\,x\right )}^7+{\mathrm {tan}\left (c+d\,x\right )}^6\,84{}\mathrm {i}+48\,{\mathrm {tan}\left (c+d\,x\right )}^5+{\mathrm {tan}\left (c+d\,x\right )}^4\,15{}\mathrm {i}+40\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,26{}\mathrm {i}-8\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \] Input:

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^8,x)
 

Output:

(33*x)/(2048*a^8) - ((tan(c + d*x)*66953i)/(215040*a^8) + 17/(105*a^8) - ( 
9097*tan(c + d*x)^2)/(26880*a^8) + (tan(c + d*x)^3*4279i)/(43008*a^8) - (9 
9*tan(c + d*x)^4)/(112*a^8) - (tan(c + d*x)^5*42537i)/(35840*a^8) + (341*t 
an(c + d*x)^6)/(640*a^8) - (tan(c + d*x)^7*1969i)/(5120*a^8) + (11*tan(c + 
 d*x)^8)/(16*a^8) + (tan(c + d*x)^9*869i)/(2048*a^8) - (33*tan(c + d*x)^10 
)/(256*a^8) - (tan(c + d*x)^11*33i)/(2048*a^8))/(d*(40*tan(c + d*x)^3 - ta 
n(c + d*x)^2*26i - 8*tan(c + d*x) + tan(c + d*x)^4*15i + 48*tan(c + d*x)^5 
 + tan(c + d*x)^6*84i - 48*tan(c + d*x)^7 + tan(c + d*x)^8*15i - 40*tan(c 
+ d*x)^9 - tan(c + d*x)^10*26i + 8*tan(c + d*x)^11 + tan(c + d*x)^12*1i + 
1i))
 

Reduce [F]

\[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{8}-8 \tan \left (d x +c \right )^{7} i -28 \tan \left (d x +c \right )^{6}+56 \tan \left (d x +c \right )^{5} i +70 \tan \left (d x +c \right )^{4}-56 \tan \left (d x +c \right )^{3} i -28 \tan \left (d x +c \right )^{2}+8 \tan \left (d x +c \right ) i +1}d x}{a^{8}} \] Input:

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x)
 

Output:

int(cos(c + d*x)**4/(tan(c + d*x)**8 - 8*tan(c + d*x)**7*i - 28*tan(c + d* 
x)**6 + 56*tan(c + d*x)**5*i + 70*tan(c + d*x)**4 - 56*tan(c + d*x)**3*i - 
 28*tan(c + d*x)**2 + 8*tan(c + d*x)*i + 1),x)/a**8