\(\int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) [175]

Optimal result

Integrand size = 31, antiderivative size = 125 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {5 x}{32 a^3}-\frac {1}{32 a^3 d (i-\cot (c+d x))}+\frac {i}{16 a^3 d (i+\cot (c+d x))^4}-\frac {1}{3 a^3 d (i+\cot (c+d x))^3}-\frac {23 i}{32 a^3 d (i+\cot (c+d x))^2}+\frac {13}{16 a^3 d (i+\cot (c+d x))} \] Output:

5/32*x/a^3-1/32/a^3/d/(I-cot(d*x+c))+1/16*I/a^3/d/(I+cot(d*x+c))^4-1/3/a^3 
/d/(I+cot(d*x+c))^3-23/32*I/a^3/d/(I+cot(d*x+c))^2+13/16/a^3/d/(I+cot(d*x+ 
c))
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {120 c+120 d x+108 i \cos (2 (c+d x))+60 i \cos (4 (c+d x))+20 i \cos (6 (c+d x))+3 i \cos (8 (c+d x))+132 \sin (2 (c+d x))+60 \sin (4 (c+d x))+20 \sin (6 (c+d x))+3 \sin (8 (c+d x))}{768 a^3 d} \] Input:

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

Output:

(120*c + 120*d*x + (108*I)*Cos[2*(c + d*x)] + (60*I)*Cos[4*(c + d*x)] + (2 
0*I)*Cos[6*(c + d*x)] + (3*I)*Cos[8*(c + d*x)] + 132*Sin[2*(c + d*x)] + 60 
*Sin[4*(c + d*x)] + 20*Sin[6*(c + d*x)] + 3*Sin[8*(c + d*x)])/(768*a^3*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 3567, 27, 516, 99, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3567

\(\displaystyle -\frac {\int \frac {\cot ^5(c+d x)}{a^3 (\cot (c+d x)+i)^3 \left (\cot ^2(c+d x)+1\right )^2}d\cot (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\cot ^5(c+d x)}{(\cot (c+d x)+i)^3 \left (\cot ^2(c+d x)+1\right )^2}d\cot (c+d x)}{a^3 d}\)

\(\Big \downarrow \) 516

\(\displaystyle -\frac {\int \frac {\cot ^5(c+d x)}{(\cot (c+d x)-i)^2 (\cot (c+d x)+i)^5}d\cot (c+d x)}{a^3 d}\)

\(\Big \downarrow \) 99

\(\displaystyle -\frac {\int \left (\frac {1}{32 (\cot (c+d x)-i)^2}+\frac {13}{16 (\cot (c+d x)+i)^2}-\frac {23 i}{16 (\cot (c+d x)+i)^3}-\frac {1}{(\cot (c+d x)+i)^4}+\frac {i}{4 (\cot (c+d x)+i)^5}+\frac {5}{32 \left (\cot ^2(c+d x)+1\right )}\right )d\cot (c+d x)}{a^3 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {5}{32} \arctan (\cot (c+d x))+\frac {1}{32 (-\cot (c+d x)+i)}-\frac {13}{16 (\cot (c+d x)+i)}+\frac {23 i}{32 (\cot (c+d x)+i)^2}+\frac {1}{3 (\cot (c+d x)+i)^3}-\frac {i}{16 (\cot (c+d x)+i)^4}}{a^3 d}\)

Input:

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

Output:

-(((5*ArcTan[Cot[c + d*x]])/32 + 1/(32*(I - Cot[c + d*x])) - (I/16)/(I + C 
ot[c + d*x])^4 + 1/(3*(I + Cot[c + d*x])^3) + ((23*I)/32)/(I + Cot[c + d*x 
])^2 - 13/(16*(I + Cot[c + d*x])))/(a^3*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 516

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free 
Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || 
(GtQ[a, 0] && GtQ[c, 0] &&  !IntegerQ[n]))
 

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 3567

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si 
n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[x^m*((b 
+ a*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Cot[c + d*x]], x] /; FreeQ[{a, 
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[ 
n, 0] && GtQ[m, 1])
 

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {5 x}{32 a^{3}}+\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 a^{3} d}+\frac {5 i {\mathrm e}^{-6 i \left (d x +c \right )}}{192 a^{3} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{256 a^{3} d}+\frac {9 i \cos \left (2 d x +2 c \right )}{64 a^{3} d}+\frac {11 \sin \left (2 d x +2 c \right )}{64 a^{3} d}\)	\(97\)
derivativedivides	\(\frac {-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{64}+\frac {i}{16 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{64}+\frac {1}{32 \tan \left (d x +c \right )+32 i}}{d \,a^{3}}\)	\(102\)
default	\(\frac {-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{64}+\frac {i}{16 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{64}+\frac {1}{32 \tan \left (d x +c \right )+32 i}}{d \,a^{3}}\)	\(102\)
orering	\(\text {Expression too large to display}\)	\(3858\)

Input:

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

Output:

5/32*x/a^3+5/64*I/a^3/d*exp(-4*I*(d*x+c))+5/192*I/a^3/d*exp(-6*I*(d*x+c))+ 
1/256*I/a^3/d*exp(-8*I*(d*x+c))+9/64*I/a^3/d*cos(2*d*x+2*c)+11/64/a^3/d*si 
n(2*d*x+2*c)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {{\left (120 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 12 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 60 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{768 \, a^{3} d} \] Input:

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

1/768*(120*d*x*e^(8*I*d*x + 8*I*c) - 12*I*e^(10*I*d*x + 10*I*c) + 120*I*e^ 
(6*I*d*x + 6*I*c) + 60*I*e^(4*I*d*x + 4*I*c) + 20*I*e^(2*I*d*x + 2*I*c) + 
3*I)*e^(-8*I*d*x - 8*I*c)/(a^3*d)
 

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 100663296 i a^{12} d^{4} e^{22 i c} e^{2 i d x} + 1006632960 i a^{12} d^{4} e^{18 i c} e^{- 2 i d x} + 503316480 i a^{12} d^{4} e^{16 i c} e^{- 4 i d x} + 167772160 i a^{12} d^{4} e^{14 i c} e^{- 6 i d x} + 25165824 i a^{12} d^{4} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{6442450944 a^{15} d^{5}} & \text {for}\: a^{15} d^{5} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 8 i c}}{32 a^{3}} - \frac {5}{32 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{32 a^{3}} \] Input:

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

Output:

Piecewise(((-100663296*I*a**12*d**4*exp(22*I*c)*exp(2*I*d*x) + 1006632960* 
I*a**12*d**4*exp(18*I*c)*exp(-2*I*d*x) + 503316480*I*a**12*d**4*exp(16*I*c 
)*exp(-4*I*d*x) + 167772160*I*a**12*d**4*exp(14*I*c)*exp(-6*I*d*x) + 25165 
824*I*a**12*d**4*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(6442450944*a**15 
*d**5), Ne(a**15*d**5*exp(20*I*c), 0)), (x*((exp(10*I*c) + 5*exp(8*I*c) + 
10*exp(6*I*c) + 10*exp(4*I*c) + 5*exp(2*I*c) + 1)*exp(-8*I*c)/(32*a**3) - 
5/(32*a**3)), True)) + 5*x/(32*a**3)
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {-\frac {60 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {60 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {12 \, {\left (5 \, \tan \left (d x + c\right ) + 7 i\right )}}{a^{3} {\left (i \, \tan \left (d x + c\right ) - 1\right )}} + \frac {-125 i \, \tan \left (d x + c\right )^{4} - 596 \, \tan \left (d x + c\right )^{3} + 1110 i \, \tan \left (d x + c\right )^{2} + 996 \, \tan \left (d x + c\right ) - 405 i}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{768 \, d} \] Input:

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

Output:

-1/768*(-60*I*log(tan(d*x + c) + I)/a^3 + 60*I*log(tan(d*x + c) - I)/a^3 - 
 12*(5*tan(d*x + c) + 7*I)/(a^3*(I*tan(d*x + c) - 1)) + (-125*I*tan(d*x + 
c)^4 - 596*tan(d*x + c)^3 + 1110*I*tan(d*x + c)^2 + 996*tan(d*x + c) - 405 
*I)/(a^3*(tan(d*x + c) - I)^4))/d
 

Mupad [B] (verification not implemented)

Time = 21.58 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {5\,x}{32\,a^3}+\frac {-\frac {27\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{16}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,33{}\mathrm {i}}{8}+\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,9{}\mathrm {i}}{8}+\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,9{}\mathrm {i}}{8}+\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,33{}\mathrm {i}}{8}-\frac {27\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^8} \] Input:

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

Output:

(5*x)/(32*a^3) + ((31*tan(c/2 + (d*x)/2)^3)/6 - (tan(c/2 + (d*x)/2)^2*33i) 
/8 - (27*tan(c/2 + (d*x)/2))/16 - (tan(c/2 + (d*x)/2)^4*9i)/8 + (89*tan(c/ 
2 + (d*x)/2)^5)/24 + (tan(c/2 + (d*x)/2)^6*9i)/8 + (31*tan(c/2 + (d*x)/2)^ 
7)/6 + (tan(c/2 + (d*x)/2)^8*33i)/8 - (27*tan(c/2 + (d*x)/2)^9)/16)/(a^3*d 
*(tan(c/2 + (d*x)/2) + 1i)^2*(tan(c/2 + (d*x)/2)*1i + 1)^8)
 

Reduce [F]

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

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

Output:

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