\(\int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\) [152]

Optimal result

Integrand size = 31, antiderivative size = 75 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {3 x}{8 a}+\frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \]

[Out]

3/8*x/a+1/4*I*cos(d*x+c)^4/a/d+3/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+c)^3*sin(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3171, 3169, 2715, 8, 2645, 30} \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {i \cos ^4(c+d x)}{4 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}+\frac {3 x}{8 a} \]

[In]

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

[Out]

(3*x)/(8*a) + ((I/4)*Cos[c + d*x]^4)/(a*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin[c + d
*x])/(4*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3171

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \int \cos ^3(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2} \\ & = -\frac {i \int \left (i a \cos ^4(c+d x)+a \cos ^3(c+d x) \sin (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {i \int \cos ^3(c+d x) \sin (c+d x) \, dx}{a}+\frac {\int \cos ^4(c+d x) \, dx}{a} \\ & = \frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac {3 \int \cos ^2(c+d x) \, dx}{4 a}+\frac {i \text {Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{a d} \\ & = \frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac {3 \int 1 \, dx}{8 a} \\ & = \frac {3 x}{8 a}+\frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {12 c+12 d x+4 i \cos (2 (c+d x))+i \cos (4 (c+d x))+8 \sin (2 (c+d x))+\sin (4 (c+d x))}{32 a d} \]

[In]

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

[Out]

(12*c + 12*d*x + (4*I)*Cos[2*(c + d*x)] + I*Cos[4*(c + d*x)] + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])/(32*a*d)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(\frac {4 i \cos \left (2 d x +2 c \right )+i \cos \left (4 d x +4 c \right )+12 d x -21 i+8 \sin \left (2 d x +2 c \right )+\sin \left (4 d x +4 c \right )}{32 a d}\)	\(60\)
risch	\(\frac {3 x}{8 a}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a d}+\frac {i \cos \left (2 d x +2 c \right )}{8 a d}+\frac {\sin \left (2 d x +2 c \right )}{4 a d}\)	\(61\)
derivativedivides	\(\frac {-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}+\frac {3 i \ln \left (\tan \left (d x +c \right )+i\right )}{16}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\)	\(75\)
default	\(\frac {-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}+\frac {3 i \ln \left (\tan \left (d x +c \right )+i\right )}{16}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\)	\(75\)

[In]

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

[Out]

1/32*(4*I*cos(2*d*x+2*c)+I*cos(4*d*x+4*c)+12*d*x-21*I+8*sin(2*d*x+2*c)+sin(4*d*x+4*c))/a/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {{\left (12 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a d} \]

[In]

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

[Out]

1/32*(12*d*x*e^(4*I*d*x + 4*I*c) - 2*I*e^(6*I*d*x + 6*I*c) + 6*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)
/(a*d)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\begin {cases} \frac {\left (- 512 i a^{2} d^{2} e^{8 i c} e^{2 i d x} + 1536 i a^{2} d^{2} e^{4 i c} e^{- 2 i d x} + 256 i a^{2} d^{2} e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{8192 a^{3} d^{3}} & \text {for}\: a^{3} d^{3} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 4 i c}}{8 a} - \frac {3}{8 a}\right ) & \text {otherwise} \end {cases} + \frac {3 x}{8 a} \]

[In]

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

[Out]

Piecewise(((-512*I*a**2*d**2*exp(8*I*c)*exp(2*I*d*x) + 1536*I*a**2*d**2*exp(4*I*c)*exp(-2*I*d*x) + 256*I*a**2*
d**2*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(8192*a**3*d**3), Ne(a**3*d**3*exp(6*I*c), 0)), (x*((exp(6*I*c) + 3
*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-4*I*c)/(8*a) - 3/(8*a)), True)) + 3*x/(8*a)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right ) + 5 i\right )}}{a {\left (-i \, \tan \left (d x + c\right ) + 1\right )}} + \frac {-9 i \, \tan \left (d x + c\right )^{2} - 26 \, \tan \left (d x + c\right ) + 21 i}{a {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{32 \, d} \]

[In]

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

[Out]

-1/32*(-6*I*log(tan(d*x + c) + I)/a + 6*I*log(tan(d*x + c) - I)/a + 2*(3*tan(d*x + c) + 5*I)/(a*(-I*tan(d*x +
c) + 1)) + (-9*I*tan(d*x + c)^2 - 26*tan(d*x + c) + 21*I)/(a*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 26.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {3\,x}{8\,a}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,1{}\mathrm {i}}{2}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^4} \]

[In]

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

[Out]

(3*x)/(8*a) - ((5*tan(c/2 + (d*x)/2))/4 + (tan(c/2 + (d*x)/2)^2*1i)/2 - tan(c/2 + (d*x)/2)^3/2 - (tan(c/2 + (d
*x)/2)^4*1i)/2 + (5*tan(c/2 + (d*x)/2)^5)/4)/(a*d*(tan(c/2 + (d*x)/2) + 1i)^2*(tan(c/2 + (d*x)/2)*1i + 1)^4)