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

Optimal. Leaf size=131 \[ \frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}+\frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

[Out]

1/8*x/a^3+1/6*I*cos(d*x+c)^3/d/(a*cos(d*x+c)+I*a*sin(d*x+c))^3+1/8*I*cos(d*x+c)^2/a/d/(a*cos(d*x+c)+I*a*sin(d*
x+c))^2+1/8*I*cos(d*x+c)/d/(a^3*cos(d*x+c)+I*a^3*sin(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3082, 8} \[ \frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}+\frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(8*a^3) + ((I/6)*Cos[c + d*x]^3)/(d*(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3) + ((I/8)*Cos[c + d*x]^2)/(a*d*(a*
Cos[c + d*x] + I*a*Sin[c + d*x])^2) + ((I/8)*Cos[c + d*x])/(d*(a^3*Cos[c + d*x] + I*a^3*Sin[c + d*x]))

Rule 8

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

Rule 3082

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Simp[(b*(a*Cos[c + d*x] + b*Sin[c + d*x])^n)/(2*a*d*n*Cos[c + d*x]^n), x] + Dist[1/(2*a), Int[(a*Cos[c
+ d*x] + b*Sin[c + d*x])^(n + 1)/Cos[c + d*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && E
qQ[a^2 + b^2, 0] && LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx}{2 a}\\ &=\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {\int \frac {\cos (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx}{4 a^2}\\ &=\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3}\\ &=\frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 84, normalized size = 0.64 \[ \frac {18 \sin (2 (c+d x))+9 \sin (4 (c+d x))+2 \sin (6 (c+d x))+18 i \cos (2 (c+d x))+9 i \cos (4 (c+d x))+2 i \cos (6 (c+d x))+12 c+12 d x}{96 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*c + 12*d*x + (18*I)*Cos[2*(c + d*x)] + (9*I)*Cos[4*(c + d*x)] + (2*I)*Cos[6*(c + d*x)] + 18*Sin[2*(c + d*x
)] + 9*Sin[4*(c + d*x)] + 2*Sin[6*(c + d*x)])/(96*a^3*d)

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fricas [A]  time = 0.62, size = 54, normalized size = 0.41 \[ \frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(12*d*x*e^(6*I*d*x + 6*I*c) + 18*I*e^(4*I*d*x + 4*I*c) + 9*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I
*c)/(a^3*d)

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giac [A]  time = 0.25, size = 80, normalized size = 0.61 \[ -\frac {\frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {-11 i \, \tan \left (d x + c\right )^{3} - 45 \, \tan \left (d x + c\right )^{2} + 69 i \, \tan \left (d x + c\right ) + 51}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*I*log(tan(d*x + c) - I)/a^3 - 6*I*log(I*tan(d*x + c) - 1)/a^3 + (-11*I*tan(d*x + c)^3 - 45*tan(d*x +
c)^2 + 69*I*tan(d*x + c) + 51)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [A]  time = 0.19, size = 98, normalized size = 0.75 \[ \frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16 a^{3} d}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16 a^{3} d}-\frac {i}{8 a^{3} d \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 a^{3} d \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 a^{3} d \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*I/a^3/d*ln(tan(d*x+c)+I)-1/16*I/a^3/d*ln(tan(d*x+c)-I)-1/8*I/a^3/d/(tan(d*x+c)-I)^2-1/6/a^3/d/(tan(d*x+c)
-I)^3+1/8/a^3/d/(tan(d*x+c)-I)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 3.49, size = 96, normalized size = 0.73 \[ \frac {x}{8\,a^3}+\frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,9{}\mathrm {i}}{2}-\frac {41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,9{}\mathrm {i}}{2}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a^3\,d\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(8*a^3) + ((7*tan(c/2 + (d*x)/2))/4 + (tan(c/2 + (d*x)/2)^2*9i)/2 - (41*tan(c/2 + (d*x)/2)^3)/6 - (tan(c/2 +
(d*x)/2)^4*9i)/2 + (7*tan(c/2 + (d*x)/2)^5)/4)/(a^3*d*(tan(c/2 + (d*x)/2)*1i + 1)^6)

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sympy [A]  time = 0.31, size = 160, normalized size = 1.22 \[ \begin {cases} - \frac {\left (- 4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{8 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(-4608*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) - 2304*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 512*I*a
**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*((e
xp(6*I*c) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) - 1/(8*a**3)), True)) + x/(8*a**3)

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