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

Optimal. Leaf size=32 \[ \frac {i \cot ^2(c+d x)}{2 a^3 d (\cot (c+d x)+i)^2} \]

[Out]

1/2*I*cot(d*x+c)^2/a^3/d/(I+cot(d*x+c))^2

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {3088, 37} \[ \frac {i \cot ^2(c+d x)}{2 a^3 d (\cot (c+d x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Cot[c + d*x]^2)/(a^3*d*(I + Cot[c + d*x])^2)

Rule 37

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

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[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])

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x}{(i a+a x)^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {i \cot ^2(c+d x)}{2 a^3 d (i+\cot (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.06, size = 77, normalized size = 2.41 \[ \frac {\sin (2 (c+d x))}{4 a^3 d}+\frac {\sin (4 (c+d x))}{8 a^3 d}+\frac {i \cos (2 (c+d x))}{4 a^3 d}+\frac {i \cos (4 (c+d x))}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*Cos[2*(c + d*x)])/(a^3*d) + ((I/8)*Cos[4*(c + d*x)])/(a^3*d) + Sin[2*(c + d*x)]/(4*a^3*d) + Sin[4*(c +
d*x)]/(8*a^3*d)

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fricas [A]  time = 0.67, size = 30, normalized size = 0.94 \[ \frac {{\left (2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{8 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a^3*d)

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giac [B]  time = 0.97, size = 57, normalized size = 1.78 \[ -\frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 23, normalized size = 0.72 \[ \frac {i}{2 d \,a^{3} \left (i \tan \left (d x +c \right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/d/a^3/(I*tan(d*x+c)+1)^2

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maxima [A]  time = 0.37, size = 51, normalized size = 1.59 \[ \frac {i \, \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + \sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )}{8 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(I*cos(4*d*x + 4*c) + 2*I*cos(2*d*x + 2*c) + sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))/(a^3*d)

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mupad [B]  time = 0.75, size = 100, normalized size = 3.12 \[ -\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}{a^3\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,6{}\mathrm {i}-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*1i - 1i))/(a^3*d*(4*tan(c/2 + (d*x)/2)^3 - t
an(c/2 + (d*x)/2)^2*6i - 4*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^4*1i + 1i))

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sympy [A]  time = 0.22, size = 97, normalized size = 3.03 \[ \begin {cases} \frac {\left (8 i a^{3} d e^{4 i c} e^{- 2 i d x} + 4 i a^{3} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{6} d^{2}} & \text {for}\: 32 a^{6} d^{2} e^{6 i c} \neq 0 \\\frac {x \left (e^{2 i c} + 1\right ) e^{- 4 i c}}{2 a^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((8*I*a**3*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**3*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(32*a**6*d*
*2), Ne(32*a**6*d**2*exp(6*I*c), 0)), (x*(exp(2*I*c) + 1)*exp(-4*I*c)/(2*a**3), True))

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