3.255 \(\int \frac{1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=31 \[ \frac{i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.0153039, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {3071} \[ \frac{i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3071

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

Rubi steps

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

Mathematica [A]  time = 0.0468053, size = 31, normalized size = 1. \[ \frac{i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.108, size = 57, normalized size = 1.8 \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ({\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}-4/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03654, size = 39, normalized size = 1.26 \begin{align*} \frac{i \, \cos \left (3 \, d x + 3 \, c\right ) + \sin \left (3 \, d x + 3 \, c\right )}{3 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.99961, size = 49, normalized size = 1.58 \begin{align*} \frac{i \, e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/3*I*e^(-3*I*d*x - 3*I*c)/(a^3*d)

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Sympy [A]  time = 0.181533, size = 46, normalized size = 1.48 \begin{align*} \begin{cases} \frac{i e^{- 3 i c} e^{- 3 i d x}}{3 a^{3} d} & \text{for}\: 3 a^{3} d e^{3 i c} \neq 0 \\\frac{x e^{- 3 i c}}{a^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Piecewise((I*exp(-3*I*c)*exp(-3*I*d*x)/(3*a**3*d), Ne(3*a**3*d*exp(3*I*c), 0)), (x*exp(-3*I*c)/a**3, True))

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Giac [A]  time = 1.10743, size = 49, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{3 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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