\(\int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) [255]

Optimal result

Integrand size = 22, antiderivative size = 31 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3150} \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

[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 3150

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*} \text {integral}& = \frac {i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i}{3 d (a \cos (c+d x)+i a \sin (c+d x))^3} \]

[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)

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{3 d \,a^{3}}\)	\(19\)
derivativedivides	\(\frac {\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{d \,a^{3}}\)	\(57\)
default	\(\frac {\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{d \,a^{3}}\)	\(57\)
norman	\(\frac {-\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {2 i}{3 a d}+\frac {6 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \, e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{3} d} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {i e^{- 3 i c} e^{- 3 i d x}}{3 a^{3} d} & \text {for}\: a^{3} d e^{3 i c} \neq 0 \\\frac {x e^{- 3 i c}}{a^{3}} & \text {otherwise} \end {cases} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \, \cos \left (3 \, d x + 3 \, c\right ) + \sin \left (3 \, d x + 3 \, c\right )}{3 \, a^{3} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{3 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 28.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.19 \[ \int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}-\mathrm {i}\right )}{3\,a^3\,d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )} \]

[In]

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

[Out]

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