3.168 \(\int \frac {1}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {i}{2 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, 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} \[ \frac {i}{2 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(-2),x]

[Out]

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

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))^2} \, dx &=\frac {i}{2 d (a \cos (c+d x)+i a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 31, normalized size = 1.00 \[ \frac {i}{2 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(-2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 17, normalized size = 0.55 \[ \frac {i \, e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*I*e^(-2*I*d*x - 2*I*c)/(a^2*d)

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giac [A]  time = 0.19, size = 30, normalized size = 0.97 \[ -\frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 23, normalized size = 0.74 \[ \frac {i}{d \,a^{2} \left (i \tan \left (d x +c \right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 22, normalized size = 0.71 \[ \frac {1}{{\left (a^{2} \tan \left (d x + c\right ) - i \, a^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/((a^2*tan(d*x + c) - I*a^2)*d)

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mupad [B]  time = 0.61, size = 31, normalized size = 1.00 \[ -\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2,x)

[Out]

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

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sympy [A]  time = 0.14, size = 46, normalized size = 1.48 \[ \begin {cases} \frac {i e^{- 2 i c} e^{- 2 i d x}}{2 a^{2} d} & \text {for}\: 2 a^{2} d e^{2 i c} \neq 0 \\\frac {x e^{- 2 i c}}{a^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Piecewise((I*exp(-2*I*c)*exp(-2*I*d*x)/(2*a**2*d), Ne(2*a**2*d*exp(2*I*c), 0)), (x*exp(-2*I*c)/a**2, True))

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