\(\int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) [166]

Optimal result

Integrand size = 31, antiderivative size = 89 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {x}{4 a^2}+\frac {i \cos ^2(c+d x)}{4 d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{4 d \left (a^2 \cos (c+d x)+i a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3161, 8} \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {i \cos (c+d x)}{4 d \left (a^2 \cos (c+d x)+i a^2 \sin (c+d x)\right )}+\frac {x}{4 a^2}+\frac {i \cos ^2(c+d x)}{4 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

[In]

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

[Out]

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

Rule 8

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

Rule 3161

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] &&
EqQ[a^2 + b^2, 0] && LtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {4 c+4 d x+4 i \cos (2 (c+d x))+i \cos (4 (c+d x))+4 \sin (2 (c+d x))+\sin (4 (c+d x))}{16 a^2 d} \]

[In]

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

[Out]

(4*c + 4*d*x + (4*I)*Cos[2*(c + d*x)] + I*Cos[4*(c + d*x)] + 4*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])/(16*a^2*d)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {x}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}\)	\(44\)
derivativedivides	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}}{d \,a^{2}}\)	\(62\)
default	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}}{d \,a^{2}}\)	\(62\)

[In]

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

[Out]

1/4*x/a^2+1/4*I/a^2/d*exp(-2*I*(d*x+c))+1/16*I/a^2/d*exp(-4*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.48 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {{\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

[In]

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

[Out]

1/16*(4*d*x*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {1}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a^{2}} \]

[In]

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

[Out]

Piecewise(((16*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((exp(4*I*c) + 2*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) - 1/(4*a**2)), Tr
ue)) + x/(4*a**2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {-\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {2 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {-3 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right ) + 11 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*(-2*I*log(tan(d*x + c) + I)/a^2 + 2*I*log(tan(d*x + c) - I)/a^2 + (-3*I*tan(d*x + c)^2 - 10*tan(d*x + c)
+ 11*I)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 24.89 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {x}{4\,a^2}+\frac {-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{a^2\,d\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^4} \]

[In]

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

[Out]

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