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

Optimal. Leaf size=89 \[ \frac{i \cos (c+d x)}{4 d \left (a^2 \cos (c+d x)+i a^2 \sin (c+d x)\right )}+\frac{x}{4 a^2}+\frac{i \cos ^2(c+d x)}{4 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

[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]))

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Rubi [A]  time = 0.0820806, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {3082, 8} \[ \frac{i \cos (c+d x)}{4 d \left (a^2 \cos (c+d x)+i a^2 \sin (c+d x)\right )}+\frac{x}{4 a^2}+\frac{i \cos ^2(c+d x)}{4 d (a \cos (c+d x)+i a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[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 3082

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] && E
qQ[a^2 + b^2, 0] && LtQ[n, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx &=\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]  time = 0.105191, size = 60, normalized size = 0.67 \[ \frac{4 \sin (2 (c+d x))+\sin (4 (c+d x))+4 i \cos (2 (c+d x))+i \cos (4 (c+d x))+4 c+4 d x}{16 a^2 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.127, size = 79, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{2}}}-{\frac{{\frac{i}{4}}}{d{a}^{2} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{1}{4\,d{a}^{2} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.466714, size = 126, normalized size = 1.42 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.407223, size = 119, normalized size = 1.34 \begin{align*} \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}\: 64 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(64*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)),
True)) + x/(4*a**2)

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Giac [A]  time = 1.15136, size = 97, normalized size = 1.09 \begin{align*} -\frac{\frac{2 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{2 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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