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

Optimal. Leaf size=71 \[ -\frac{\sin ^3(c+d x)}{5 a^2 d}+\frac{3 \sin (c+d x)}{5 a^2 d}+\frac{2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

(3*Sin[c + d*x])/(5*a^2*d) - Sin[c + d*x]^3/(5*a^2*d) + (((2*I)/5)*Cos[c + d*x]^3)/(d*(a^2 + I*a^2*Tan[c + d*x
]))

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Rubi [A]  time = 0.048421, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3500, 2633} \[ -\frac{\sin ^3(c+d x)}{5 a^2 d}+\frac{3 \sin (c+d x)}{5 a^2 d}+\frac{2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sin[c + d*x])/(5*a^2*d) - Sin[c + d*x]^3/(5*a^2*d) + (((2*I)/5)*Cos[c + d*x]^3)/(d*(a^2 + I*a^2*Tan[c + d*x
]))

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{3 \int \cos ^3(c+d x) \, dx}{5 a^2}\\ &=\frac{2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 a^2 d}\\ &=\frac{3 \sin (c+d x)}{5 a^2 d}-\frac{\sin ^3(c+d x)}{5 a^2 d}+\frac{2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.2722, size = 68, normalized size = 0.96 \[ \frac{\sec (c+d x) (4 i \cos (2 (c+d x))+5 \tan (c+d x)-3 \sin (3 (c+d x)) \sec (c+d x)-12 i)}{20 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(-12*I + (4*I)*Cos[2*(c + d*x)] - 3*Sec[c + d*x]*Sin[3*(c + d*x)] + 5*Tan[c + d*x]))/(20*a^2*d*(
-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.087, size = 108, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{{a}^{2}d} \left ({\frac{-i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+{\frac{5/4\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}+2/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}-3/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3}+{\frac{7}{8\,\tan \left ( 1/2\,dx+c/2 \right ) -8\,i}}+1/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +i \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2*(-I/(tan(1/2*d*x+1/2*c)-I)^4+5/4*I/(tan(1/2*d*x+1/2*c)-I)^2+2/5/(tan(1/2*d*x+1/2*c)-I)^5-3/2/(tan(1/2*
d*x+1/2*c)-I)^3+7/8/(tan(1/2*d*x+1/2*c)-I)+1/8/(tan(1/2*d*x+1/2*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)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.24008, size = 161, normalized size = 2.27 \begin{align*} \frac{{\left (-5 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(-5*I*e^(6*I*d*x + 6*I*c) + 15*I*e^(4*I*d*x + 4*I*c) + 5*I*e^(2*I*d*x + 2*I*c) + I)*e^(-5*I*d*x - 5*I*c)/
(a^2*d)

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Sympy [A]  time = 0.722779, size = 165, normalized size = 2.32 \begin{align*} \begin{cases} \frac{\left (- 2560 i a^{6} d^{3} e^{10 i c} e^{i d x} + 7680 i a^{6} d^{3} e^{8 i c} e^{- i d x} + 2560 i a^{6} d^{3} e^{6 i c} e^{- 3 i d x} + 512 i a^{6} d^{3} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{20480 a^{8} d^{4}} & \text{for}\: 20480 a^{8} d^{4} e^{9 i c} \neq 0 \\\frac{x \left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 5 i c}}{8 a^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2560*I*a**6*d**3*exp(10*I*c)*exp(I*d*x) + 7680*I*a**6*d**3*exp(8*I*c)*exp(-I*d*x) + 2560*I*a**6*d
**3*exp(6*I*c)*exp(-3*I*d*x) + 512*I*a**6*d**3*exp(4*I*c)*exp(-5*I*d*x))*exp(-9*I*c)/(20480*a**8*d**4), Ne(204
80*a**8*d**4*exp(9*I*c), 0)), (x*(exp(6*I*c) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-5*I*c)/(8*a**2), True))

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Giac [A]  time = 1.15062, size = 126, normalized size = 1.77 \begin{align*} \frac{\frac{5}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right )}} + \frac{35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 90 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 70 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{5}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5/(a^2*(tan(1/2*d*x + 1/2*c) + I)) + (35*tan(1/2*d*x + 1/2*c)^4 - 90*I*tan(1/2*d*x + 1/2*c)^3 - 120*tan(
1/2*d*x + 1/2*c)^2 + 70*I*tan(1/2*d*x + 1/2*c) + 21)/(a^2*(tan(1/2*d*x + 1/2*c) - I)^5))/d