3.145 \(\int \frac{\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=101 \[ -\frac{4 \sin ^3(c+d x)}{35 a^3 d}+\frac{12 \sin (c+d x)}{35 a^3 d}+\frac{8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3} \]

[Out]

(12*Sin[c + d*x])/(35*a^3*d) - (4*Sin[c + d*x]^3)/(35*a^3*d) + ((I/7)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^
3) + (((8*I)/35)*Cos[c + d*x]^3)/(d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.0810969, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3502, 3500, 2633} \[ -\frac{4 \sin ^3(c+d x)}{35 a^3 d}+\frac{12 \sin (c+d x)}{35 a^3 d}+\frac{8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(12*Sin[c + d*x])/(35*a^3*d) - (4*Sin[c + d*x]^3)/(35*a^3*d) + ((I/7)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^
3) + (((8*I)/35)*Cos[c + d*x]^3)/(d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 3502

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

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))^3} \, dx &=\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}+\frac{4 \int \frac{\cos (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{7 a}\\ &=\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}+\frac{8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{12 \int \cos ^3(c+d x) \, dx}{35 a^3}\\ &=\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}+\frac{8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{12 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{35 a^3 d}\\ &=\frac{12 \sin (c+d x)}{35 a^3 d}-\frac{4 \sin ^3(c+d x)}{35 a^3 d}+\frac{i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}+\frac{8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.177119, size = 76, normalized size = 0.75 \[ -\frac{\sec ^3(c+d x) (56 i \sin (2 (c+d x))-20 i \sin (4 (c+d x))+84 \cos (2 (c+d x))-15 \cos (4 (c+d x))+35)}{280 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-(Sec[c + d*x]^3*(35 + 84*Cos[2*(c + d*x)] - 15*Cos[4*(c + d*x)] + (56*I)*Sin[2*(c + d*x)] - (20*I)*Sin[4*(c +
d*x)]))/(280*a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.094, size = 141, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ({\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{6}}}-{\frac{9/2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+{\frac{{\frac{17\,i}{8}}}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}-4/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-7}+{\frac{19}{5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{5}}}-{\frac{15}{4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{3}}}+{\frac{15}{16\,\tan \left ( 1/2\,dx+c/2 \right ) -16\,i}}+1/16\, \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))^3,x)

[Out]

2/d/a^3*(2*I/(tan(1/2*d*x+1/2*c)-I)^6-9/2*I/(tan(1/2*d*x+1/2*c)-I)^4+17/8*I/(tan(1/2*d*x+1/2*c)-I)^2-4/7/(tan(
1/2*d*x+1/2*c)-I)^7+19/5/(tan(1/2*d*x+1/2*c)-I)^5-15/4/(tan(1/2*d*x+1/2*c)-I)^3+15/16/(tan(1/2*d*x+1/2*c)-I)+1
/16/(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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.23384, size = 205, normalized size = 2.03 \begin{align*} \frac{{\left (-35 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 140 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 28 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{560 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/560*(-35*I*e^(8*I*d*x + 8*I*c) + 140*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) + 28*I*e^(2*I*d*x + 2*
I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a^3*d)

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Sympy [A]  time = 1.05874, size = 199, normalized size = 1.97 \begin{align*} \begin{cases} \frac{\left (- 71680 i a^{12} d^{4} e^{17 i c} e^{i d x} + 286720 i a^{12} d^{4} e^{15 i c} e^{- i d x} + 143360 i a^{12} d^{4} e^{13 i c} e^{- 3 i d x} + 57344 i a^{12} d^{4} e^{11 i c} e^{- 5 i d x} + 10240 i a^{12} d^{4} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{1146880 a^{15} d^{5}} & \text{for}\: 1146880 a^{15} d^{5} e^{16 i c} \neq 0 \\\frac{x \left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 7 i c}}{16 a^{3}} & \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))**3,x)

[Out]

Piecewise(((-71680*I*a**12*d**4*exp(17*I*c)*exp(I*d*x) + 286720*I*a**12*d**4*exp(15*I*c)*exp(-I*d*x) + 143360*
I*a**12*d**4*exp(13*I*c)*exp(-3*I*d*x) + 57344*I*a**12*d**4*exp(11*I*c)*exp(-5*I*d*x) + 10240*I*a**12*d**4*exp
(9*I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(1146880*a**15*d**5), Ne(1146880*a**15*d**5*exp(16*I*c), 0)), (x*(exp(8*I*
c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*exp(-7*I*c)/(16*a**3), True))

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Giac [A]  time = 1.1653, size = 161, normalized size = 1.59 \begin{align*} \frac{\frac{35}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right )}} + \frac{525 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1960 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4025 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 4480 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3143 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1176 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 243}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{7}}}{280 \, 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))^3,x, algorithm="giac")

[Out]

1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) + I)) + (525*tan(1/2*d*x + 1/2*c)^6 - 1960*I*tan(1/2*d*x + 1/2*c)^5 - 402
5*tan(1/2*d*x + 1/2*c)^4 + 4480*I*tan(1/2*d*x + 1/2*c)^3 + 3143*tan(1/2*d*x + 1/2*c)^2 - 1176*I*tan(1/2*d*x +
1/2*c) - 243)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^7))/d