3.128 \(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=89 \[ \frac {\sin ^5(c+d x)}{7 a^2 d}-\frac {10 \sin ^3(c+d x)}{21 a^2 d}+\frac {5 \sin (c+d x)}{7 a^2 d}+\frac {2 i \cos ^5(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

5/7*sin(d*x+c)/a^2/d-10/21*sin(d*x+c)^3/a^2/d+1/7*sin(d*x+c)^5/a^2/d+2/7*I*cos(d*x+c)^5/d/(a^2+I*a^2*tan(d*x+c
))

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Rubi [A]  time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3500, 2633} \[ \frac {\sin ^5(c+d x)}{7 a^2 d}-\frac {10 \sin ^3(c+d x)}{21 a^2 d}+\frac {5 \sin (c+d x)}{7 a^2 d}+\frac {2 i \cos ^5(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(5*Sin[c + d*x])/(7*a^2*d) - (10*Sin[c + d*x]^3)/(21*a^2*d) + Sin[c + d*x]^5/(7*a^2*d) + (((2*I)/7)*Cos[c + d*
x]^5)/(d*(a^2 + I*a^2*Tan[c + d*x]))

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]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac {2 i \cos ^5(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {5 \int \cos ^5(c+d x) \, dx}{7 a^2}\\ &=\frac {2 i \cos ^5(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {5 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{7 a^2 d}\\ &=\frac {5 \sin (c+d x)}{7 a^2 d}-\frac {10 \sin ^3(c+d x)}{21 a^2 d}+\frac {\sin ^5(c+d x)}{7 a^2 d}+\frac {2 i \cos ^5(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 95, normalized size = 1.07 \[ \frac {i \sec ^2(c+d x) (-70 i \sin (c+d x)+63 i \sin (3 (c+d x))+5 i \sin (5 (c+d x))-140 \cos (c+d x)+42 \cos (3 (c+d x))+2 \cos (5 (c+d x)))}{336 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((I/336)*Sec[c + d*x]^2*(-140*Cos[c + d*x] + 42*Cos[3*(c + d*x)] + 2*Cos[5*(c + d*x)] - (70*I)*Sin[c + d*x] +
(63*I)*Sin[3*(c + d*x)] + (5*I)*Sin[5*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2)

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fricas [A]  time = 0.60, size = 74, normalized size = 0.83 \[ \frac {{\left (-7 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 105 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 210 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{672 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/672*(-7*I*e^(10*I*d*x + 10*I*c) - 105*I*e^(8*I*d*x + 8*I*c) + 210*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x +
4*I*c) + 21*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-7*I*d*x - 7*I*c)/(a^2*d)

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giac [A]  time = 0.93, size = 145, normalized size = 1.63 \[ \frac {\frac {7 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{3}} + \frac {273 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1155 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2870 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2037 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 791 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 152}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/168*(7*(9*tan(1/2*d*x + 1/2*c)^2 + 15*I*tan(1/2*d*x + 1/2*c) - 8)/(a^2*(tan(1/2*d*x + 1/2*c) + I)^3) + (273*
tan(1/2*d*x + 1/2*c)^6 - 1155*I*tan(1/2*d*x + 1/2*c)^5 - 2450*tan(1/2*d*x + 1/2*c)^4 + 2870*I*tan(1/2*d*x + 1/
2*c)^3 + 2037*tan(1/2*d*x + 1/2*c)^2 - 791*I*tan(1/2*d*x + 1/2*c) - 152)/(a^2*(tan(1/2*d*x + 1/2*c) - I)^7))/d

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maple [B]  time = 0.43, size = 174, normalized size = 1.96 \[ \frac {-\frac {i}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}+\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}-\frac {5 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {23 i}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {55}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/d/a^2*(-1/16*I/(tan(1/2*d*x+1/2*c)+I)^2-1/24/(tan(1/2*d*x+1/2*c)+I)^3+3/16/(tan(1/2*d*x+1/2*c)+I)+I/(tan(1/2
*d*x+1/2*c)-I)^6-5/2*I/(tan(1/2*d*x+1/2*c)-I)^4+23/16*I/(tan(1/2*d*x+1/2*c)-I)^2-2/7/(tan(1/2*d*x+1/2*c)-I)^7+
2/(tan(1/2*d*x+1/2*c)-I)^5-55/24/(tan(1/2*d*x+1/2*c)-I)^3+13/16/(tan(1/2*d*x+1/2*c)-I))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.87, size = 161, normalized size = 1.81 \[ \frac {\left (-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,42{}\mathrm {i}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,56{}\mathrm {i}+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,28{}\mathrm {i}+76\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,24{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-6{}\mathrm {i}\right )\,2{}\mathrm {i}}{21\,a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^3\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

((3*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2*24i + 76*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*28i + 42*ta
n(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x)/2)^6*56i + 28*tan(c/2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8*42i - 21*tan(c/
2 + (d*x)/2)^9 - 6i)*2i)/(21*a^2*d*(tan(c/2 + (d*x)/2) + 1i)^3*(tan(c/2 + (d*x)/2)*1i + 1)^7)

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sympy [A]  time = 0.59, size = 233, normalized size = 2.62 \[ \begin {cases} \frac {\left (- 176160768 i a^{10} d^{5} e^{19 i c} e^{3 i d x} - 2642411520 i a^{10} d^{5} e^{17 i c} e^{i d x} + 5284823040 i a^{10} d^{5} e^{15 i c} e^{- i d x} + 1761607680 i a^{10} d^{5} e^{13 i c} e^{- 3 i d x} + 528482304 i a^{10} d^{5} e^{11 i c} e^{- 5 i d x} + 75497472 i a^{10} d^{5} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{16911433728 a^{12} d^{6}} & \text {for}\: 16911433728 a^{12} d^{6} e^{16 i c} \neq 0 \\\frac {x \left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 7 i c}}{32 a^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-176160768*I*a**10*d**5*exp(19*I*c)*exp(3*I*d*x) - 2642411520*I*a**10*d**5*exp(17*I*c)*exp(I*d*x)
+ 5284823040*I*a**10*d**5*exp(15*I*c)*exp(-I*d*x) + 1761607680*I*a**10*d**5*exp(13*I*c)*exp(-3*I*d*x) + 528482
304*I*a**10*d**5*exp(11*I*c)*exp(-5*I*d*x) + 75497472*I*a**10*d**5*exp(9*I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(169
11433728*a**12*d**6), Ne(16911433728*a**12*d**6*exp(16*I*c), 0)), (x*(exp(10*I*c) + 5*exp(8*I*c) + 10*exp(6*I*
c) + 10*exp(4*I*c) + 5*exp(2*I*c) + 1)*exp(-7*I*c)/(32*a**2), True))

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