3.33 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=69 \[ -\frac{a^2 \sin ^3(c+d x)}{5 d}+\frac{3 a^2 \sin (c+d x)}{5 d}-\frac{2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0491498, antiderivative size = 69, normalized size of antiderivative = 1., 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 = {3496, 2633} \[ -\frac{a^2 \sin ^3(c+d x)}{5 d}+\frac{3 a^2 \sin (c+d x)}{5 d}-\frac{2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && 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 \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac{1}{5} \left (3 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=-\frac{2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{3 a^2 \sin (c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x)}{5 d}-\frac{2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.401438, size = 72, normalized size = 1.04 \[ \frac{a^2 (\sin (2 (c+d x))-i \cos (2 (c+d x))) (-5 i \sin (c+d x)+3 i \sin (3 (c+d x))+10 \cos (c+d x)-2 \cos (3 (c+d x)))}{20 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(10*Cos[c + d*x] - 2*Cos[3*(c + d*x)] - (5*I)*Sin[c + d*x] + (
3*I)*Sin[3*(c + d*x)]))/(20*d)

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Maple [A]  time = 0.057, size = 91, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{2\,i}{5}}{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-2/5*I*a^2*cos(d*x+c)^5+1/5*a^2*(8/3+
cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.07611, size = 107, normalized size = 1.55 \begin{align*} -\frac{6 i \, a^{2} \cos \left (d x + c\right )^{5} -{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} -{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/15*(6*I*a^2*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^2 - (3*sin(d*x + c)^5 - 10*sin(d*x + c
)^3 + 15*sin(d*x + c))*a^2)/d

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Fricas [A]  time = 1.10349, size = 169, normalized size = 2.45 \begin{align*} \frac{{\left (-i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 15 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.981247, size = 155, normalized size = 2.25 \begin{align*} \begin{cases} \frac{\left (- 512 i a^{2} d^{3} e^{6 i c} e^{5 i d x} - 2560 i a^{2} d^{3} e^{4 i c} e^{3 i d x} - 7680 i a^{2} d^{3} e^{2 i c} e^{i d x} + 2560 i a^{2} d^{3} e^{- i d x}\right ) e^{- i c}}{20480 d^{4}} & \text{for}\: 20480 d^{4} e^{i c} \neq 0 \\\frac{x \left (a^{2} e^{6 i c} + 3 a^{2} e^{4 i c} + 3 a^{2} e^{2 i c} + a^{2}\right ) e^{- i c}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

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

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Giac [B]  time = 1.28452, size = 828, normalized size = 12. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/160*(45*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 90*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c)
+ 1) + 45*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 40*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) -
1) + 80*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 40*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1)
- 45*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 90*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1
) - 45*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 40*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1
) - 80*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 40*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1)
- 5*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 10*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c))
- 5*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c))
+ 10*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) +
4*I*a^2*e^(10*I*d*x + 8*I*c) + 28*I*a^2*e^(8*I*d*x + 6*I*c) + 104*I*a^2*e^(6*I*d*x + 4*I*c) + 120*I*a^2*e^(4*
I*d*x + 2*I*c) + 20*I*a^2*e^(2*I*d*x) - 20*I*a^2*e^(-2*I*c))/(d*e^(5*I*d*x + 3*I*c) + 2*d*e^(3*I*d*x + I*c) +
d*e^(I*d*x - I*c))