3.49 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=88 \[ -\frac{a^3 \sin ^3(c+d x)}{15 d}+\frac{a^3 \sin (c+d x)}{5 d}-\frac{i a^3 \cos ^3(c+d x)}{15 d}-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[Out]

((-I/15)*a^3*Cos[c + d*x]^3)/d + (a^3*Sin[c + d*x])/(5*d) - (a^3*Sin[c + d*x]^3)/(15*d) - (((2*I)/5)*a*Cos[c +
d*x]^5*(a + I*a*Tan[c + d*x])^2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.0710485, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3496, 3486, 2633} \[ -\frac{a^3 \sin ^3(c+d x)}{15 d}+\frac{a^3 \sin (c+d x)}{5 d}-\frac{i a^3 \cos ^3(c+d x)}{15 d}-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-I/15)*a^3*Cos[c + d*x]^3)/d + (a^3*Sin[c + d*x])/(5*d) - (a^3*Sin[c + d*x]^3)/(15*d) - (((2*I)/5)*a*Cos[c +
d*x]^5*(a + I*a*Tan[c + d*x])^2)/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 3486

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

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))^3 \, dx &=-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{1}{5} a^2 \int \cos ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac{i a^3 \cos ^3(c+d x)}{15 d}-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{1}{5} a^3 \int \cos ^3(c+d x) \, dx\\ &=-\frac{i a^3 \cos ^3(c+d x)}{15 d}-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac{a^3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=-\frac{i a^3 \cos ^3(c+d x)}{15 d}+\frac{a^3 \sin (c+d x)}{5 d}-\frac{a^3 \sin ^3(c+d x)}{15 d}-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.447172, size = 55, normalized size = 0.62 \[ \frac{a^3 (-6 i \sin (2 (c+d x))+9 \cos (2 (c+d x))+5) (\sin (3 (c+d x))-i \cos (3 (c+d x)))}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*(5 + 9*Cos[2*(c + d*x)] - (6*I)*Sin[2*(c + d*x)])*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]))/(30*d)

________________________________________________________________________________________

Maple [A]  time = 0.06, size = 126, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( -i{a}^{3} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) -3\,{a}^{3} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,i}{5}}{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{{a}^{3}\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))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.20822, size = 142, normalized size = 1.61 \begin{align*} -\frac{9 i \, a^{3} \cos \left (d x + c\right )^{5} + i \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 3 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{3} -{\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^{3}}{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))^3,x, algorithm="maxima")

[Out]

-1/15*(9*I*a^3*cos(d*x + c)^5 + I*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 - 3*(3*sin(d*x + c)^5 - 5*sin(d*x
+ c)^3)*a^3 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.1175, size = 131, normalized size = 1.49 \begin{align*} \frac{-3 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{3} e^{\left (i \, d x + i \, c\right )}}{60 \, 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))^3,x, algorithm="fricas")

[Out]

1/60*(-3*I*a^3*e^(5*I*d*x + 5*I*c) - 10*I*a^3*e^(3*I*d*x + 3*I*c) - 15*I*a^3*e^(I*d*x + I*c))/d

________________________________________________________________________________________

Sympy [A]  time = 0.727356, size = 117, normalized size = 1.33 \begin{align*} \begin{cases} \frac{- 24 i a^{3} d^{2} e^{5 i c} e^{5 i d x} - 80 i a^{3} d^{2} e^{3 i c} e^{3 i d x} - 120 i a^{3} d^{2} e^{i c} e^{i d x}}{480 d^{3}} & \text{for}\: 480 d^{3} \neq 0 \\x \left (\frac{a^{3} e^{5 i c}}{4} + \frac{a^{3} e^{3 i c}}{2} + \frac{a^{3} e^{i c}}{4}\right ) & \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))**3,x)

[Out]

Piecewise(((-24*I*a**3*d**2*exp(5*I*c)*exp(5*I*d*x) - 80*I*a**3*d**2*exp(3*I*c)*exp(3*I*d*x) - 120*I*a**3*d**2
*exp(I*c)*exp(I*d*x))/(480*d**3), Ne(480*d**3, 0)), (x*(a**3*exp(5*I*c)/4 + a**3*exp(3*I*c)/2 + a**3*exp(I*c)/
4), True))

________________________________________________________________________________________

Giac [B]  time = 1.50648, size = 1254, normalized size = 14.25 \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))^3,x, algorithm="giac")

[Out]

1/7680*(1785*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 7140*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 7140*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 10710*a^3*e^(4*I*d*x)*log(I*e^(I*d*x +
I*c) + 1) + 1785*a^3*e^(-4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1530*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*
c) - 1) + 6120*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 6120*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*
x + I*c) - 1) + 9180*a^3*e^(4*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1530*a^3*e^(-4*I*c)*log(I*e^(I*d*x + I*c) -
1) - 1785*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 7140*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x +
I*c) + 1) - 7140*a^3*e^(2*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 10710*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x
+ I*c) + 1) - 1785*a^3*e^(-4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1530*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x +
I*c) - 1) - 6120*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6120*a^3*e^(2*I*d*x - 2*I*c)*log(-I*e^
(I*d*x + I*c) - 1) - 9180*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 1530*a^3*e^(-4*I*c)*log(-I*e^(I*d*x +
I*c) - 1) - 255*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1020*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*
d*x) + e^(-I*c)) - 1020*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1530*a^3*e^(4*I*d*x)*log(I*e^(I*
d*x) + e^(-I*c)) - 255*a^3*e^(-4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 255*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*
x) + e^(-I*c)) + 1020*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 1020*a^3*e^(2*I*d*x - 2*I*c)*log(
-I*e^(I*d*x) + e^(-I*c)) + 1530*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) + 255*a^3*e^(-4*I*c)*log(-I*e^(I*
d*x) + e^(-I*c)) - 384*I*a^3*e^(13*I*d*x + 9*I*c) - 2816*I*a^3*e^(11*I*d*x + 7*I*c) - 9344*I*a^3*e^(9*I*d*x +
5*I*c) - 16896*I*a^3*e^(7*I*d*x + 3*I*c) - 17024*I*a^3*e^(5*I*d*x + I*c) - 8960*I*a^3*e^(3*I*d*x - I*c) - 1920
*I*a^3*e^(I*d*x - 3*I*c))/(d*e^(8*I*d*x + 4*I*c) + 4*d*e^(6*I*d*x + 2*I*c) + 4*d*e^(2*I*d*x - 2*I*c) + 6*d*e^(
4*I*d*x) + d*e^(-4*I*c))