3.66 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=55 \[ \frac{i a^7}{2 d (a-i a \tan (c+d x))^2}-\frac{2 i a^8}{3 d (a-i a \tan (c+d x))^3} \]

[Out]

(((-2*I)/3)*a^8)/(d*(a - I*a*Tan[c + d*x])^3) + ((I/2)*a^7)/(d*(a - I*a*Tan[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0485425, antiderivative size = 55, 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 = {3487, 43} \[ \frac{i a^7}{2 d (a-i a \tan (c+d x))^2}-\frac{2 i a^8}{3 d (a-i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(((-2*I)/3)*a^8)/(d*(a - I*a*Tan[c + d*x])^3) + ((I/2)*a^7)/(d*(a - I*a*Tan[c + d*x])^2)

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{a+x}{(a-x)^4} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \left (\frac{2 a}{(a-x)^4}-\frac{1}{(a-x)^3}\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{2 i a^8}{3 d (a-i a \tan (c+d x))^3}+\frac{i a^7}{2 d (a-i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.485156, size = 50, normalized size = 0.91 \[ \frac{a^5 (5 \cos (c+d x)-i \sin (c+d x)) (\sin (5 (c+d x))-i \cos (5 (c+d x)))}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*(5*Cos[c + d*x] - I*Sin[c + d*x])*((-I)*Cos[5*(c + d*x)] + Sin[5*(c + d*x)]))/(24*d)

________________________________________________________________________________________

Maple [B]  time = 0.084, size = 231, normalized size = 4.2 \begin{align*}{\frac{1}{d} \left ({\frac{i}{6}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}+5\,{a}^{5} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1/8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -10\,i{a}^{5} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) -10\,{a}^{5} \left ( -1/6\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{\frac{5\,i}{6}}{a}^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{a}^{5} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x)

[Out]

1/d*(1/6*I*a^5*sin(d*x+c)^6+5*a^5*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*
sin(d*x+c)+1/16*d*x+1/16*c)-10*I*a^5*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)-10*a^5*(-1/6*cos(d*x+c
)^5*sin(d*x+c)+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-5/6*I*a^5*cos(d*x+c)^6+a^5*(1/6*
(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

________________________________________________________________________________________

Maxima [B]  time = 1.6701, size = 126, normalized size = 2.29 \begin{align*} \frac{-24 i \, a^{5} \tan \left (d x + c\right )^{4} - 80 \, a^{5} \tan \left (d x + c\right )^{3} + 96 i \, a^{5} \tan \left (d x + c\right )^{2} + 48 \, a^{5} \tan \left (d x + c\right ) - 8 i \, a^{5}}{48 \,{\left (\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

1/48*(-24*I*a^5*tan(d*x + c)^4 - 80*a^5*tan(d*x + c)^3 + 96*I*a^5*tan(d*x + c)^2 + 48*a^5*tan(d*x + c) - 8*I*a
^5)/((tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1)*d)

________________________________________________________________________________________

Fricas [A]  time = 1.12146, size = 93, normalized size = 1.69 \begin{align*} \frac{-2 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/24*(-2*I*a^5*e^(6*I*d*x + 6*I*c) - 3*I*a^5*e^(4*I*d*x + 4*I*c))/d

________________________________________________________________________________________

Sympy [A]  time = 0.604502, size = 82, normalized size = 1.49 \begin{align*} \begin{cases} \frac{- 8 i a^{5} d e^{6 i c} e^{6 i d x} - 12 i a^{5} d e^{4 i c} e^{4 i d x}}{96 d^{2}} & \text{for}\: 96 d^{2} \neq 0 \\x \left (\frac{a^{5} e^{6 i c}}{2} + \frac{a^{5} e^{4 i c}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**5,x)

[Out]

Piecewise(((-8*I*a**5*d*exp(6*I*c)*exp(6*I*d*x) - 12*I*a**5*d*exp(4*I*c)*exp(4*I*d*x))/(96*d**2), Ne(96*d**2,
0)), (x*(a**5*exp(6*I*c)/2 + a**5*exp(4*I*c)/2), True))

________________________________________________________________________________________

Giac [B]  time = 1.58836, size = 252, normalized size = 4.58 \begin{align*} \frac{-32 i \, a^{5} e^{\left (18 i \, d x + 12 i \, c\right )} - 240 i \, a^{5} e^{\left (16 i \, d x + 10 i \, c\right )} - 768 i \, a^{5} e^{\left (14 i \, d x + 8 i \, c\right )} - 1360 i \, a^{5} e^{\left (12 i \, d x + 6 i \, c\right )} - 1440 i \, a^{5} e^{\left (10 i \, d x + 4 i \, c\right )} - 912 i \, a^{5} e^{\left (8 i \, d x + 2 i \, c\right )} - 48 i \, a^{5} e^{\left (4 i \, d x - 2 i \, c\right )} - 320 i \, a^{5} e^{\left (6 i \, d x\right )}}{384 \,{\left (d e^{\left (12 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 4 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 2 i \, c\right )} + 15 \, d e^{\left (4 i \, d x - 2 i \, c\right )} + 6 \, d e^{\left (2 i \, d x - 4 i \, c\right )} + 20 \, d e^{\left (6 i \, d x\right )} + d e^{\left (-6 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

1/384*(-32*I*a^5*e^(18*I*d*x + 12*I*c) - 240*I*a^5*e^(16*I*d*x + 10*I*c) - 768*I*a^5*e^(14*I*d*x + 8*I*c) - 13
60*I*a^5*e^(12*I*d*x + 6*I*c) - 1440*I*a^5*e^(10*I*d*x + 4*I*c) - 912*I*a^5*e^(8*I*d*x + 2*I*c) - 48*I*a^5*e^(
4*I*d*x - 2*I*c) - 320*I*a^5*e^(6*I*d*x))/(d*e^(12*I*d*x + 6*I*c) + 6*d*e^(10*I*d*x + 4*I*c) + 15*d*e^(8*I*d*x
+ 2*I*c) + 15*d*e^(4*I*d*x - 2*I*c) + 6*d*e^(2*I*d*x - 4*I*c) + 20*d*e^(6*I*d*x) + d*e^(-6*I*c))