3.43 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=90 \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3}-\frac{i a^5}{8 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{a^3 x}{8} \]

[Out]

(a^3*x)/8 - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3) - ((I/8)*a^5)/(d*(a - I*a*Tan[c + d*x])^2) - ((I/8)*a^4)/
(d*(a - I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0667252, antiderivative size = 90, 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 = {3487, 44, 206} \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3}-\frac{i a^5}{8 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*x)/8 - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3) - ((I/8)*a^5)/(d*(a - I*a*Tan[c + d*x])^2) - ((I/8)*a^4)/
(d*(a - I*a*Tan[c + d*x]))

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.538751, size = 109, normalized size = 1.21 \[ \frac{a^3 (-9 \sin (c+d x)-12 i d x \sin (3 (c+d x))+2 \sin (3 (c+d x))-27 i \cos (c+d x)+2 (6 d x-i) \cos (3 (c+d x))) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{96 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*((-27*I)*Cos[c + d*x] + 2*(-I + 6*d*x)*Cos[3*(c + d*x)] - 9*Sin[c + d*x] + 2*Sin[3*(c + d*x)] - (12*I)*d*
x*Sin[3*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d*x)]))/(96*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [B]  time = 0.062, size = 156, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ( -i{a}^{3} \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 ) -3\,{a}^{3} \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{i}{2}}{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{a}^{3} \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))^3,x)

[Out]

1/d*(-I*a^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)-3*a^3*(-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)-1/2*I*a^3*cos(d*x+c)^6+a^3*(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))

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Maxima [A]  time = 1.69633, size = 142, normalized size = 1.58 \begin{align*} \frac{6 \,{\left (d x + c\right )} a^{3} + \frac{6 \, a^{3} \tan \left (d x + c\right )^{5} + 16 \, a^{3} \tan \left (d x + c\right )^{3} + 12 i \, a^{3} \tan \left (d x + c\right )^{2} + 42 \, a^{3} \tan \left (d x + c\right ) - 20 i \, a^{3}}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, 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))^3,x, algorithm="maxima")

[Out]

1/48*(6*(d*x + c)*a^3 + (6*a^3*tan(d*x + c)^5 + 16*a^3*tan(d*x + c)^3 + 12*I*a^3*tan(d*x + c)^2 + 42*a^3*tan(d
*x + c) - 20*I*a^3)/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.21181, size = 151, normalized size = 1.68 \begin{align*} \frac{12 \, a^{3} d x - 2 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )}}{96 \, 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))^3,x, algorithm="fricas")

[Out]

1/96*(12*a^3*d*x - 2*I*a^3*e^(6*I*d*x + 6*I*c) - 9*I*a^3*e^(4*I*d*x + 4*I*c) - 18*I*a^3*e^(2*I*d*x + 2*I*c))/d

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Sympy [A]  time = 0.731826, size = 133, normalized size = 1.48 \begin{align*} \frac{a^{3} x}{8} + \begin{cases} \frac{- 512 i a^{3} d^{2} e^{6 i c} e^{6 i d x} - 2304 i a^{3} d^{2} e^{4 i c} e^{4 i d x} - 4608 i a^{3} d^{2} e^{2 i c} e^{2 i d x}}{24576 d^{3}} & \text{for}\: 24576 d^{3} \neq 0 \\x \left (\frac{a^{3} e^{6 i c}}{8} + \frac{3 a^{3} e^{4 i c}}{8} + \frac{3 a^{3} e^{2 i c}}{8}\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))**3,x)

[Out]

a**3*x/8 + Piecewise(((-512*I*a**3*d**2*exp(6*I*c)*exp(6*I*d*x) - 2304*I*a**3*d**2*exp(4*I*c)*exp(4*I*d*x) - 4
608*I*a**3*d**2*exp(2*I*c)*exp(2*I*d*x))/(24576*d**3), Ne(24576*d**3, 0)), (x*(a**3*exp(6*I*c)/8 + 3*a**3*exp(
4*I*c)/8 + 3*a**3*exp(2*I*c)/8), True))

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Giac [B]  time = 1.35198, size = 617, normalized size = 6.86 \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)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/384*(48*a^3*d*x*e^(8*I*d*x + 4*I*c) + 192*a^3*d*x*e^(6*I*d*x + 2*I*c) + 192*a^3*d*x*e^(2*I*d*x - 2*I*c) + 28
8*a^3*d*x*e^(4*I*d*x) + 48*a^3*d*x*e^(-4*I*c) - 12*I*a^3*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 48
*I*a^3*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 48*I*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c)
+ 1) - 72*I*a^3*e^(4*I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a^3*e^(-4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) +
12*I*a^3*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 48*I*a^3*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x) + e
^(-2*I*c)) + 48*I*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 72*I*a^3*e^(4*I*d*x)*log(e^(2*I*d*x)
+ e^(-2*I*c)) + 12*I*a^3*e^(-4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - 8*I*a^3*e^(14*I*d*x + 10*I*c) - 68*I*a^3*
e^(12*I*d*x + 8*I*c) - 264*I*a^3*e^(10*I*d*x + 6*I*c) - 536*I*a^3*e^(8*I*d*x + 4*I*c) - 584*I*a^3*e^(6*I*d*x +
2*I*c) - 72*I*a^3*e^(2*I*d*x - 2*I*c) - 324*I*a^3*e^(4*I*d*x))/(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))