\(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [43]

Optimal result

Integrand size = 24, antiderivative size = 90 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 x}{8}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3}-\frac {i a^5}{8 d (a-i a \tan (c+d x))^2}-\frac {i a^4}{8 d (a-i a \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3568, 46, 212} \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3}-\frac {i a^5}{8 d (a-i a \tan (c+d x))^2}-\frac {i a^4}{8 d (a-i a \tan (c+d x))}+\frac {a^3 x}{8} \]

[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 46

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] && Lt
Q[m + n + 2, 0])

Rule 212

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

Rule 3568

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^7\right ) \text {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 ) \text {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 ) \text {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] (verified)

Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.72 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \left (-10+9 i \tan (c+d x)+3 \tan ^2(c+d x)+3 \arctan (\tan (c+d x)) (i+\tan (c+d x))^3\right )}{24 d (i+\tan (c+d x))^3} \]

[In]

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

[Out]

(a^3*(-10 + (9*I)*Tan[c + d*x] + 3*Tan[c + d*x]^2 + 3*ArcTan[Tan[c + d*x]]*(I + Tan[c + d*x])^3))/(24*d*(I + T
an[c + d*x])^3)

Maple [A] (verified)

Time = 47.77 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {a^{3} x}{8}-\frac {i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}}{48 d}-\frac {3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{32 d}-\frac {3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}\)	\(62\)
derivativedivides	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )-3 a^{3} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {i a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2}+a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(156\)
default	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )-3 a^{3} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {i a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2}+a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(156\)

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/8*a^3*x-1/48*I/d*a^3*exp(6*I*(d*x+c))-3/32*I/d*a^3*exp(4*I*(d*x+c))-3/16*I/d*a^3*exp(2*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.61 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {12 \, a^{3} d x - 2 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )}}{96 \, d} \]

[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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.46 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} x}{8} + \begin {cases} \frac {- 512 i a^{3} d^{2} e^{6 i c} e^{6 i d x} - 2304 i a^{3} d^{2} e^{4 i c} e^{4 i d x} - 4608 i a^{3} d^{2} e^{2 i c} e^{2 i d x}}{24576 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (\frac {a^{3} e^{6 i c}}{8} + \frac {3 a^{3} e^{4 i c}}{8} + \frac {3 a^{3} e^{2 i c}}{8}\right ) & \text {otherwise} \end {cases} \]

[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(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))

Maxima [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.17 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {3 \, {\left (d x + c\right )} a^{3} + \frac {3 \, a^{3} \tan \left (d x + c\right )^{5} + 8 \, a^{3} \tan \left (d x + c\right )^{3} + 6 i \, a^{3} \tan \left (d x + c\right )^{2} + 21 \, a^{3} \tan \left (d x + c\right ) - 10 i \, a^{3}}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*(3*(d*x + c)*a^3 + (3*a^3*tan(d*x + c)^5 + 8*a^3*tan(d*x + c)^3 + 6*I*a^3*tan(d*x + c)^2 + 21*a^3*tan(d*x
+ c) - 10*I*a^3)/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (70) = 140\).

Time = 0.81 (sec) , antiderivative size = 457, normalized size of antiderivative = 5.08 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {12 \, a^{3} d x e^{\left (8 i \, d x + 4 i \, c\right )} + 48 \, a^{3} d x e^{\left (6 i \, d x + 2 i \, c\right )} + 48 \, a^{3} d x e^{\left (2 i \, d x - 2 i \, c\right )} + 72 \, a^{3} d x e^{\left (4 i \, d x\right )} + 12 \, a^{3} d x e^{\left (-4 i \, c\right )} - 3 i \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 18 i \, a^{3} e^{\left (4 i \, d x\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 3 i \, a^{3} e^{\left (-4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 i \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) + 12 i \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) + 12 i \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) + 18 i \, a^{3} e^{\left (4 i \, d x\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) + 3 i \, a^{3} e^{\left (-4 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - 2 i \, a^{3} e^{\left (14 i \, d x + 10 i \, c\right )} - 17 i \, a^{3} e^{\left (12 i \, d x + 8 i \, c\right )} - 66 i \, a^{3} e^{\left (10 i \, d x + 6 i \, c\right )} - 134 i \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} - 146 i \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} - 18 i \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} - 81 i \, a^{3} e^{\left (4 i \, d x\right )}}{96 \, {\left (d e^{\left (8 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 2 i \, c\right )} + 4 \, d e^{\left (2 i \, d x - 2 i \, c\right )} + 6 \, d e^{\left (4 i \, d x\right )} + d e^{\left (-4 i \, c\right )}\right )}} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(12*a^3*d*x*e^(8*I*d*x + 4*I*c) + 48*a^3*d*x*e^(6*I*d*x + 2*I*c) + 48*a^3*d*x*e^(2*I*d*x - 2*I*c) + 72*a^
3*d*x*e^(4*I*d*x) + 12*a^3*d*x*e^(-4*I*c) - 3*I*a^3*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a^
3*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1)
- 18*I*a^3*e^(4*I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) - 3*I*a^3*e^(-4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 3*I*a
^3*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 12*I*a^3*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*
c)) + 12*I*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 18*I*a^3*e^(4*I*d*x)*log(e^(2*I*d*x) + e^(-
2*I*c)) + 3*I*a^3*e^(-4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - 2*I*a^3*e^(14*I*d*x + 10*I*c) - 17*I*a^3*e^(12*I*
d*x + 8*I*c) - 66*I*a^3*e^(10*I*d*x + 6*I*c) - 134*I*a^3*e^(8*I*d*x + 4*I*c) - 146*I*a^3*e^(6*I*d*x + 2*I*c) -
18*I*a^3*e^(2*I*d*x - 2*I*c) - 81*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))

Mupad [B] (verification not implemented)

Time = 4.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3\,x}{8}-\frac {\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8}+\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{8}-\frac {5\,a^3}{12}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(a^3*x)/8 - ((a^3*tan(c + d*x)*3i)/8 - (5*a^3)/12 + (a^3*tan(c + d*x)^2)/8)/(d*(3*tan(c + d*x) - tan(c + d*x)^
2*3i - tan(c + d*x)^3 + 1i))