\(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [26]

Optimal result

Integrand size = 24, antiderivative size = 117 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 x}{4}-\frac {i a^5}{12 d (a-i a \tan (c+d x))^3}-\frac {i a^4}{8 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3}{16 d (a-i a \tan (c+d x))}+\frac {i a^3}{16 d (a+i a \tan (c+d x))} \]

[Out]

1/4*a^2*x-1/12*I*a^5/d/(a-I*a*tan(d*x+c))^3-1/8*I*a^4/d/(a-I*a*tan(d*x+c))^2-3/16*I*a^3/d/(a-I*a*tan(d*x+c))+1
/16*I*a^3/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 117, 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} \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i a^5}{12 d (a-i a \tan (c+d x))^3}-\frac {i a^4}{8 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3}{16 d (a-i a \tan (c+d x))}+\frac {i a^3}{16 d (a+i a \tan (c+d x))}+\frac {a^2 x}{4} \]

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*x)/4 - ((I/12)*a^5)/(d*(a - I*a*Tan[c + d*x])^3) - ((I/8)*a^4)/(d*(a - I*a*Tan[c + d*x])^2) - (((3*I)/16)
*a^3)/(d*(a - I*a*Tan[c + d*x])) + ((I/16)*a^3)/(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)^2} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^7\right ) \text {Subst}\left (\int \left (\frac {1}{4 a^2 (a-x)^4}+\frac {1}{4 a^3 (a-x)^3}+\frac {3}{16 a^4 (a-x)^2}+\frac {1}{16 a^4 (a+x)^2}+\frac {1}{4 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^5}{12 d (a-i a \tan (c+d x))^3}-\frac {i a^4}{8 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3}{16 d (a-i a \tan (c+d x))}+\frac {i a^3}{16 d (a+i a \tan (c+d x))}-\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{4 d} \\ & = \frac {a^2 x}{4}-\frac {i a^5}{12 d (a-i a \tan (c+d x))^3}-\frac {i a^4}{8 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3}{16 d (a-i a \tan (c+d x))}+\frac {i a^3}{16 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \left (4 i-\tan (c+d x)+6 i \tan ^2(c+d x)+3 \tan ^3(c+d x)+3 \arctan (\tan (c+d x)) (-i+\tan (c+d x)) (i+\tan (c+d x))^3\right )}{12 d (-i+\tan (c+d x)) (i+\tan (c+d x))^3} \]

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*(4*I - Tan[c + d*x] + (6*I)*Tan[c + d*x]^2 + 3*Tan[c + d*x]^3 + 3*ArcTan[Tan[c + d*x]]*(-I + Tan[c + d*x]
)*(I + Tan[c + d*x])^3))/(12*d*(-I + Tan[c + d*x])*(I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 26.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {a^{2} x}{4}-\frac {i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}}{96 d}-\frac {i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}}{16 d}-\frac {5 i a^{2} \cos \left (2 d x +2 c \right )}{32 d}+\frac {7 a^{2} \sin \left (2 d x +2 c \right )}{32 d}\)	\(79\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {i a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{3}+a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(121\)
default	\(\frac {-a^{2} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {i a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{3}+a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(121\)

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4*a^2*x-1/96*I/d*a^2*exp(6*I*(d*x+c))-1/16*I/d*a^2*exp(4*I*(d*x+c))-5/32*I/d*a^2*cos(2*d*x+2*c)+7/32/d*a^2*s
in(2*d*x+2*c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {{\left (24 \, a^{2} d x e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 6 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 18 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{96 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/96*(24*a^2*d*x*e^(2*I*d*x + 2*I*c) - I*a^2*e^(8*I*d*x + 8*I*c) - 6*I*a^2*e^(6*I*d*x + 6*I*c) - 18*I*a^2*e^(4
*I*d*x + 4*I*c) + 3*I*a^2)*e^(-2*I*d*x - 2*I*c)/d

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.58 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^{2} x}{4} + \begin {cases} \frac {\left (- 8192 i a^{2} d^{3} e^{8 i c} e^{6 i d x} - 49152 i a^{2} d^{3} e^{6 i c} e^{4 i d x} - 147456 i a^{2} d^{3} e^{4 i c} e^{2 i d x} + 24576 i a^{2} d^{3} e^{- 2 i d x}\right ) e^{- 2 i c}}{786432 d^{4}} & \text {for}\: d^{4} e^{2 i c} \neq 0 \\x \left (- \frac {a^{2}}{4} + \frac {\left (a^{2} e^{8 i c} + 4 a^{2} e^{6 i c} + 6 a^{2} e^{4 i c} + 4 a^{2} e^{2 i c} + a^{2}\right ) e^{- 2 i c}}{16}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*x/4 + Piecewise(((-8192*I*a**2*d**3*exp(8*I*c)*exp(6*I*d*x) - 49152*I*a**2*d**3*exp(6*I*c)*exp(4*I*d*x) -
147456*I*a**2*d**3*exp(4*I*c)*exp(2*I*d*x) + 24576*I*a**2*d**3*exp(-2*I*d*x))*exp(-2*I*c)/(786432*d**4), Ne(d
**4*exp(2*I*c), 0)), (x*(-a**2/4 + (a**2*exp(8*I*c) + 4*a**2*exp(6*I*c) + 6*a**2*exp(4*I*c) + 4*a**2*exp(2*I*c
) + a**2)*exp(-2*I*c)/16), True))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {3 \, {\left (d x + c\right )} a^{2} + \frac {3 \, a^{2} \tan \left (d x + c\right )^{5} + 8 \, a^{2} \tan \left (d x + c\right )^{3} + 9 \, a^{2} \tan \left (d x + c\right ) - 4 i \, a^{2}}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*(d*x + c)*a^2 + (3*a^2*tan(d*x + c)^5 + 8*a^2*tan(d*x + c)^3 + 9*a^2*tan(d*x + c) - 4*I*a^2)/(tan(d*x
+ c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.44 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {24 \, a^{2} d x e^{\left (6 i \, d x + 4 i \, c\right )} + 48 \, a^{2} d x e^{\left (4 i \, d x + 2 i \, c\right )} + 24 \, a^{2} d x e^{\left (2 i \, d x\right )} - i \, a^{2} e^{\left (12 i \, d x + 10 i \, c\right )} - 8 i \, a^{2} e^{\left (10 i \, d x + 8 i \, c\right )} - 31 i \, a^{2} e^{\left (8 i \, d x + 6 i \, c\right )} - 42 i \, a^{2} e^{\left (6 i \, d x + 4 i \, c\right )} - 15 i \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} + 6 i \, a^{2} e^{\left (2 i \, d x\right )} + 3 i \, a^{2} e^{\left (-2 i \, c\right )}}{96 \, {\left (d e^{\left (6 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (4 i \, d x + 2 i \, c\right )} + d e^{\left (2 i \, d x\right )}\right )}} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/96*(24*a^2*d*x*e^(6*I*d*x + 4*I*c) + 48*a^2*d*x*e^(4*I*d*x + 2*I*c) + 24*a^2*d*x*e^(2*I*d*x) - I*a^2*e^(12*I
*d*x + 10*I*c) - 8*I*a^2*e^(10*I*d*x + 8*I*c) - 31*I*a^2*e^(8*I*d*x + 6*I*c) - 42*I*a^2*e^(6*I*d*x + 4*I*c) -
15*I*a^2*e^(4*I*d*x + 2*I*c) + 6*I*a^2*e^(2*I*d*x) + 3*I*a^2*e^(-2*I*c))/(d*e^(6*I*d*x + 4*I*c) + 2*d*e^(4*I*d
*x + 2*I*c) + d*e^(2*I*d*x))

Mupad [B] (verification not implemented)

Time = 3.64 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,x}{4}+\frac {\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{4}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2}-\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{12}+\frac {a^2\,1{}\mathrm {i}}{3}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^3\,2{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-1\right )} \]

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(a^2*x)/4 + ((a^2*1i)/3 - (a^2*tan(c + d*x))/12 + (a^2*tan(c + d*x)^2*1i)/2 + (a^2*tan(c + d*x)^3)/4)/(d*(tan(
c + d*x)*2i + tan(c + d*x)^3*2i + tan(c + d*x)^4 - 1))