\(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx\) [84]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=-8 a^8 x+\frac {8 i a^8 \log (\cos (c+d x))}{d}+\frac {a^8 \tan (c+d x)}{d}-\frac {16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac {24 i a^9}{d (a-i a \tan (c+d x))} \]

[Out]

-8*a^8*x+8*I*a^8*ln(cos(d*x+c))/d+a^8*tan(d*x+c)/d-16/3*I*a^11/d/(a-I*a*tan(d*x+c))^3+16*I*a^10/d/(a-I*a*tan(d
*x+c))^2-24*I*a^9/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac {24 i a^9}{d (a-i a \tan (c+d x))}+\frac {a^8 \tan (c+d x)}{d}+\frac {8 i a^8 \log (\cos (c+d x))}{d}-8 a^8 x \]

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

-8*a^8*x + ((8*I)*a^8*Log[Cos[c + d*x]])/d + (a^8*Tan[c + d*x])/d - (((16*I)/3)*a^11)/(d*(a - I*a*Tan[c + d*x]
)^3) + ((16*I)*a^10)/(d*(a - I*a*Tan[c + d*x])^2) - ((24*I)*a^9)/(d*(a - I*a*Tan[c + d*x]))

Rule 45

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])

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 {(a+x)^4}{(a-x)^4} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^7\right ) \text {Subst}\left (\int \left (1+\frac {16 a^4}{(a-x)^4}-\frac {32 a^3}{(a-x)^3}+\frac {24 a^2}{(a-x)^2}-\frac {8 a}{a-x}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -8 a^8 x+\frac {8 i a^8 \log (\cos (c+d x))}{d}+\frac {a^8 \tan (c+d x)}{d}-\frac {16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac {24 i a^9}{d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {i a^7 \left (8 a \log (i+\tan (c+d x))+i a \tan (c+d x)+\frac {8 i a \left (-5+12 i \tan (c+d x)+9 \tan ^2(c+d x)\right )}{3 (i+\tan (c+d x))^3}\right )}{d} \]

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

((-I)*a^7*(8*a*Log[I + Tan[c + d*x]] + I*a*Tan[c + d*x] + (((8*I)/3)*a*(-5 + (12*I)*Tan[c + d*x] + 9*Tan[c + d
*x]^2))/(I + Tan[c + d*x])^3))/d

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (105 ) = 210\).

Time = 0.84 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.80

\[\frac {32 i a^{8} \left (\sin ^{6}\left (d x +c \right )\right )}{3 d}+\frac {14 i a^{8} \left (\cos ^{4}\left (d x +c \right )\right )}{3 d}+\frac {28 i a^{8} \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{3 d}+\frac {8 i a^{8} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{8} \left (\sin ^{7}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{d}+\frac {2 i a^{8} \left (\sin ^{4}\left (d x +c \right )\right )}{d}-\frac {4 i a^{8} \left (\cos ^{6}\left (d x +c \right )\right )}{3 d}-\frac {35 a^{8} \left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}-\frac {233 a^{8} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{24 d}+\frac {29 a^{8} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6 d}+\frac {4 i a^{8} \left (\sin ^{2}\left (d x +c \right )\right )}{d}-8 a^{8} x +\frac {a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {35 a^{8} \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{6 d}+\frac {175 a^{8} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}+\frac {111 a^{8} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{8 d}-\frac {8 a^{8} c}{d}\]

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x)

[Out]

32/3*I/d*a^8*sin(d*x+c)^6+14/3*I/d*a^8*cos(d*x+c)^4+28/3*I/d*a^8*sin(d*x+c)^2*cos(d*x+c)^4+8*I*a^8*ln(cos(d*x+
c))/d+1/d*a^8*sin(d*x+c)^7*cos(d*x+c)+2*I/d*a^8*sin(d*x+c)^4-4/3*I/d*a^8*cos(d*x+c)^6-35/3/d*a^8*sin(d*x+c)^3*
cos(d*x+c)^3-233/24/d*a^8*sin(d*x+c)*cos(d*x+c)^3+29/6/d*a^8*cos(d*x+c)^5*sin(d*x+c)+4*I/d*a^8*sin(d*x+c)^2-8*
a^8*x+1/d*a^8*sin(d*x+c)^9/cos(d*x+c)+35/6/d*a^8*cos(d*x+c)*sin(d*x+c)^5+175/24/d*a^8*cos(d*x+c)*sin(d*x+c)^3+
111/8/d*a^8*sin(d*x+c)*cos(d*x+c)-8/d*a^8*c

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {2 \, {\left (i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 2 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{8} + 12 \, {\left (-i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.51 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {2 i a^{8}}{d e^{2 i c} e^{2 i d x} + d} + \frac {8 i a^{8} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \begin {cases} \frac {- 2 i a^{8} d^{2} e^{6 i c} e^{6 i d x} + 6 i a^{8} d^{2} e^{4 i c} e^{4 i d x} - 18 i a^{8} d^{2} e^{2 i c} e^{2 i d x}}{3 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (4 a^{8} e^{6 i c} - 8 a^{8} e^{4 i c} + 12 a^{8} e^{2 i c}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**8,x)

[Out]

2*I*a**8/(d*exp(2*I*c)*exp(2*I*d*x) + d) + 8*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + Piecewise(((-2*I*a**8*
d**2*exp(6*I*c)*exp(6*I*d*x) + 6*I*a**8*d**2*exp(4*I*c)*exp(4*I*d*x) - 18*I*a**8*d**2*exp(2*I*c)*exp(2*I*d*x))
/(3*d**3), Ne(d**3, 0)), (x*(4*a**8*exp(6*I*c) - 8*a**8*exp(4*I*c) + 12*a**8*exp(2*I*c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.28 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {24 \, {\left (d x + c\right )} a^{8} + 12 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 3 \, a^{8} \tan \left (d x + c\right ) - \frac {8 \, {\left (9 \, a^{8} \tan \left (d x + c\right )^{5} - 15 i \, a^{8} \tan \left (d x + c\right )^{4} + 4 \, a^{8} \tan \left (d x + c\right )^{3} - 12 i \, a^{8} \tan \left (d x + c\right )^{2} + 3 \, a^{8} \tan \left (d x + c\right ) - 5 i \, a^{8}\right )}}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/3*(24*(d*x + c)*a^8 + 12*I*a^8*log(tan(d*x + c)^2 + 1) - 3*a^8*tan(d*x + c) - 8*(9*a^8*tan(d*x + c)^5 - 15*
I*a^8*tan(d*x + c)^4 + 4*a^8*tan(d*x + c)^3 - 12*I*a^8*tan(d*x + c)^2 + 3*a^8*tan(d*x + c) - 5*I*a^8)/(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. 799 vs. \(2 (98) = 196\).

Time = 1.09 (sec) , antiderivative size = 799, normalized size of antiderivative = 7.01 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/3*(-12*I*a^8*e^(28*I*d*x + 14*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 168*I*a^8*e^(26*I*d*x + 12*I*c)*log(e^(2*
I*d*x + 2*I*c) + 1) - 1092*I*a^8*e^(24*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 4368*I*a^8*e^(22*I*d*x +
8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12012*I*a^8*e^(20*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24024*I
*a^8*e^(18*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 36036*I*a^8*e^(16*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) - 36036*I*a^8*e^(12*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24024*I*a^8*e^(10*I*d*x - 4*I*c)*lo
g(e^(2*I*d*x + 2*I*c) + 1) - 12012*I*a^8*e^(8*I*d*x - 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 4368*I*a^8*e^(6*I*
d*x - 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 1092*I*a^8*e^(4*I*d*x - 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 168
*I*a^8*e^(2*I*d*x - 12*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 41184*I*a^8*e^(14*I*d*x)*log(e^(2*I*d*x + 2*I*c) +
1) - 12*I*a^8*e^(-14*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + I*a^8*e^(34*I*d*x + 20*I*c) + 11*I*a^8*e^(32*I*d*x +
18*I*c) + 58*I*a^8*e^(30*I*d*x + 16*I*c) + 217*I*a^8*e^(28*I*d*x + 14*I*c) + 725*I*a^8*e^(26*I*d*x + 12*I*c) +
2236*I*a^8*e^(24*I*d*x + 10*I*c) + 5772*I*a^8*e^(22*I*d*x + 8*I*c) + 11583*I*a^8*e^(20*I*d*x + 6*I*c) + 17589
*I*a^8*e^(18*I*d*x + 4*I*c) + 20020*I*a^8*e^(16*I*d*x + 2*I*c) + 10231*I*a^8*e^(12*I*d*x - 2*I*c) + 4147*I*a^8
*e^(10*I*d*x - 4*I*c) + 872*I*a^8*e^(8*I*d*x - 6*I*c) - 80*I*a^8*e^(6*I*d*x - 8*I*c) - 111*I*a^8*e^(4*I*d*x -
10*I*c) - 30*I*a^8*e^(2*I*d*x - 12*I*c) + 16874*I*a^8*e^(14*I*d*x) - 3*I*a^8*e^(-14*I*c))/(d*e^(28*I*d*x + 14*
I*c) + 14*d*e^(26*I*d*x + 12*I*c) + 91*d*e^(24*I*d*x + 10*I*c) + 364*d*e^(22*I*d*x + 8*I*c) + 1001*d*e^(20*I*d
*x + 6*I*c) + 2002*d*e^(18*I*d*x + 4*I*c) + 3003*d*e^(16*I*d*x + 2*I*c) + 3003*d*e^(12*I*d*x - 2*I*c) + 2002*d
*e^(10*I*d*x - 4*I*c) + 1001*d*e^(8*I*d*x - 6*I*c) + 364*d*e^(6*I*d*x - 8*I*c) + 91*d*e^(4*I*d*x - 10*I*c) + 1
4*d*e^(2*I*d*x - 12*I*c) + 3432*d*e^(14*I*d*x) + d*e^(-14*I*c))

Mupad [B] (verification not implemented)

Time = 4.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8\,\mathrm {tan}\left (c+d\,x\right )}{d}-\frac {24\,a^8\,{\mathrm {tan}\left (c+d\,x\right )}^2+a^8\,\mathrm {tan}\left (c+d\,x\right )\,32{}\mathrm {i}-\frac {40\,a^8}{3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}-\frac {a^8\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \]

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^8,x)

[Out]

(a^8*tan(c + d*x))/d - (a^8*log(tan(c + d*x) + 1i)*8i)/d - (a^8*tan(c + d*x)*32i - (40*a^8)/3 + 24*a^8*tan(c +
d*x)^2)/(d*(3*tan(c + d*x) - tan(c + d*x)^2*3i - tan(c + d*x)^3 + 1i))