\(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx\) [93]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {63 a^8 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {63 i a^8 \sec (c+d x)}{2 d}+\frac {6 i a^3 \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{5 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}-\frac {42 i a^2 \cos (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^3}{5 d}-\frac {21 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{2 d} \] Output:

-63/2*a^8*arctanh(sin(d*x+c))/d-63/2*I*a^8*sec(d*x+c)/d+6/5*I*a^3*cos(d*x+ 
c)^3*(a+I*a*tan(d*x+c))^5/d-2/5*I*a*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^7/d-42 
/5*I*a^2*cos(d*x+c)*(a^2+I*a^2*tan(d*x+c))^3/d-21/2*I*sec(d*x+c)*(a^8+I*a^ 
8*tan(d*x+c))/d
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1162\) vs. \(2(173)=346\).

Time = 7.94 (sec) , antiderivative size = 1162, normalized size of antiderivative = 6.72 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^8,x]
 

Output:

(63*Cos[8*c]*Cos[c + d*x]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*( 
a + I*a*Tan[c + d*x])^8)/(2*d*(Cos[d*x] + I*Sin[d*x])^8) - (63*Cos[8*c]*Co 
s[c + d*x]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(a + I*a*Tan[c + 
 d*x])^8)/(2*d*(Cos[d*x] + I*Sin[d*x])^8) + (Cos[5*d*x]*Cos[c + d*x]^8*((( 
-8*I)/5)*Cos[3*c] - (8*Sin[3*c])/5)*(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] 
 + I*Sin[d*x])^8) + (Cos[3*d*x]*Cos[c + d*x]^8*((8*I)*Cos[5*c] + 8*Sin[5*c 
])*(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8) + (Cos[d*x]*Cos 
[c + d*x]^8*((-48*I)*Cos[7*c] - 48*Sin[7*c])*(a + I*a*Tan[c + d*x])^8)/(d* 
(Cos[d*x] + I*Sin[d*x])^8) + (Cos[c + d*x]^8*Sec[c]*((-8*I)*Cos[8*c] - 8*S 
in[8*c])*(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8) - (((63*I 
)/2)*Cos[c + d*x]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sin[8*c]* 
(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8) + (((63*I)/2)*Cos[ 
c + d*x]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sin[8*c]*(a + I*a* 
Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8) + (Cos[c + d*x]^8*(48*Cos[7 
*c] - (48*I)*Sin[7*c])*Sin[d*x]*(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I 
*Sin[d*x])^8) + (Cos[c + d*x]^8*(-8*Cos[5*c] + (8*I)*Sin[5*c])*Sin[3*d*x]* 
(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8) + (Cos[c + d*x]^8* 
((8*Cos[3*c])/5 - ((8*I)/5)*Sin[3*c])*Sin[5*d*x]*(a + I*a*Tan[c + d*x])^8) 
/(d*(Cos[d*x] + I*Sin[d*x])^8) + (Cos[c + d*x]^8*(Cos[8*c]/4 - (I/4)*Sin[8 
*c])*(a + I*a*Tan[c + d*x])^8)/(d*(Cos[d*x] + I*Sin[d*x])^8*(Cos[c/2 + ...
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3977, 3042, 3977, 3042, 3977, 3042, 3979, 3042, 3967, 3042, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^8}{\sec (c+d x)^5}dx\)

\(\Big \downarrow \) 3977

\(\displaystyle -\frac {9}{5} a^2 \int \cos ^3(c+d x) (i \tan (c+d x) a+a)^6dx-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{5} a^2 \int \frac {(i \tan (c+d x) a+a)^6}{\sec (c+d x)^3}dx-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3977

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \int \cos (c+d x) (i \tan (c+d x) a+a)^4dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \int \frac {(i \tan (c+d x) a+a)^4}{\sec (c+d x)}dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3977

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3979

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3967

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \left (\frac {3}{2} a \left (a \int \sec (c+d x)dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \left (\frac {3}{2} a \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {9}{5} a^2 \left (-\frac {7}{3} a^2 \left (-5 a^2 \left (\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac {3}{2} a \left (\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d}\right )\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^5}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^7}{5 d}\)

Input:

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^8,x]
 

Output:

(((-2*I)/5)*a*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^7)/d - (9*a^2*((((-2*I 
)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^5)/d - (7*a^2*(((-2*I)*a*Cos[ 
c + d*x]*(a + I*a*Tan[c + d*x])^3)/d - 5*a^2*((3*a*((a*ArcTanh[Sin[c + d*x 
]])/d + (I*a*Sec[c + d*x])/d))/2 + ((I/2)*Sec[c + d*x]*(a^2 + I*a^2*Tan[c 
+ d*x]))/d)))/3))/5
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 3977

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( 
n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m))   Int[(d*Sec[e + f*x]) 
^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] 
&& EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || 
 (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & 
& LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) 
&& IntegerQ[2*m]
 

rule 3979

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n 
 - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1))   Int[(d*Se 
c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, 
 m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ 
[2*m, 2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (152 ) = 304\).

Time = 2.78 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.86

\[\frac {a^{8} \sin \left (d x +c \right )^{9}}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{8} \sin \left (d x +c \right )^{7}}{2 d}+\frac {203 a^{8} \sin \left (d x +c \right )^{5}}{10 d}+\frac {21 a^{8} \sin \left (d x +c \right )^{3}}{2 d}+\frac {283 a^{8} \sin \left (d x +c \right )}{10 d}-\frac {63 a^{8} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {112 i a^{8} \cos \left (d x +c \right )^{3}}{15 d}-\frac {832 i a^{8} \cos \left (d x +c \right )}{15 d}-\frac {104 i a^{8} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}}{5 d}-\frac {8 i a^{8} \cos \left (d x +c \right )^{5}}{5 d}+\frac {29 a^{8} \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5 d}-\frac {8 a^{8} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{5 d}-\frac {8 i a^{8} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}}{d}-\frac {8 i a^{8} \sin \left (d x +c \right )^{8}}{d \cos \left (d x +c \right )}-\frac {416 i a^{8} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}}{15 d}+\frac {56 i a^{8} \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5 d}\]

Input:

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^8,x)
 

Output:

1/2/d*a^8*sin(d*x+c)^9/cos(d*x+c)^2+1/2*a^8*sin(d*x+c)^7/d+203/10*a^8*sin( 
d*x+c)^5/d+21/2*a^8*sin(d*x+c)^3/d+283/10*a^8*sin(d*x+c)/d-63/2/d*a^8*ln(s 
ec(d*x+c)+tan(d*x+c))+112/15*I/d*a^8*cos(d*x+c)^3-832/15*I/d*a^8*cos(d*x+c 
)-104/5*I/d*a^8*cos(d*x+c)*sin(d*x+c)^4-8/5*I/d*a^8*cos(d*x+c)^5+29/5/d*a^ 
8*sin(d*x+c)*cos(d*x+c)^4-8/5/d*a^8*cos(d*x+c)^2*sin(d*x+c)-8*I/d*a^8*cos( 
d*x+c)*sin(d*x+c)^6-8*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)-416/15*I/d*a^8*cos(d 
*x+c)*sin(d*x+c)^2+56/5*I/d*a^8*cos(d*x+c)^3*sin(d*x+c)^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.10 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {-16 i \, a^{8} e^{\left (9 i \, d x + 9 i \, c\right )} + 48 i \, a^{8} e^{\left (7 i \, d x + 7 i \, c\right )} - 336 i \, a^{8} e^{\left (5 i \, d x + 5 i \, c\right )} - 1050 i \, a^{8} e^{\left (3 i \, d x + 3 i \, c\right )} - 630 i \, a^{8} e^{\left (i \, d x + i \, c\right )} - 315 \, {\left (a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 315 \, {\left (a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{10 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
 

Output:

1/10*(-16*I*a^8*e^(9*I*d*x + 9*I*c) + 48*I*a^8*e^(7*I*d*x + 7*I*c) - 336*I 
*a^8*e^(5*I*d*x + 5*I*c) - 1050*I*a^8*e^(3*I*d*x + 3*I*c) - 630*I*a^8*e^(I 
*d*x + I*c) - 315*(a^8*e^(4*I*d*x + 4*I*c) + 2*a^8*e^(2*I*d*x + 2*I*c) + a 
^8)*log(e^(I*d*x + I*c) + I) + 315*(a^8*e^(4*I*d*x + 4*I*c) + 2*a^8*e^(2*I 
*d*x + 2*I*c) + a^8)*log(e^(I*d*x + I*c) - I))/(d*e^(4*I*d*x + 4*I*c) + 2* 
d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.36 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {63 a^{8} \left (\frac {\log {\left (e^{i d x} - i e^{- i c} \right )}}{2} - \frac {\log {\left (e^{i d x} + i e^{- i c} \right )}}{2}\right )}{d} + \frac {- 17 i a^{8} e^{3 i c} e^{3 i d x} - 15 i a^{8} e^{i c} e^{i d x}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} + \begin {cases} \frac {- 8 i a^{8} d^{2} e^{5 i c} e^{5 i d x} + 40 i a^{8} d^{2} e^{3 i c} e^{3 i d x} - 240 i a^{8} d^{2} e^{i c} e^{i d x}}{5 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (8 a^{8} e^{5 i c} - 24 a^{8} e^{3 i c} + 48 a^{8} e^{i c}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**8,x)
 

Output:

63*a**8*(log(exp(I*d*x) - I*exp(-I*c))/2 - log(exp(I*d*x) + I*exp(-I*c))/2 
)/d + (-17*I*a**8*exp(3*I*c)*exp(3*I*d*x) - 15*I*a**8*exp(I*c)*exp(I*d*x)) 
/(d*exp(4*I*c)*exp(4*I*d*x) + 2*d*exp(2*I*c)*exp(2*I*d*x) + d) + Piecewise 
(((-8*I*a**8*d**2*exp(5*I*c)*exp(5*I*d*x) + 40*I*a**8*d**2*exp(3*I*c)*exp( 
3*I*d*x) - 240*I*a**8*d**2*exp(I*c)*exp(I*d*x))/(5*d**3), Ne(d**3, 0)), (x 
*(8*a**8*exp(5*I*c) - 24*a**8*exp(3*I*c) + 48*a**8*exp(I*c)), True))
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (143) = 286\).

Time = 0.05 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.88 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {96 i \, a^{8} \cos \left (d x + c\right )^{5} - 840 \, a^{8} \sin \left (d x + c\right )^{5} + 224 i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{8} + 224 i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a^{8} + 96 i \, {\left (\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} + \frac {5}{\cos \left (d x + c\right )} + 15 \, \cos \left (d x + c\right )\right )} a^{8} - {\left (12 \, \sin \left (d x + c\right )^{5} + 40 \, \sin \left (d x + c\right )^{3} - \frac {30 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 180 \, \sin \left (d x + c\right )\right )} a^{8} - 56 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a^{8} - 112 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{8} - 4 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{8}}{60 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
 

Output:

-1/60*(96*I*a^8*cos(d*x + c)^5 - 840*a^8*sin(d*x + c)^5 + 224*I*(3*cos(d*x 
 + c)^5 - 5*cos(d*x + c)^3)*a^8 + 224*I*(3*cos(d*x + c)^5 - 10*cos(d*x + c 
)^3 + 15*cos(d*x + c))*a^8 + 96*I*(cos(d*x + c)^5 - 5*cos(d*x + c)^3 + 5/c 
os(d*x + c) + 15*cos(d*x + c))*a^8 - (12*sin(d*x + c)^5 + 40*sin(d*x + c)^ 
3 - 30*sin(d*x + c)/(sin(d*x + c)^2 - 1) - 105*log(sin(d*x + c) + 1) + 105 
*log(sin(d*x + c) - 1) + 180*sin(d*x + c))*a^8 - 56*(6*sin(d*x + c)^5 + 10 
*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30 
*sin(d*x + c))*a^8 - 112*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^8 - 4*(3* 
sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^8)/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. 2849 vs. \(2 (143) = 286\).

Time = 0.97 (sec) , antiderivative size = 2849, normalized size of antiderivative = 16.47 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
 

Output:

1/655360*(42021645*a^8*e^(28*I*d*x + 14*I*c)*log(I*e^(I*d*x + I*c) + 1) + 
588303030*a^8*e^(26*I*d*x + 12*I*c)*log(I*e^(I*d*x + I*c) + 1) + 382396969 
5*a^8*e^(24*I*d*x + 10*I*c)*log(I*e^(I*d*x + I*c) + 1) + 15295878780*a^8*e 
^(22*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 42063666645*a^8*e^(20*I*d 
*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 84127333290*a^8*e^(18*I*d*x + 4*I 
*c)*log(I*e^(I*d*x + I*c) + 1) + 126190999935*a^8*e^(16*I*d*x + 2*I*c)*log 
(I*e^(I*d*x + I*c) + 1) + 126190999935*a^8*e^(12*I*d*x - 2*I*c)*log(I*e^(I 
*d*x + I*c) + 1) + 84127333290*a^8*e^(10*I*d*x - 4*I*c)*log(I*e^(I*d*x + I 
*c) + 1) + 42063666645*a^8*e^(8*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) 
+ 15295878780*a^8*e^(6*I*d*x - 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 3823969 
695*a^8*e^(4*I*d*x - 10*I*c)*log(I*e^(I*d*x + I*c) + 1) + 588303030*a^8*e^ 
(2*I*d*x - 12*I*c)*log(I*e^(I*d*x + I*c) + 1) + 144218285640*a^8*e^(14*I*d 
*x)*log(I*e^(I*d*x + I*c) + 1) + 42021645*a^8*e^(-14*I*c)*log(I*e^(I*d*x + 
 I*c) + 1) + 21376575*a^8*e^(28*I*d*x + 14*I*c)*log(I*e^(I*d*x + I*c) - 1) 
 + 299272050*a^8*e^(26*I*d*x + 12*I*c)*log(I*e^(I*d*x + I*c) - 1) + 194526 
8325*a^8*e^(24*I*d*x + 10*I*c)*log(I*e^(I*d*x + I*c) - 1) + 7781073300*a^8 
*e^(22*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 21397951575*a^8*e^(20*I 
*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 42795903150*a^8*e^(18*I*d*x + 4 
*I*c)*log(I*e^(I*d*x + I*c) - 1) + 64193854725*a^8*e^(16*I*d*x + 2*I*c)*lo 
g(I*e^(I*d*x + I*c) - 1) + 64193854725*a^8*e^(12*I*d*x - 2*I*c)*log(I*e...
 

Mupad [B] (verification not implemented)

Time = 4.58 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.62 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {63\,a^8\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {65\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,309{}\mathrm {i}-761\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1109{}\mathrm {i}+\frac {7351\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1223{}\mathrm {i}-\frac {4407\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,431{}\mathrm {i}+\frac {496\,a^8}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,5{}\mathrm {i}-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,20{}\mathrm {i}+26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,26{}\mathrm {i}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,12{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \] Input:

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^8,x)
 

Output:

(a^8*tan(c/2 + (d*x)/2)^3*1223i - (4407*a^8*tan(c/2 + (d*x)/2)^2)/5 + (735 
1*a^8*tan(c/2 + (d*x)/2)^4)/5 - a^8*tan(c/2 + (d*x)/2)^5*1109i - 761*a^8*t 
an(c/2 + (d*x)/2)^6 + a^8*tan(c/2 + (d*x)/2)^7*309i + 65*a^8*tan(c/2 + (d* 
x)/2)^8 + (496*a^8)/5 - a^8*tan(c/2 + (d*x)/2)*431i)/(d*(5*tan(c/2 + (d*x) 
/2) - tan(c/2 + (d*x)/2)^2*12i - 20*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x) 
/2)^4*26i + 26*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^6*20i - 12*tan(c/ 
2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8*5i + tan(c/2 + (d*x)/2)^9 + 1i)) - ( 
63*a^8*atanh(tan(c/2 + (d*x)/2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^{8} \left (-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i +128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i -288 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i +496 \cos \left (d x +c \right ) i +315 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-315 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-315 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+315 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+256 \sin \left (d x +c \right )^{7}-256 \sin \left (d x +c \right )^{5}+320 \sin \left (d x +c \right )^{3}+496 \sin \left (d x +c \right )^{2} i -325 \sin \left (d x +c \right )-496 i \right )}{10 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^8,x)
 

Output:

(a**8*( - 256*cos(c + d*x)*sin(c + d*x)**6*i + 128*cos(c + d*x)*sin(c + d* 
x)**4*i - 288*cos(c + d*x)*sin(c + d*x)**2*i + 496*cos(c + d*x)*i + 315*lo 
g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 315*log(tan((c + d*x)/2) - 1) - 
315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 315*log(tan((c + d*x)/2) + 
 1) + 256*sin(c + d*x)**7 - 256*sin(c + d*x)**5 + 320*sin(c + d*x)**3 + 49 
6*sin(c + d*x)**2*i - 325*sin(c + d*x) - 496*i))/(10*d*(sin(c + d*x)**2 - 
1))