\(\int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [292]

Optimal result

Integrand size = 26, antiderivative size = 223 \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {63 i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{128 \sqrt {2} d}+\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \] Output:

63/256*I*a^(1/2)*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c)) 
^(1/2))*2^(1/2)/d+21/64*I*a*cos(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+9/40*I*a 
*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-63/128*I*cos(d*x+c)*(a+I*a*tan(d* 
x+c))^(1/2)/d-21/80*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d-1/5*I*cos(d* 
x+c)^5*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.68 \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {i e^{-5 i (c+d x)} \left (-10-95 e^{2 i (c+d x)}+203 e^{4 i (c+d x)}+344 e^{6 i (c+d x)}+64 e^{8 i (c+d x)}+8 e^{10 i (c+d x)}-315 e^{4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{1280 d} \] Input:

Integrate[Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

((-1/1280*I)*(-10 - 95*E^((2*I)*(c + d*x)) + 203*E^((4*I)*(c + d*x)) + 344 
*E^((6*I)*(c + d*x)) + 64*E^((8*I)*(c + d*x)) + 8*E^((10*I)*(c + d*x)) - 3 
15*E^((4*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^(( 
2*I)*(c + d*x))]])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((5*I)*(c + d*x)))
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3978, 3042, 3983, 3042, 3978, 3042, 3983, 3042, 3971, 3042, 3970, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

\(\displaystyle \frac {9}{10} a \int \frac {\cos ^3(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \int \frac {1}{\sec (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9}{10} a \left (\frac {7 \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \left (\frac {7 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3971

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {9}{10} a \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\)

Input:

Int[Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

((-1/5*I)*Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]])/d + (9*a*(((I/4)*Cos[ 
c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (7*(((-1/3*I)*Cos[c + d*x]^3* 
Sqrt[a + I*a*Tan[c + d*x]])/d + (5*a*(((I/2)*Cos[c + d*x])/(d*Sqrt[a + I*a 
*Tan[c + d*x]]) + (3*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*S 
qrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x]*Sqrt[a + I*a*Ta 
n[c + d*x]])/d))/(4*a)))/6))/(8*a)))/10
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3970

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S 
ymbol] :> Simp[-2*(a/(b*f))   Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ 
Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 
]
 

rule 3971

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/ 
(a*f*m)), x] + Simp[a/(2*d^2)   Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + 
 f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && 
 EqQ[m/2 + n, 0] && GtQ[n, 0]
 

rule 3978

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/( 
a*f*m)), x] + Simp[a*((m + n)/(m*d^2))   Int[(d*Sec[e + f*x])^(m + 2)*(a + 
b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b 
^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 3983

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n))   Int[(d*Sec[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] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* 
n]
 

Maple [A] (verified)

Time = 4.48 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {\left (-315 i \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (315 \cos \left (d x +c \right )+315\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \left (-315 \cos \left (d x +c \right )-315\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )-315 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (288 \cos \left (d x +c \right )^{2}+420\right )+i \cos \left (d x +c \right ) \left (32 \cos \left (d x +c \right )^{4}+84 \cos \left (d x +c \right )^{2}-630\right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{1280 d}\)	\(361\)

Input:

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/1280/d*(-315*I*sin(d*x+c)*arctanh(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c 
)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))*2^(1/2))*(-cos(d*x+c)/(cos 
(d*x+c)+1))^(1/2)+(315*cos(d*x+c)+315)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
arctanh(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc 
(d*x+c)-cot(d*x+c))*2^(1/2))+I*(-315*cos(d*x+c)-315)*(-cos(d*x+c)/(cos(d*x 
+c)+1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))-315 
*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)*2^(1/2))*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)^2*(288*cos(d*x+c)^2+42 
0)+I*cos(d*x+c)*(32*cos(d*x+c)^4+84*cos(d*x+c)^2-630))*(a*(1+I*tan(d*x+c)) 
)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.20 \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (315 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d}\right ) - 315 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 64 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 344 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 203 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 95 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{1280 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/1280*(315*sqrt(1/2)*d*sqrt(-a/d^2)*e^(4*I*d*x + 4*I*c)*log(63/64*(sqrt(2 
)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))* 
sqrt(-a/d^2) + I*a)*e^(-I*d*x - I*c)/d) - 315*sqrt(1/2)*d*sqrt(-a/d^2)*e^( 
4*I*d*x + 4*I*c)*log(-63/64*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + d) 
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) - I*a)*e^(-I*d*x - I*c)/d) 
 + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(10*I*d*x + 10*I*c) - 
 64*I*e^(8*I*d*x + 8*I*c) - 344*I*e^(6*I*d*x + 6*I*c) - 203*I*e^(4*I*d*x + 
 4*I*c) + 95*I*e^(2*I*d*x + 2*I*c) + 10*I))*e^(-4*I*d*x - 4*I*c)/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Timed out
 

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. 2220 vs. \(2 (168) = 336\).

Time = 0.40 (sec) , antiderivative size = 2220, normalized size of antiderivative = 9.96 \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/5120*(20*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1 
/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin( 
4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(3/4)*((-3*I*sqrt(2)*cos(4*d*x + 4*c 
) - 3*sqrt(2)*sin(4*d*x + 4*c) - 8*I*sqrt(2))*cos(3/2*arctan2(sin(1/2*arct 
an2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c) 
, cos(4*d*x + 4*c))) + 1)) + (3*sqrt(2)*cos(4*d*x + 4*c) - 3*I*sqrt(2)*sin 
(4*d*x + 4*c) + 8*sqrt(2))*sin(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c 
), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)) 
) + 1)))*sqrt(a) + 4*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) 
^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*ar 
ctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*(8*(-I*sqrt(2)*cos(1 
/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 - I*sqrt(2)*sin(1/2*arct 
an2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 - 2*I*sqrt(2)*cos(1/2*arctan2(s 
in(4*d*x + 4*c), cos(4*d*x + 4*c))) - I*sqrt(2))*cos(5/2*arctan2(sin(1/2*a 
rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4 
*c), cos(4*d*x + 4*c))) + 1)) + 5*(5*I*sqrt(2)*cos(4*d*x + 4*c) + 20*I*sqr 
t(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 5*sqrt(2)*sin( 
4*d*x + 4*c) + 20*sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4* 
c))) - 48*I*sqrt(2))*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos 
(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) +...
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^5\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \] Input:

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(1/2),x)
 

Output:

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(1/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{5}d x \right ) \] Input:

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x)
 

Output:

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