\(\int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [339]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 300 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {429 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} \sqrt {a} d}+\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}}-\frac {13 i a^7}{48 d (a+i a \tan (c+d x))^{7/2} \left (a^2-i a^2 \tan (c+d x)\right )^2}-\frac {143 i a^7}{192 d (a+i a \tan (c+d x))^{7/2} \left (a^4-i a^4 \tan (c+d x)\right )} \] Output:

-429/2048*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/ 
a^(1/2)/d+429/896*I*a^3/d/(a+I*a*tan(d*x+c))^(7/2)-1/6*I*a^6/d/(a-I*a*tan( 
d*x+c))^3/(a+I*a*tan(d*x+c))^(7/2)+429/1280*I*a^2/d/(a+I*a*tan(d*x+c))^(5/ 
2)+143/512*I*a/d/(a+I*a*tan(d*x+c))^(3/2)+429/1024*I/d/(a+I*a*tan(d*x+c))^ 
(1/2)-13/48*I*a^7/d/(a+I*a*tan(d*x+c))^(7/2)/(a^2-I*a^2*tan(d*x+c))^2-143/ 
192*I*a^7/d/(a+I*a*tan(d*x+c))^(7/2)/(a^4-I*a^4*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.18 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a^3 \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},4,-\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{56 d (a+i a \tan (c+d x))^{7/2}} \] Input:

Integrate[Cos[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

((I/56)*a^3*Hypergeometric2F1[-7/2, 4, -5/2, (1 + I*Tan[c + d*x])/2])/(d*( 
a + I*a*Tan[c + d*x])^(7/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3968, 52, 52, 52, 61, 61, 61, 61, 73, 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.

\(\displaystyle \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^6 \sqrt {a+i a \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i a^7 \int \frac {1}{(a-i a \tan (c+d x))^4 (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {13 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\frac {\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {i a^7 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {\frac {\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}\right )}{d}\)

Input:

Int[Cos[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

((-I)*a^7*(1/(6*a*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(7/2)) + 
 (13*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(7/2)) + (11* 
(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(7/2)) + (9*(-1/7*1/ 
(a*(a + I*a*Tan[c + d*x])^(7/2)) + (-1/5*1/(a*(a + I*a*Tan[c + d*x])^(5/2) 
) + (-1/3*1/(a*(a + I*a*Tan[c + d*x])^(3/2)) + ((I*ArcTan[(Sqrt[a]*Tan[c + 
 d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + I*a*Tan[c + d*x]]))/(2* 
a))/(2*a))/(2*a)))/(4*a)))/(8*a)))/(12*a)))/d
 

Defintions of rubi rules used

rule 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 3968
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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]
 
Maple [A] (verified)

Time = 7.72 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.17

method result size
default \(\frac {45045 i \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+45045 \left (\cos \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+45045 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+52 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (1155+640 \cos \left (d x +c \right )^{5}+640 \cos \left (d x +c \right )^{4}+792 \cos \left (d x +c \right )^{3}+792 \cos \left (d x +c \right )^{2}+1155 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 i \cos \left (d x +c \right ) \left (45045-1280 \cos \left (d x +c \right )^{6}-1280 \cos \left (d x +c \right )^{5}-2288 \cos \left (d x +c \right )^{4}-2288 \cos \left (d x +c \right )^{3}-6006 \cos \left (d x +c \right )^{2}-6006 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{215040 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) \(350\)

Input:

int(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/215040/d*(45045*I*sin(d*x+c)*arctanh(1/2/(-cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)*(cot(d*x+c)-csc(d*x+c)+I))+45045*(cos(d*x+c)+1)*arctanh(1/2/(-cos(d*x+ 
c)/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))+45045*I*2^(1/2)*(-2*co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)+52*sin(d*x+c)*cos(d*x+c)*(1155+640*cos(d*x+ 
c)^5+640*cos(d*x+c)^4+792*cos(d*x+c)^3+792*cos(d*x+c)^2+1155*cos(d*x+c))*( 
-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*cos(d*x+c)*(45045-1280*cos(d*x+c)^6- 
1280*cos(d*x+c)^5-2288*cos(d*x+c)^4-2288*cos(d*x+c)^3-6006*cos(d*x+c)^2-60 
06*cos(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(cos(d*x+c)+1)/(-cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-280 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 2870 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 16345 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 27029 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 49792 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11072 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2304 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 240 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{215040 \, a d} \] Input:

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/215040*(-45045*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(7*I*d*x + 7*I*c)*log(4 
*(sqrt(2)*sqrt(1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2 
*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 45045 
*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(7*I*d*x + 7*I*c)*log(-4*(sqrt(2)*sqrt( 
1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t(1/(a*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2 
*I*d*x + 2*I*c) + 1))*(-280*I*e^(14*I*d*x + 14*I*c) - 2870*I*e^(12*I*d*x + 
 12*I*c) - 16345*I*e^(10*I*d*x + 10*I*c) + 27029*I*e^(8*I*d*x + 8*I*c) + 4 
9792*I*e^(6*I*d*x + 6*I*c) + 11072*I*e^(4*I*d*x + 4*I*c) + 2304*I*e^(2*I*d 
*x + 2*I*c) + 240*I))*e^(-7*I*d*x - 7*I*c)/(a*d)
 

Sympy [F]

\[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{6}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:

integrate(cos(d*x+c)**6/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(cos(c + d*x)**6/sqrt(I*a*(tan(c + d*x) - I)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (45045 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a - 240240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 396396 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 164736 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 36608 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 19968 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 15360 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3}}\right )}}{430080 \, a d} \] Input:

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/430080*I*(45045*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x 
 + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*(45045*(I* 
a*tan(d*x + c) + a)^6*a - 240240*(I*a*tan(d*x + c) + a)^5*a^2 + 396396*(I* 
a*tan(d*x + c) + a)^4*a^3 - 164736*(I*a*tan(d*x + c) + a)^3*a^4 - 36608*(I 
*a*tan(d*x + c) + a)^2*a^5 - 19968*(I*a*tan(d*x + c) + a)*a^6 - 15360*a^7) 
/((I*a*tan(d*x + c) + a)^(13/2) - 6*(I*a*tan(d*x + c) + a)^(11/2)*a + 12*( 
I*a*tan(d*x + c) + a)^(9/2)*a^2 - 8*(I*a*tan(d*x + c) + a)^(7/2)*a^3))/(a* 
d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^6/(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. \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^6}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6}}{\tan \left (d x +c \right )^{2}+1}d x \right )}{a} \] Input:

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