\(\int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [353]

Optimal result

Integrand size = 26, antiderivative size = 329 \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {715 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2048 \sqrt {2} a^{3/2} d}+\frac {715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}+\frac {715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac {715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac {715 i}{2048 a d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^7}{16 d (a+i a \tan (c+d x))^{9/2} \left (a^2-i a^2 \tan (c+d x)\right )^2}-\frac {65 i a^7}{64 d (a+i a \tan (c+d x))^{9/2} \left (a^4-i a^4 \tan (c+d x)\right )} \] Output:

-715/4096*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/ 
a^(3/2)/d+715/1152*I*a^3/d/(a+I*a*tan(d*x+c))^(9/2)-1/6*I*a^6/d/(a-I*a*tan 
(d*x+c))^3/(a+I*a*tan(d*x+c))^(9/2)+715/1792*I*a^2/d/(a+I*a*tan(d*x+c))^(7 
/2)+143/512*I*a/d/(a+I*a*tan(d*x+c))^(5/2)+715/3072*I/d/(a+I*a*tan(d*x+c)) 
^(3/2)+715/2048*I/a/d/(a+I*a*tan(d*x+c))^(1/2)-5/16*I*a^7/d/(a+I*a*tan(d*x 
+c))^(9/2)/(a^2-I*a^2*tan(d*x+c))^2-65/64*I*a^7/d/(a+I*a*tan(d*x+c))^(9/2) 
/(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 = 0.47 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.16 \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i a^3 \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},4,-\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{72 d (a+i a \tan (c+d x))^{9/2}} \] Input:

Integrate[Cos[c + d*x]^6/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((I/72)*a^3*Hypergeometric2F1[-9/2, 4, -7/2, (1 + I*Tan[c + d*x])/2])/(d*( 
a + I*a*Tan[c + d*x])^(9/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3968, 52, 52, 52, 61, 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.

\(\displaystyle \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^6 (a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i a^7 \int \frac {1}{(a-i a \tan (c+d x))^4 (i \tan (c+d x) a+a)^{11/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

Int[Cos[c + d*x]^6/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-I)*a^7*(1/(6*a*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(9/2)) + 
 (5*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(9/2)) + (13*( 
1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + (11*(-1/9*1/ 
(a*(a + I*a*Tan[c + d*x])^(9/2)) + (-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))/(2*a)))/(4*a) 
))/(8*a)))/(4*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 = 9.95 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.40

method	result	size
default	\(\frac {\left (-90090 \cos \left (d x +c \right )-45045\right ) \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+i \left (90090 \cos \left (d x +c \right )^{2}+45045 \cos \left (d x +c \right )-45045\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+45045 i \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+45045 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-35840 \cos \left (d x +c \right )^{6}-35840 \cos \left (d x +c \right )^{5}-45760 \cos \left (d x +c \right )^{4}-45760 \cos \left (d x +c \right )^{3}-72072 \cos \left (d x +c \right )^{2}-72072 \cos \left (d x +c \right )+90090\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\cos \left (d x +c \right ) \left (-7168 \cos \left (d x +c \right )^{7}-7168 \cos \left (d x +c \right )^{6}-12480 \cos \left (d x +c \right )^{5}-12480 \cos \left (d x +c \right )^{4}-30888 \cos \left (d x +c \right )^{3}-30888 \cos \left (d x +c \right )^{2}+150150 \cos \left (d x +c \right )+60060\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{258048 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-i \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right )-i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) a}\)	\(462\)

Input:

int(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/258048/d*((-90090*cos(d*x+c)-45045)*sin(d*x+c)*arctanh(1/2*(-cot(d*x+c)+ 
csc(d*x+c)-I)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+I*(90090*cos(d*x+c)^2+45 
045*cos(d*x+c)-45045)*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-I)/(-cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2))+45045*I*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)+45045*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2) 
+I*sin(d*x+c)*cos(d*x+c)*(-35840*cos(d*x+c)^6-35840*cos(d*x+c)^5-45760*cos 
(d*x+c)^4-45760*cos(d*x+c)^3-72072*cos(d*x+c)^2-72072*cos(d*x+c)+90090)*(- 
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)*(-7168*cos(d*x+c)^7-7168*cos(d 
*x+c)^6-12480*cos(d*x+c)^5-12480*cos(d*x+c)^4-30888*cos(d*x+c)^3-30888*cos 
(d*x+c)^2+150150*cos(d*x+c)+60060)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(a* 
(1+I*tan(d*x+c)))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(-I*cos(d*x+c)^ 
2+cos(d*x+c)*sin(d*x+c)-I*cos(d*x+c)+sin(d*x+c))/a
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-168 i \, e^{\left (16 i \, d x + 16 i \, c\right )} - 1974 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 13209 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 33301 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 57632 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 17344 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 5440 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1136 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 112 i\right )}\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{258048 \, a^{2} d} \] Input:

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/258048*(-45045*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(9*I*d*x + 9*I*c)*l 
og(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I 
*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c 
)) + 45045*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(9*I*d*x + 9*I*c)*log(-4* 
(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*sqrt(1/(a^3*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))*(-168*I*e^(16*I*d*x + 16*I*c) - 
1974*I*e^(14*I*d*x + 14*I*c) - 13209*I*e^(12*I*d*x + 12*I*c) + 33301*I*e^( 
10*I*d*x + 10*I*c) + 57632*I*e^(8*I*d*x + 8*I*c) + 17344*I*e^(6*I*d*x + 6* 
I*c) + 5440*I*e^(4*I*d*x + 4*I*c) + 1136*I*e^(2*I*d*x + 2*I*c) + 112*I))*e 
^(-9*I*d*x - 9*I*c)/(a^2*d)
 

Sympy [F]

\[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\cos ^{6}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cos(d*x+c)**6/(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral(cos(c + d*x)**6/(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \, {\left (\frac {45045 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} - 240240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 396396 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 164736 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 36608 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 19968 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 15360 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 14336 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}}\right )}}{516096 \, a d} \] Input:

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

1/516096*I*(45045*sqrt(2)*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)))/sqrt(a) + 4*(45045*(I* 
a*tan(d*x + c) + a)^7 - 240240*(I*a*tan(d*x + c) + a)^6*a + 396396*(I*a*ta 
n(d*x + c) + a)^5*a^2 - 164736*(I*a*tan(d*x + c) + a)^4*a^3 - 36608*(I*a*t 
an(d*x + c) + a)^3*a^4 - 19968*(I*a*tan(d*x + c) + a)^2*a^5 - 15360*(I*a*t 
an(d*x + c) + a)*a^6 - 14336*a^7)/((I*a*tan(d*x + c) + a)^(15/2) - 6*(I*a* 
tan(d*x + c) + a)^(13/2)*a + 12*(I*a*tan(d*x + c) + a)^(11/2)*a^2 - 8*(I*a 
*tan(d*x + c) + a)^(9/2)*a^3))/(a*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),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. \[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^6}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:

int(cos(c + d*x)^6/(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int(cos(c + d*x)^6/(a + a*tan(c + d*x)*1i)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{6}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i +\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a} \] Input:

int(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

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