Integrand size = 26, antiderivative size = 266 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {231 i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}} \]
-231/1024*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^(1/2)/ d*2^(1/2)+231/512*I*a/d/(a+I*a*tan(d*x+c))^(1/2)+231/640*I*a^3/d/(a+I*a*ta n(d*x+c))^(5/2)-1/6*I*a^6/d/(a-I*a*tan(d*x+c))^3/(a+I*a*tan(d*x+c))^(5/2)- 11/48*I*a^5/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(5/2)-33/64*I*a^4/d/ (a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2)+77/256*I*a^2/d/(a+I*a*tan(d*x+ c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.20 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},4,-\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{40 d (a+i a \tan (c+d x))^{5/2}} \]
((I/40)*a^3*Hypergeometric2F1[-5/2, 4, -3/2, (1 + I*Tan[c + d*x])/2])/(d*( a + I*a*Tan[c + d*x])^(5/2))
Time = 0.35 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3968, 52, 52, 52, 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.
| \(\displaystyle \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sec (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i a^7 \int \frac {1}{(a-i a \tan (c+d x))^4 (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {i a^7 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{12 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
((-I)*a^7*(1/(6*a*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(5/2)) + (11*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(5/2)) + (9*( 1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (7*(-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*Sq rt[a + I*a*Tan[c + d*x]]))/(2*a))/(2*a)))/(4*a)))/(8*a)))/(12*a)))/d
3.3.85.3.1 Defintions of rubi rules used
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]
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]
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]
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])
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (214 ) = 428\).
Time = 108.03 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.65
| method | result | size |
| default | \(-\frac {i \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2816 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-256 \left (\cos ^{6}\left (d x +c \right )\right )+3696 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sin \left (d x +c \right )-528 \left (\cos ^{4}\left (d x +c \right )\right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+6930 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )-3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )-3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-2310 \left (\cos ^{2}\left (d x +c \right )\right )\right )}{15360 d}\) | \(438\) |
-1/15360*I/d*(a*(1+I*tan(d*x+c)))^(1/2)*(2816*I*cos(d*x+c)^5*sin(d*x+c)-25 6*cos(d*x+c)^6+3696*I*cos(d*x+c)^3*sin(d*x+c)+3465*I*(-cos(d*x+c)/(cos(d*x +c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2))*cos(d*x+c)+3465*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-co s(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-528*cos(d*x+c)^4+3465*I*(-cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c) /(cos(d*x+c)+1))^(1/2))+6930*I*cos(d*x+c)*sin(d*x+c)+3465*(-cos(d*x+c)/(co s(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x +c)+1))^(1/2))*sin(d*x+c)-3465*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3465*(-cos(d*x+c)/(cos(d*x+c )+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2310*cos(d*x+c)^2)
Time = 0.25 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.12 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {{\left (3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-40 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 350 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1645 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1433 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3184 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 464 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{15360 \, d} \]
-1/15360*(3465*sqrt(1/2)*d*sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2 )*sqrt(1/2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3465*sqrt(1/2)*d *sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*d*e^(2*I*d *x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-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)) *(-40*I*e^(12*I*d*x + 12*I*c) - 350*I*e^(10*I*d*x + 10*I*c) - 1645*I*e^(8* I*d*x + 8*I*c) + 1433*I*e^(6*I*d*x + 6*I*c) + 3184*I*e^(4*I*d*x + 4*I*c) + 464*I*e^(2*I*d*x + 2*I*c) + 48*I))*e^(-5*I*d*x - 5*I*c)/d
\[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cos ^{6}{\left (c + d x \right )}\, dx \]
Time = 0.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.86 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (3465 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 18480 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 30492 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 12672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 2816 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 1536 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3}}\right )}}{30720 \, a d} \]
1/30720*I*(3465*sqrt(2)*a^(3/2)*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))) + 4*(3465*(I*a*t an(d*x + c) + a)^5*a^2 - 18480*(I*a*tan(d*x + c) + a)^4*a^3 + 30492*(I*a*t an(d*x + c) + a)^3*a^4 - 12672*(I*a*tan(d*x + c) + a)^2*a^5 - 2816*(I*a*ta n(d*x + c) + a)*a^6 - 1536*a^7)/((I*a*tan(d*x + c) + a)^(11/2) - 6*(I*a*ta n(d*x + c) + a)^(9/2)*a + 12*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 8*(I*a*tan (d*x + c) + a)^(5/2)*a^3))/(a*d)
\[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{6} \,d x } \]
Timed out. \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]