\(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [331]

Optimal result

Integrand size = 26, antiderivative size = 268 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {11 i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} d}+\frac {11 i a^4 \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}-\frac {11 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 d}-\frac {11 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{120 d}-\frac {11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac {11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d} \] Output:

11/128*I*a^(7/2)*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+11/96*I*a^4*cos(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-11/64* 
I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-11/120*I*a^3*cos(d*x+c)^3*(a+I 
*a*tan(d*x+c))^(1/2)/d-11/140*I*a^2*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2)/ 
d-11/126*I*a*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(5/2)/d-1/9*I*cos(d*x+c)^9*(a 
+I*a*tan(d*x+c))^(7/2)/d
 

Mathematica [A] (verified)

Time = 3.55 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.70 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^3 e^{-3 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-315+4303 e^{2 i (c+d x)}+7034 e^{4 i (c+d x)}+3754 e^{6 i (c+d x)}+1798 e^{8 i (c+d x)}+530 e^{10 i (c+d x)}+70 e^{12 i (c+d x)}-3465 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{20160 \sqrt {2} d} \] Input:

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((-1/20160*I)*a^3*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]* 
(-315 + 4303*E^((2*I)*(c + d*x)) + 7034*E^((4*I)*(c + d*x)) + 3754*E^((6*I 
)*(c + d*x)) + 1798*E^((8*I)*(c + d*x)) + 530*E^((10*I)*(c + d*x)) + 70*E^ 
((12*I)*(c + d*x)) - 3465*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x)) 
]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*E^((3*I)*(c + d*x)))
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3042, 3978, 3042, 3978, 3042, 3978, 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.

\(\displaystyle \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

\(\displaystyle \frac {11}{18} a \int \cos ^7(c+d x) (i \tan (c+d x) a+a)^{5/2}dx-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \int \frac {(i \tan (c+d x) a+a)^{5/2}}{\sec (c+d x)^7}dx-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \int \cos ^5(c+d x) (i \tan (c+d x) a+a)^{3/2}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\sec (c+d x)^5}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3971

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {11}{18} a \left (\frac {9}{14} a \left (\frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\right )-\frac {i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\)

Input:

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((-1/9*I)*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^(7/2))/d + (11*a*(((-1/7*I 
)*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2))/d + (9*a*(((-1/5*I)*Cos[c + 
 d*x]^5*(a + I*a*Tan[c + d*x])^(3/2))/d + (7*a*(((-1/3*I)*Cos[c + d*x]^3*S 
qrt[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]*Sq 
rt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x]*Sqrt[a + I*a*Tan 
[c + d*x]])/d))/(4*a)))/6))/10))/14))/18
 

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 [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2004 vs. \(2 (219 ) = 438\).

Time = 6.88 (sec) , antiderivative size = 2005, normalized size of antiderivative = 7.48

\[\text {Expression too large to display}\]

Input:

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

-1/40320/d*cos(d*x+c)^3*((27720*cos(d*x+c)^3+13860*cos(d*x+c)^2-13860*cos( 
d*x+c)-3465)*sin(d*x+c)*tan(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar 
ctanh(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*(83160*cos(d*x+c)^3+41580*cos(d*x+c)^2-41580* 
cos(d*x+c)-10395)*sin(d*x+c)*tan(d*x+c)*(-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))+I*cos(d*x+c)^2*( 
-7840*cos(d*x+c)^4+50424*cos(d*x+c)^2-25410)+(83160*cos(d*x+c)^4+41580*cos 
(d*x+c)^3-83160*cos(d*x+c)^2-31185*cos(d*x+c)+10395)*tan(d*x+c)^2*(-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))+(-83160*cos(d*x+c) 
^3-41580*cos(d*x+c)^2+41580*cos(d*x+c)+10395)*sin(d*x+c)*tan(d*x+c)*(-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*sin(d*x+c)^2*( 
23520*cos(d*x+c)^4-151272*cos(d*x+c)^2+76230)+I*sin(d*x+c)^2*(-36960*cos(d 
*x+c)^4+127512*cos(d*x+c)^2-20790)+(-27720*cos(d*x+c)^4-13860*cos(d*x+c)^3 
+27720*cos(d*x+c)^2+10395*cos(d*x+c)-3465)*(-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*(27720*cos(d*x+c)^4+13860*cos(d*x+c)^3- 
27720*cos(d*x+c)^2-10395*cos(d*x+c)+3465)*tan(d*x+c)^3*(-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...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.17 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {{\left (3465 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {11 \, {\left (-i \, a^{4} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\right ) - 3465 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {11 \, {\left (-i \, a^{4} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\right ) - \sqrt {2} {\left (-70 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 530 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 1798 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 3754 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 7034 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4303 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 315 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{40320 \, d} \] Input:

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-1/40320*(3465*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c)*log(-11/32*( 
-I*a^4 + sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt 
(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 3465*sqrt(1/2)*sqrt(- 
a^7/d^2)*d*e^(2*I*d*x + 2*I*c)*log(-11/32*(-I*a^4 - sqrt(2)*sqrt(1/2)*sqrt 
(-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))* 
e^(-I*d*x - I*c)/d) - sqrt(2)*(-70*I*a^3*e^(12*I*d*x + 12*I*c) - 530*I*a^3 
*e^(10*I*d*x + 10*I*c) - 1798*I*a^3*e^(8*I*d*x + 8*I*c) - 3754*I*a^3*e^(6* 
I*d*x + 6*I*c) - 7034*I*a^3*e^(4*I*d*x + 4*I*c) - 4303*I*a^3*e^(2*I*d*x + 
2*I*c) + 315*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c 
)/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

Timed out
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),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. \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \] Input:

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(7/2),x)
 

Output:

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(7/2), x)
 

Reduce [F]

\[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{9} \tan \left (d x +c \right )^{3}d x \right ) i -3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{9} \tan \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{9} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{9}d x \right ) \] Input:

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

sqrt(a)*a**3*( - int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**9*tan(c + d*x) 
**3,x)*i - 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**9*tan(c + d*x)**2, 
x) + 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**9*tan(c + d*x),x)*i + in 
t(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**9,x))