\(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [330]

Optimal result

Integrand size = 26, antiderivative size = 196 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{12 d}-\frac {i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{10 d}-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d} \] Output:

1/16*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-1/8*I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/12*I*a^2 
*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d-1/10*I*a*cos(d*x+c)^5*(a+I*a*tan( 
d*x+c))^(5/2)/d-1/7*I*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(7/2)/d
 

Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.67 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^3 e^{-i (c+d x)} \left (176+298 e^{2 i (c+d x)}+188 e^{4 i (c+d x)}+81 e^{6 i (c+d x)}+15 e^{8 i (c+d x)}-105 \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{1680 d} \] Input:

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

Output:

((-1/1680*I)*a^3*(176 + 298*E^((2*I)*(c + d*x)) + 188*E^((4*I)*(c + d*x)) 
+ 81*E^((6*I)*(c + d*x)) + 15*E^((8*I)*(c + d*x)) - 105*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^(I*(c + d*x)))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3971, 3042, 3971, 3042, 3971, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3971

\(\displaystyle \frac {1}{2} a \int \cos ^5(c+d x) (i \tan (c+d x) a+a)^{5/2}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a \int \frac {(i \tan (c+d x) a+a)^{5/2}}{\sec (c+d x)^5}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\)

\(\Big \downarrow \) 3971

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3971

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3971

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3970

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\)

Input:

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

Output:

((-1/7*I)*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(7/2))/d + (a*(((-1/5*I)*C 
os[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2))/d + (a*(((-1/3*I)*Cos[c + d*x] 
^3*(a + I*a*Tan[c + d*x])^(3/2))/d + (a*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c 
 + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d 
*x]*Sqrt[a + I*a*Tan[c + d*x]])/d))/2))/2))/2
 

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]
 

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. 1925 vs. \(2 (159 ) = 318\).

Time = 3.98 (sec) , antiderivative size = 1926, normalized size of antiderivative = 9.83

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

Input:

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

Output:

-1/1680/d*cos(d*x+c)^3*((840*cos(d*x+c)^3+420*cos(d*x+c)^2-420*cos(d*x+c)- 
105)*sin(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))*tan(d*x+c)^3+I*(840*cos(d*x+c)^4+420*cos(d*x+c)^3-840*cos(d*x+c)^2-3 
15*cos(d*x+c)+105)*(-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))*tan(d*x+c)^3+I*(2520*cos(d*x+c)^3+1260*cos(d*x+c)^2-1260*cos(d*x+ 
c)-315)*sin(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))*tan(d*x+c)+(2520*cos(d*x+c)^4+1260*co 
s(d*x+c)^3-2520*cos(d*x+c)^2-945*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))*tan(d*x+c)^2+(-2520*cos(d*x+c)^3-12 
60*cos(d*x+c)^2+1260*cos(d*x+c)+315)*sin(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))*tan(d*x+c)+I*(-840*cos(d*x+c)^3-420* 
cos(d*x+c)^2+420*cos(d*x+c)+105)*sin(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))*tan(d*x+c)^3 
+I*sin(d*x+c)^2*(3864*cos(d*x+c)^2-630)+(-840*cos(d*x+c)^4-420*cos(d*x+c)^ 
3+840*cos(d*x+c)^2+315*cos(d*x+c)-105)*(-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)*(...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.32 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{4} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, d}\right ) - 105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{4} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, d}\right ) + \sqrt {2} {\left (-15 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 81 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 188 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 298 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 176 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1680 \, d} \] Input:

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

Output:

1/1680*(105*sqrt(1/2)*sqrt(-a^7/d^2)*d*log(1/4*(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) - 105*sqrt(1/2)*sqrt(-a^7/d^2)*d*log(1/4*(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)*(-15*I*a^3*e^(8*I*d*x 
+ 8*I*c) - 81*I*a^3*e^(6*I*d*x + 6*I*c) - 188*I*a^3*e^(4*I*d*x + 4*I*c) - 
298*I*a^3*e^(2*I*d*x + 2*I*c) - 176*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 
)))/d
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**7*(a+I*a*tan(d*x+c))**(7/2),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. 1253 vs. \(2 (149) = 298\).

Time = 0.43 (sec) , antiderivative size = 1253, normalized size of antiderivative = 6.39 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/6720*(20*(7*I*sqrt(2)*a^3*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c) + 1)) - 7*sqrt(2)*a^3*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
2*c) + 1)) + 3*(I*sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + I*sqrt(2)*a^3*sin(2*d*x 
 + 2*c)^2 + 2*I*sqrt(2)*a^3*cos(2*d*x + 2*c) + I*sqrt(2)*a^3)*cos(7/2*arct 
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 3*(sqrt(2)*a^3*cos(2*d*x + 
2*c)^2 + sqrt(2)*a^3*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*a^3*cos(2*d*x + 2*c) + 
 sqrt(2)*a^3)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(c 
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqr 
t(a) + 84*(5*I*sqrt(2)*a^3*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 
*c) + 1)) - 5*sqrt(2)*a^3*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* 
c) + 1)) + (I*sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + I*sqrt(2)*a^3*sin(2*d*x + 2 
*c)^2 + 2*I*sqrt(2)*a^3*cos(2*d*x + 2*c) + I*sqrt(2)*a^3)*cos(5/2*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (sqrt(2)*a^3*cos(2*d*x + 2*c)^2 
 + sqrt(2)*a^3*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*a^3*cos(2*d*x + 2*c) + sqrt( 
2)*a^3)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d 
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 
 105*(2*sqrt(2)*a^3*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c 
os(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 
*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1)...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^7*(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)**7*tan(c + d*x) 
**3,x)*i - 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**7*tan(c + d*x)**2, 
x) + 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**7*tan(c + d*x),x)*i + in 
t(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**7,x))