\(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [390]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \sec ^3(c+d x) \left (-3+e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-3 \cos (2 (c+d x))+i \sin (2 (c+d x))\right )}{16 a^3 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((I/16)*Sec[c + d*x]^3*(-3 + E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d* 
x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - 3*Cos[2*(c + d*x)] + I*Sin[2 
*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.36, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3982, 3042, 3983, 3042, 3983, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^3}{(a+i a \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

\(\displaystyle \frac {2 i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \left (\frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\right )}{3 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \left (\frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\right )}{3 a}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {2 i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \left (\frac {3 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\right )}{3 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \left (\frac {3 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\right )}{3 a}\)

\(\Big \downarrow \) 3970

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \left (\frac {3 \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\right )}{3 a}\)

Input:

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

(((2*I)/3)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) - (2*(((I/4)*S 
ec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (3*(((I/2)*ArcTanh[(Sqrt[a 
]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) 
 + ((I/2)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(3/2))))/(8*a)))/(3*a)
 

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 3982

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) 
 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] && LtQ[n, 0] && GtQ[m, 1] &&  !IL 
tQ[m + n, 0] && NeQ[m + n - 1, 0] && 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. 468 vs. \(2 (100 ) = 200\).

Time = 8.44 (sec) , antiderivative size = 469, normalized size of antiderivative = 3.75

method	result	size
default	\(\frac {\tan \left (d x +c \right ) \sec \left (d x +c \right )^{2} \left (-1+8 \cos \left (d x +c \right )^{3}+4 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (-4-8 \cos \left (d x +c \right )+8 \sec \left (d x +c \right )+3 \sec \left (d x +c \right )^{2}-\sec \left (d x +c \right )^{3}\right )+i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (8 \sin \left (d x +c \right )-4 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (8 \cos \left (d x +c \right )-8 \sec \left (d x +c \right )+\sec \left (d x +c \right )^{3}\right )+i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-16 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+12 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-16 \cos \left (d x +c \right )+4+20 \sec \left (d x +c \right )+2 \sec \left (d x +c \right )^{2}\right )}{16 d \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3} \left (\cos \left (d x +c \right )+1\right )}\)	\(469\)

Input:

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/d/(-tan(d*x+c)+I)^3/(a*(1+I*tan(d*x+c)))^(1/2)/(-cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)/a^3/(cos(d*x+c)+1)*(tan(d*x+c)*sec(d*x+c)^2*(-1+8*cos(d*x+c)^ 
3+4*cos(d*x+c)^2-4*cos(d*x+c))*arctanh(1/2/(cot(d*x+c)^2-2*cot(d*x+c)*csc( 
d*x+c)+csc(d*x+c)^2-1)^(1/2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2))+I*arctanh( 
1/2/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(I-cot(d*x 
+c)+csc(d*x+c))*2^(1/2))*(-4-8*cos(d*x+c)+8*sec(d*x+c)+3*sec(d*x+c)^2-sec( 
d*x+c)^3)+I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(8*sin(d*x+c)-4*s 
ec(d*x+c)*tan(d*x+c))+2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(8*cos( 
d*x+c)-8*sec(d*x+c)+sec(d*x+c)^3)+I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-1 
6*sin(d*x+c)+4*tan(d*x+c)+12*sec(d*x+c)*tan(d*x+c))+(-cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*(-16*cos(d*x+c)+4+20*sec(d*x+c)+2*sec(d*x+c)^2))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (94) = 188\).

Time = 0.08 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{3} d}\right ) + i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{4} d} \] Input:

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/16*(-I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/4*(s 
qrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d* 
x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + I)*e^(-I*d*x - I*c)/(a^3*d)) + I*sqrt 
(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/4*(sqrt(2)*sqrt(1 
/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 
 1))*sqrt(1/(a^7*d^2)) + I)*e^(-I*d*x - I*c)/(a^3*d)) + sqrt(2)*sqrt(a/(e^ 
(2*I*d*x + 2*I*c) + 1))*(I*e^(4*I*d*x + 4*I*c) + 3*I*e^(2*I*d*x + 2*I*c) + 
 2*I))*e^(-4*I*d*x - 4*I*c)/(a^4*d)
 

Sympy [F]

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

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(7/2),x)
 

Output:

Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(7/2), x)
 

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. 977 vs. \(2 (94) = 188\).

Time = 0.30 (sec) , antiderivative size = 977, normalized size of antiderivative = 7.82 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

-1/64*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1 
)^(3/4)*((-I*sqrt(2)*cos(4*d*x + 4*c) - sqrt(2)*sin(4*d*x + 4*c))*cos(3/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(4*d*x + 4* 
c) - I*sqrt(2)*sin(4*d*x + 4*c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c) + 1)))*sqrt(a) + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2* 
cos(2*d*x + 2*c) + 1)^(1/4)*((-I*sqrt(2)*cos(4*d*x + 4*c) - sqrt(2)*sin(4* 
d*x + 4*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sq 
rt(2)*cos(4*d*x + 4*c) - I*sqrt(2)*sin(4*d*x + 4*c))*sin(1/2*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - (2*sqrt(2)*arctan2((cos(2*d 
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*a 
rctan2(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) - 2*sqrt(2)*arctan2((cos(2*d*x + 2*c)^ 
2 + sin(2*d*x + 2*c)^2 + 2*cos(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), co 
s(2*d*x + 2*c) + 1)) - 1) - I*sqrt(2)*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2* 
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2 
*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^3/(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. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \] Input:

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(7/2)),x)
 

Output:

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(7/2)), x)
 

Reduce [F]

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

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

( - int(sec(c + d*x)**3/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3*i + 3*sq 
rt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 3*sqrt(tan(c + d*x)*i + 1)*tan(c 
+ d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt(a)*a**3)