\(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [159]

Optimal result

Integrand size = 24, antiderivative size = 107 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15 \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {15 \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

-15/2*arctanh(sin(d*x+c))/a^4/d-15/2*sec(d*x+c)*tan(d*x+c)/a^4/d+2*I*sec(d 
*x+c)^5/a/d/(a+I*a*tan(d*x+c))^3+10*I*sec(d*x+c)^3/d/(a^4+I*a^4*tan(d*x+c) 
)
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(988\) vs. \(2(107)=214\).

Time = 6.50 (sec) , antiderivative size = 988, normalized size of antiderivative = 9.23 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx =\text {Too large to display} \] Input:

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(15*Cos[4*c]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*( 
Cos[d*x] + I*Sin[d*x])^4)/(2*d*(a + I*a*Tan[c + d*x])^4) - (15*Cos[4*c]*Lo 
g[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*(Cos[d*x] + I*Si 
n[d*x])^4)/(2*d*(a + I*a*Tan[c + d*x])^4) + (Cos[d*x]*Sec[c + d*x]^4*((8*I 
)*Cos[3*c] - 8*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d* 
x])^4) + (Sec[c]*Sec[c + d*x]^4*((4*I)*Cos[4*c] - 4*Sin[4*c])*(Cos[d*x] + 
I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d*x])^4) + (((15*I)/2)*Log[Cos[c/2 + (d 
*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*Sin[4*c]*(Cos[d*x] + I*Sin[d*x 
])^4)/(d*(a + I*a*Tan[c + d*x])^4) - (((15*I)/2)*Log[Cos[c/2 + (d*x)/2] + 
Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*Sin[4*c]*(Cos[d*x] + I*Sin[d*x])^4)/(d* 
(a + I*a*Tan[c + d*x])^4) + (Sec[c + d*x]^4*(8*Cos[3*c] + (8*I)*Sin[3*c])* 
(Cos[d*x] + I*Sin[d*x])^4*Sin[d*x])/(d*(a + I*a*Tan[c + d*x])^4) + (Sec[c 
+ d*x]^4*(Cos[4*c]/4 + (I/4)*Sin[4*c])*(Cos[d*x] + I*Sin[d*x])^4)/(d*(Cos[ 
c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d*x])^4) + (Sec[c 
+ d*x]^4*(-1/4*Cos[4*c] - (I/4)*Sin[4*c])*(Cos[d*x] + I*Sin[d*x])^4)/(d*(C 
os[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d*x])^4) + (4*S 
ec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])^4*(Cos[4*c - (d*x)/2]/2 - Cos[4*c + 
(d*x)/2]/2 + (I/2)*Sin[4*c - (d*x)/2] - (I/2)*Sin[4*c + (d*x)/2]))/(d*(Cos 
[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c 
 + d*x])^4) + (4*Sec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])^4*(-1/2*Cos[4*c...
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3981, 3042, 3981, 3042, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^7}{(a+i a \tan (c+d x))^4}dx\)

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

((2*I)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^3) - (5*(((-2*I)*Sec[c 
+ d*x]^3)/(d*(a^2 + I*a^2*Tan[c + d*x])) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) 
 + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2))/a^2
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3981

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
 f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) 
Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] 
 && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {8 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{4} d}+\frac {i \left (7 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{4} d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{4} d}\)	\(107\)
derivativedivides	\(\frac {\frac {16}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \left (\frac {1}{4}+2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {2 \left (\frac {1}{4}-2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}}{a^{4} d}\)	\(118\)
default	\(\frac {\frac {16}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \left (\frac {1}{4}+2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {2 \left (\frac {1}{4}-2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}}{a^{4} d}\)	\(118\)

Input:

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

8*I/a^4/d*exp(-I*(d*x+c))+I/d/a^4/(exp(2*I*(d*x+c))+1)^2*(7*exp(3*I*(d*x+c 
))+9*exp(I*(d*x+c)))-15/2/a^4/d*ln(exp(I*(d*x+c))+I)+15/2/a^4/d*ln(exp(I*( 
d*x+c))-I)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i}{2 \, {\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

-1/2*(15*(e^(5*I*d*x + 5*I*c) + 2*e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*l 
og(e^(I*d*x + I*c) + I) - 15*(e^(5*I*d*x + 5*I*c) + 2*e^(3*I*d*x + 3*I*c) 
+ e^(I*d*x + I*c))*log(e^(I*d*x + I*c) - I) - 30*I*e^(4*I*d*x + 4*I*c) - 5 
0*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^4*d*e^(3* 
I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))
 

Sympy [F]

\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Integral(sec(c + d*x)**7/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c 
+ d*x)**2 + 4*I*tan(c + d*x) + 1), x)/a**4
 

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. 457 vs. \(2 (95) = 190\).

Time = 0.16 (sec) , antiderivative size = 457, normalized size of antiderivative = 4.27 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {30 \, {\left (\cos \left (5 \, d x + 5 \, c\right ) + 2 \, \cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right ) + 2 i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 30 \, {\left (\cos \left (5 \, d x + 5 \, c\right ) + 2 \, \cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right ) + 2 i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (i \, \cos \left (5 \, d x + 5 \, c\right ) + 2 i \, \cos \left (3 \, d x + 3 \, c\right ) + i \, \cos \left (d x + c\right ) - \sin \left (5 \, d x + 5 \, c\right ) - 2 \, \sin \left (3 \, d x + 3 \, c\right ) - \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (-i \, \cos \left (5 \, d x + 5 \, c\right ) - 2 i \, \cos \left (3 \, d x + 3 \, c\right ) - i \, \cos \left (d x + c\right ) + \sin \left (5 \, d x + 5 \, c\right ) + 2 \, \sin \left (3 \, d x + 3 \, c\right ) + \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 60 \, \cos \left (4 \, d x + 4 \, c\right ) + 100 \, \cos \left (2 \, d x + 2 \, c\right ) + 60 i \, \sin \left (4 \, d x + 4 \, c\right ) + 100 i \, \sin \left (2 \, d x + 2 \, c\right ) + 32}{-4 \, {\left (i \, a^{4} \cos \left (5 \, d x + 5 \, c\right ) + 2 i \, a^{4} \cos \left (3 \, d x + 3 \, c\right ) + i \, a^{4} \cos \left (d x + c\right ) - a^{4} \sin \left (5 \, d x + 5 \, c\right ) - 2 \, a^{4} \sin \left (3 \, d x + 3 \, c\right ) - a^{4} \sin \left (d x + c\right )\right )} d} \] Input:

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

(30*(cos(5*d*x + 5*c) + 2*cos(3*d*x + 3*c) + cos(d*x + c) + I*sin(5*d*x + 
5*c) + 2*I*sin(3*d*x + 3*c) + I*sin(d*x + c))*arctan2(cos(d*x + c), sin(d* 
x + c) + 1) + 30*(cos(5*d*x + 5*c) + 2*cos(3*d*x + 3*c) + cos(d*x + c) + I 
*sin(5*d*x + 5*c) + 2*I*sin(3*d*x + 3*c) + I*sin(d*x + c))*arctan2(cos(d*x 
 + c), -sin(d*x + c) + 1) + 15*(I*cos(5*d*x + 5*c) + 2*I*cos(3*d*x + 3*c) 
+ I*cos(d*x + c) - sin(5*d*x + 5*c) - 2*sin(3*d*x + 3*c) - sin(d*x + c))*l 
og(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 15*(-I*cos(5*d* 
x + 5*c) - 2*I*cos(3*d*x + 3*c) - I*cos(d*x + c) + sin(5*d*x + 5*c) + 2*si 
n(3*d*x + 3*c) + sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin 
(d*x + c) + 1) + 60*cos(4*d*x + 4*c) + 100*cos(2*d*x + 2*c) + 60*I*sin(4*d 
*x + 4*c) + 100*I*sin(2*d*x + 2*c) + 32)/((-4*I*a^4*cos(5*d*x + 5*c) - 8*I 
*a^4*cos(3*d*x + 3*c) - 4*I*a^4*cos(d*x + c) + 4*a^4*sin(5*d*x + 5*c) + 8* 
a^4*sin(3*d*x + 3*c) + 4*a^4*sin(d*x + c))*d)
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} - \frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} - \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {32}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}}}{2 \, d} \] Input:

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

-1/2*(15*log(tan(1/2*d*x + 1/2*c) + 1)/a^4 - 15*log(tan(1/2*d*x + 1/2*c) - 
 1)/a^4 - 2*(tan(1/2*d*x + 1/2*c)^3 - 8*I*tan(1/2*d*x + 1/2*c)^2 + tan(1/2 
*d*x + 1/2*c) + 8*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - 32/(a^4*(tan(1 
/2*d*x + 1/2*c) - I)))/d
 

Mupad [B] (verification not implemented)

Time = 2.68 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}+\frac {\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^4}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,39{}\mathrm {i}}{a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,17{}\mathrm {i}}{a^4}+\frac {24{}\mathrm {i}}{a^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,2{}\mathrm {i}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}+1\right )} \] Input:

int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^4),x)
 

Output:

((9*tan(c/2 + (d*x)/2)^3)/a^4 - (tan(c/2 + (d*x)/2)^2*39i)/a^4 + (tan(c/2 
+ (d*x)/2)^4*17i)/a^4 + 24i/a^4 - (7*tan(c/2 + (d*x)/2))/a^4)/(d*(tan(c/2 
+ (d*x)/2)*1i - 2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*2i + tan(c/2 
 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5*1i + 1)) - (15*atanh(tan(c/2 + (d*x)/ 
2)))/(a^4*d)
 

Reduce [F]

\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {too large to display} \] Input:

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x)
 

Output:

(914048*cos(c + d*x)**2*sin(c + d*x) - 229432*cos(c + d*x)**2*i - 634368*c 
os(c + d*x)*int(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**6*i - 16*cos(c 
+ d*x)*sin(c + d*x)**4*i + 9*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x) 
*i - 8*sin(c + d*x)**7 + 20*sin(c + d*x)**5 - 16*sin(c + d*x)**3 + 4*sin(c 
 + d*x)),x)*sin(c + d*x)**2*d*i + 634368*cos(c + d*x)*int(cos(c + d*x)/(8* 
cos(c + d*x)*sin(c + d*x)**6*i - 16*cos(c + d*x)*sin(c + d*x)**4*i + 9*cos 
(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)*i - 8*sin(c + d*x)**7 + 20*sin( 
c + d*x)**5 - 16*sin(c + d*x)**3 + 4*sin(c + d*x)),x)*d*i - 68096*cos(c + 
d*x)*int(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**5*i - 12*cos(c + d*x)* 
sin(c + d*x)**3*i + 4*cos(c + d*x)*sin(c + d*x)*i - 8*sin(c + d*x)**6 + 16 
*sin(c + d*x)**4 - 9*sin(c + d*x)**2 + 1),x)*sin(c + d*x)**2*d + 68096*cos 
(c + d*x)*int(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**5*i - 12*cos(c + 
d*x)*sin(c + d*x)**3*i + 4*cos(c + d*x)*sin(c + d*x)*i - 8*sin(c + d*x)**6 
 + 16*sin(c + d*x)**4 - 9*sin(c + d*x)**2 + 1),x)*d + 68096*cos(c + d*x)*i 
nt(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**5 - 12*cos(c + d*x)*sin(c + 
d*x)**3 + 4*cos(c + d*x)*sin(c + d*x) + 8*sin(c + d*x)**6*i - 16*sin(c + d 
*x)**4*i + 9*sin(c + d*x)**2*i - i),x)*sin(c + d*x)**2*d*i - 68096*cos(c + 
 d*x)*int(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**5 - 12*cos(c + d*x)*s 
in(c + d*x)**3 + 4*cos(c + d*x)*sin(c + d*x) + 8*sin(c + d*x)**6*i - 16*si 
n(c + d*x)**4*i + 9*sin(c + d*x)**2*i - i),x)*d*i - 922112*cos(c + d*x)...