\(\int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [176]

Optimal result

Integrand size = 24, antiderivative size = 205 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {1155 \text {arctanh}(\sin (c+d x))}{8 a^8 d}+\frac {1155 \sec (c+d x) \tan (c+d x)}{8 a^8 d}+\frac {385 \sec ^3(c+d x) \tan (c+d x)}{4 a^8 d}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )} \] Output:

1155/8*arctanh(sin(d*x+c))/a^8/d+1155/8*sec(d*x+c)*tan(d*x+c)/a^8/d+385/4* 
sec(d*x+c)^3*tan(d*x+c)/a^8/d+2/3*I*sec(d*x+c)^11/a/d/(a+I*a*tan(d*x+c))^7 
-22/3*I*sec(d*x+c)^9/a^3/d/(a+I*a*tan(d*x+c))^5-66*I*sec(d*x+c)^7/a^2/d/(a 
^2+I*a^2*tan(d*x+c))^3-154*I*sec(d*x+c)^5/d/(a^8+I*a^8*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. \(1704\) vs. \(2(205)=410\).

Time = 7.19 (sec) , antiderivative size = 1704, normalized size of antiderivative = 8.31 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx =\text {Too large to display} \] Input:

Integrate[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

(-1155*Cos[8*c]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^ 
8*(Cos[d*x] + I*Sin[d*x])^8)/(8*d*(a + I*a*Tan[c + d*x])^8) + (1155*Cos[8* 
c]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x] + 
 I*Sin[d*x])^8)/(8*d*(a + I*a*Tan[c + d*x])^8) + (Cos[3*d*x]*Sec[c + d*x]^ 
8*(((32*I)/3)*Cos[5*c] - (32*Sin[5*c])/3)*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a 
 + I*a*Tan[c + d*x])^8) + (Cos[d*x]*Sec[c + d*x]^8*((-160*I)*Cos[7*c] + 16 
0*Sin[7*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) - (((1 
155*I)/8)*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[ 
8*c]*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (((1155*I)/ 
8)*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[8*c]*(C 
os[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c]*Sec[c + d* 
x]^8*(((-236*I)/3)*Cos[8*c] + (236*Sin[8*c])/3)*(Cos[d*x] + I*Sin[d*x])^8) 
/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(-160*Cos[7*c] - (160*I)*S 
in[7*c])*(Cos[d*x] + I*Sin[d*x])^8*Sin[d*x])/(d*(a + I*a*Tan[c + d*x])^8) 
+ (Sec[c + d*x]^8*((32*Cos[5*c])/3 + ((32*I)/3)*Sin[5*c])*(Cos[d*x] + I*Si 
n[d*x])^8*Sin[3*d*x])/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(Cos[ 
8*c]/16 + (I/16)*Sin[8*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2 + (d*x)/ 
2] - Sin[c/2 + (d*x)/2])^4*(a + I*a*Tan[c + d*x])^8) - ((1/96 + I/96)*Sec[ 
c + d*x]^8*((-407*I)*Cos[(15*c)/2] + 343*Cos[(17*c)/2] + 407*Sin[(15*c)/2] 
 + (343*I)*Sin[(17*c)/2])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2] - Sin...
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 3981, 3042, 3981, 3042, 3981, 3042, 3981, 3042, 4255, 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.

\(\displaystyle \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{13}}{(a+i a \tan (c+d x))^8}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^6}dx}{3 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^6}dx}{3 a^2}\)

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \left (\frac {2 i \sec ^9(c+d x)}{a d (a+i a \tan (c+d x))^5}-\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{3 a^2}-\frac {2 i \sec ^5(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}-\frac {2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}\right )}{a^2}\right )}{3 a^2}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \left (\frac {2 i \sec ^9(c+d x)}{a d (a+i a \tan (c+d x))^5}-\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{3 a^2}-\frac {2 i \sec ^5(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}-\frac {2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}\right )}{a^2}\right )}{3 a^2}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \left (\frac {2 i \sec ^9(c+d x)}{a d (a+i a \tan (c+d x))^5}-\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{3 a^2}-\frac {2 i \sec ^5(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}-\frac {2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}\right )}{a^2}\right )}{3 a^2}\)

Input:

Int[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

(((2*I)/3)*Sec[c + d*x]^11)/(a*d*(a + I*a*Tan[c + d*x])^7) - (11*(((2*I)*S 
ec[c + d*x]^9)/(a*d*(a + I*a*Tan[c + d*x])^5) - (9*(((-2*I)*Sec[c + d*x]^7 
)/(a*d*(a + I*a*Tan[c + d*x])^3) + (7*((((-2*I)/3)*Sec[c + d*x]^5)/(d*(a^2 
 + I*a^2*Tan[c + d*x])) + (5*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(Ar 
cTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/(3*a^2 
)))/a^2))/a^2))/(3*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 = 1.23 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {160 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{8} d}+\frac {32 i {\mathrm e}^{-3 i \left (d x +c \right )}}{3 a^{8} d}-\frac {i \left (1545 \,{\mathrm e}^{7 i \left (d x +c \right )}+5153 \,{\mathrm e}^{5 i \left (d x +c \right )}+5855 \,{\mathrm e}^{3 i \left (d x +c \right )}+2295 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{12 d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {1155 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a^{8} d}+\frac {1155 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 a^{8} d}\)	\(147\)
derivativedivides	\(\frac {\frac {2 \left (\frac {1}{4}-\frac {4 i}{3}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (-\frac {121}{16}-2 i\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {123}{16}+38 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {2 \left (\frac {121}{16}-2 i\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\frac {1}{4}+\frac {4 i}{3}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (-\frac {123}{16}-38 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {256}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {256}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{8} d}\)	\(219\)
default	\(\frac {\frac {2 \left (\frac {1}{4}-\frac {4 i}{3}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (-\frac {121}{16}-2 i\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {123}{16}+38 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {2 \left (\frac {121}{16}-2 i\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\frac {1}{4}+\frac {4 i}{3}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (-\frac {123}{16}-38 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {256}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {256}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{8} d}\)	\(219\)

Input:

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)
 

Output:

-160*I/a^8/d*exp(-I*(d*x+c))+32/3*I/a^8/d*exp(-3*I*(d*x+c))-1/12*I/d/a^8/( 
exp(2*I*(d*x+c))+1)^4*(1545*exp(7*I*(d*x+c))+5153*exp(5*I*(d*x+c))+5855*ex 
p(3*I*(d*x+c))+2295*exp(I*(d*x+c)))-1155/8/a^8/d*ln(exp(I*(d*x+c))-I)+1155 
/8/a^8/d*ln(exp(I*(d*x+c))+I)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {3465 \, {\left (e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3465 \, {\left (e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6930 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 25410 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 33726 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 18414 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2816 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 256 i}{24 \, {\left (a^{8} d e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, a^{8} d e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, a^{8} d e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, a^{8} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{8} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
 

Output:

1/24*(3465*(e^(11*I*d*x + 11*I*c) + 4*e^(9*I*d*x + 9*I*c) + 6*e^(7*I*d*x + 
 7*I*c) + 4*e^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3*I*c))*log(e^(I*d*x + I*c) 
 + I) - 3465*(e^(11*I*d*x + 11*I*c) + 4*e^(9*I*d*x + 9*I*c) + 6*e^(7*I*d*x 
 + 7*I*c) + 4*e^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3*I*c))*log(e^(I*d*x + I* 
c) - I) - 6930*I*e^(10*I*d*x + 10*I*c) - 25410*I*e^(8*I*d*x + 8*I*c) - 337 
26*I*e^(6*I*d*x + 6*I*c) - 18414*I*e^(4*I*d*x + 4*I*c) - 2816*I*e^(2*I*d*x 
 + 2*I*c) + 256*I)/(a^8*d*e^(11*I*d*x + 11*I*c) + 4*a^8*d*e^(9*I*d*x + 9*I 
*c) + 6*a^8*d*e^(7*I*d*x + 7*I*c) + 4*a^8*d*e^(5*I*d*x + 5*I*c) + a^8*d*e^ 
(3*I*d*x + 3*I*c))
 

Sympy [F]

\[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{13}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \] Input:

integrate(sec(d*x+c)**13/(a+I*a*tan(d*x+c))**8,x)
 

Output:

Integral(sec(c + d*x)**13/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan( 
c + d*x)**6 + 56*I*tan(c + d*x)**5 + 70*tan(c + d*x)**4 - 56*I*tan(c + d*x 
)**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8
 

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. 786 vs. \(2 (179) = 358\).

Time = 0.23 (sec) , antiderivative size = 786, normalized size of antiderivative = 3.83 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
 

Output:

-(6930*(cos(11*d*x + 11*c) + 4*cos(9*d*x + 9*c) + 6*cos(7*d*x + 7*c) + 4*c 
os(5*d*x + 5*c) + cos(3*d*x + 3*c) + I*sin(11*d*x + 11*c) + 4*I*sin(9*d*x 
+ 9*c) + 6*I*sin(7*d*x + 7*c) + 4*I*sin(5*d*x + 5*c) + I*sin(3*d*x + 3*c)) 
*arctan2(cos(d*x + c), sin(d*x + c) + 1) + 6930*(cos(11*d*x + 11*c) + 4*co 
s(9*d*x + 9*c) + 6*cos(7*d*x + 7*c) + 4*cos(5*d*x + 5*c) + cos(3*d*x + 3*c 
) + I*sin(11*d*x + 11*c) + 4*I*sin(9*d*x + 9*c) + 6*I*sin(7*d*x + 7*c) + 4 
*I*sin(5*d*x + 5*c) + I*sin(3*d*x + 3*c))*arctan2(cos(d*x + c), -sin(d*x + 
 c) + 1) + 3465*(I*cos(11*d*x + 11*c) + 4*I*cos(9*d*x + 9*c) + 6*I*cos(7*d 
*x + 7*c) + 4*I*cos(5*d*x + 5*c) + I*cos(3*d*x + 3*c) - sin(11*d*x + 11*c) 
 - 4*sin(9*d*x + 9*c) - 6*sin(7*d*x + 7*c) - 4*sin(5*d*x + 5*c) - sin(3*d* 
x + 3*c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3465 
*(-I*cos(11*d*x + 11*c) - 4*I*cos(9*d*x + 9*c) - 6*I*cos(7*d*x + 7*c) - 4* 
I*cos(5*d*x + 5*c) - I*cos(3*d*x + 3*c) + sin(11*d*x + 11*c) + 4*sin(9*d*x 
 + 9*c) + 6*sin(7*d*x + 7*c) + 4*sin(5*d*x + 5*c) + sin(3*d*x + 3*c))*log( 
cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 13860*cos(10*d*x + 
 10*c) + 50820*cos(8*d*x + 8*c) + 67452*cos(6*d*x + 6*c) + 36828*cos(4*d*x 
 + 4*c) + 5632*cos(2*d*x + 2*c) + 13860*I*sin(10*d*x + 10*c) + 50820*I*sin 
(8*d*x + 8*c) + 67452*I*sin(6*d*x + 6*c) + 36828*I*sin(4*d*x + 4*c) + 5632 
*I*sin(2*d*x + 2*c) - 512)/((-48*I*a^8*cos(11*d*x + 11*c) - 192*I*a^8*cos( 
9*d*x + 9*c) - 288*I*a^8*cos(7*d*x + 7*c) - 192*I*a^8*cos(5*d*x + 5*c) ...
 

Giac [A] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {3465 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {3465 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {1024 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}} - \frac {2 \, {\left (369 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1728 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 393 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5568 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 393 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5696 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 369 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1856 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a^{8}}}{24 \, d} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
 

Output:

1/24*(3465*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 3465*log(tan(1/2*d*x + 1/2* 
c) - 1)/a^8 - 1024*(6*tan(1/2*d*x + 1/2*c)^2 - 15*I*tan(1/2*d*x + 1/2*c) - 
 7)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^3) - 2*(369*tan(1/2*d*x + 1/2*c)^7 - 1 
728*I*tan(1/2*d*x + 1/2*c)^6 - 393*tan(1/2*d*x + 1/2*c)^5 + 5568*I*tan(1/2 
*d*x + 1/2*c)^4 - 393*tan(1/2*d*x + 1/2*c)^3 - 5696*I*tan(1/2*d*x + 1/2*c) 
^2 + 369*tan(1/2*d*x + 1/2*c) + 1856*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*a^ 
8))/d
 

Mupad [B] (verification not implemented)

Time = 4.78 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.68 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {33847\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{6\,a^8}-\frac {12041\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^8}-\frac {3585\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{a^8}+\frac {3505\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4\,a^8}+\frac {4293\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,27565{}\mathrm {i}}{12\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,4575{}\mathrm {i}}{a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,25993{}\mathrm {i}}{6\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,5639{}\mathrm {i}}{3\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,1147{}\mathrm {i}}{4\,a^8}-\frac {1360{}\mathrm {i}}{3\,a^8}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,7{}\mathrm {i}+13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,18{}\mathrm {i}-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,22{}\mathrm {i}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,13{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )}+\frac {1155\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^8\,d} \] Input:

int(1/(cos(c + d*x)^13*(a + a*tan(c + d*x)*1i)^8),x)
 

Output:

((tan(c/2 + (d*x)/2)^2*27565i)/(12*a^8) - (12041*tan(c/2 + (d*x)/2)^3)/(3* 
a^8) - (tan(c/2 + (d*x)/2)^4*4575i)/a^8 + (33847*tan(c/2 + (d*x)/2)^5)/(6* 
a^8) + (tan(c/2 + (d*x)/2)^6*25993i)/(6*a^8) - (3585*tan(c/2 + (d*x)/2)^7) 
/a^8 - (tan(c/2 + (d*x)/2)^8*5639i)/(3*a^8) + (3505*tan(c/2 + (d*x)/2)^9)/ 
(4*a^8) + (tan(c/2 + (d*x)/2)^10*1147i)/(4*a^8) - 1360i/(3*a^8) + (4293*ta 
n(c/2 + (d*x)/2))/(4*a^8))/(d*(tan(c/2 + (d*x)/2)*3i - 7*tan(c/2 + (d*x)/2 
)^2 - tan(c/2 + (d*x)/2)^3*13i + 18*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x) 
/2)^5*22i - 22*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^7*18i + 13*tan(c/ 
2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^9*7i - 3*tan(c/2 + (d*x)/2)^10 - tan(c 
/2 + (d*x)/2)^11*1i + 1)) + (1155*atanh(tan(c/2 + (d*x)/2)))/(4*a^8*d)
 

Reduce [F]

\[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\text {too large to display} \] Input:

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x)
 

Output:

( - 456672017154048*cos(c + d*x)**2*sin(c + d*x)**3 - 4982530650240*cos(c 
+ d*x)**2*sin(c + d*x)**2*i + 514842517856256*cos(c + d*x)**2*sin(c + d*x) 
 + 6279604284096*cos(c + d*x)**2*i - 8168122220544*cos(c + d*x)*int(cos(c 
+ d*x)/(128*cos(c + d*x)*sin(c + d*x)**12*i - 512*cos(c + d*x)*sin(c + d*x 
)**10*i + 800*cos(c + d*x)*sin(c + d*x)**8*i - 608*cos(c + d*x)*sin(c + d* 
x)**6*i + 225*cos(c + d*x)*sin(c + d*x)**4*i - 34*cos(c + d*x)*sin(c + d*x 
)**2*i + cos(c + d*x)*i - 128*sin(c + d*x)**13 + 576*sin(c + d*x)**11 - 10 
40*sin(c + d*x)**9 + 952*sin(c + d*x)**7 - 456*sin(c + d*x)**5 + 104*sin(c 
 + d*x)**3 - 8*sin(c + d*x)),x)*sin(c + d*x)**4*d*i + 16336244441088*cos(c 
 + d*x)*int(cos(c + d*x)/(128*cos(c + d*x)*sin(c + d*x)**12*i - 512*cos(c 
+ d*x)*sin(c + d*x)**10*i + 800*cos(c + d*x)*sin(c + d*x)**8*i - 608*cos(c 
 + d*x)*sin(c + d*x)**6*i + 225*cos(c + d*x)*sin(c + d*x)**4*i - 34*cos(c 
+ d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 128*sin(c + d*x)**13 + 576*sin 
(c + d*x)**11 - 1040*sin(c + d*x)**9 + 952*sin(c + d*x)**7 - 456*sin(c + d 
*x)**5 + 104*sin(c + d*x)**3 - 8*sin(c + d*x)),x)*sin(c + d*x)**2*d*i - 81 
68122220544*cos(c + d*x)*int(cos(c + d*x)/(128*cos(c + d*x)*sin(c + d*x)** 
12*i - 512*cos(c + d*x)*sin(c + d*x)**10*i + 800*cos(c + d*x)*sin(c + d*x) 
**8*i - 608*cos(c + d*x)*sin(c + d*x)**6*i + 225*cos(c + d*x)*sin(c + d*x) 
**4*i - 34*cos(c + d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 128*sin(c + d 
*x)**13 + 576*sin(c + d*x)**11 - 1040*sin(c + d*x)**9 + 952*sin(c + d*x...