Integrand size = 24, antiderivative size = 183 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {63 \text {arctanh}(\sin (c+d x))}{2 a^8 d}-\frac {63 \sec (c+d x) \tan (c+d x)}{2 a^8 d}+\frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {42 i \sec ^5(c+d x)}{5 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac {42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]
-63/2*arctanh(sin(d*x+c))/a^8/d-63/2*sec(d*x+c)*tan(d*x+c)/a^8/d+2/5*I*sec (d*x+c)^9/a/d/(a+I*a*tan(d*x+c))^7-6/5*I*sec(d*x+c)^7/a^3/d/(a+I*a*tan(d*x +c))^5+42/5*I*sec(d*x+c)^5/a^2/d/(a^2+I*a^2*tan(d*x+c))^3+42*I*sec(d*x+c)^ 3/d/(a^8+I*a^8*tan(d*x+c))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1244\) vs. \(2(183)=366\).
Time = 7.08 (sec) , antiderivative size = 1244, normalized size of antiderivative = 6.80 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx =\text {Too large to display} \]
(63*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)/(2*d*(a + I*a*Tan[c + d*x])^8) - (63*Cos[8*c]*Lo g[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x] + I*Si n[d*x])^8)/(2*d*(a + I*a*Tan[c + d*x])^8) + (Cos[5*d*x]*Sec[c + d*x]^8*((( 8*I)/5)*Cos[3*c] - (8*Sin[3*c])/5)*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a* Tan[c + d*x])^8) + (Cos[3*d*x]*Sec[c + d*x]^8*((-8*I)*Cos[5*c] + 8*Sin[5*c ])*(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*((48*I)*Cos[7*c] - 48*Sin[7*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d* (a + I*a*Tan[c + d*x])^8) + (Sec[c]*Sec[c + d*x]^8*((8*I)*Cos[8*c] - 8*Sin [8*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (((63*I)/ 2)*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) - (((63*I)/2)*Log[Co s[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) + (Sec[c + d*x]^8*(48*Cos[7*c ] + (48*I)*Sin[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*(-8*Cos[5*c] - (8*I)*Sin[5*c])*(Cos[d*x] + I *Sin[d*x])^8*Sin[3*d*x])/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(( 8*Cos[3*c])/5 + ((8*I)/5)*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^8*Sin[5*d*x])/ (d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(Cos[8*c]/4 + (I/4)*Sin[8*c ])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/...
Time = 0.93 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3981, 3042, 3981, 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.
| \(\displaystyle \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{11}}{(a+i a \tan (c+d x))^8}dx\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^6}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^6}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \int \frac {\sec ^7(c+d x)}{(i \tan (c+d x) a+a)^4}dx}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \int \frac {\sec (c+d x)^7}{(i \tan (c+d x) a+a)^4}dx}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \int \frac {\sec ^5(c+d x)}{(i \tan (c+d x) a+a)^2}dx}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \int \frac {\sec (c+d x)^5}{(i \tan (c+d x) a+a)^2}dx}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \left (\frac {3 \int \sec ^3(c+d x)dx}{a^2}-\frac {2 i \sec ^3(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \left (\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {2 i \sec ^3(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \left (\frac {3 \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {2 i \sec ^3(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \left (\frac {3 \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {2 i \sec ^3(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac {9 \left (\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^5}-\frac {7 \left (\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \left (\frac {3 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {2 i \sec ^3(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\right )}{3 a^2}\right )}{5 a^2}\) |
(((2*I)/5)*Sec[c + d*x]^9)/(a*d*(a + I*a*Tan[c + d*x])^7) - (9*((((2*I)/3) *Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^5) - (7*(((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))/(3*a^2)))/(5*a^2)
3.2.77.3.1 Defintions of rubi rules used
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]
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.96 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.78
| method | result | size |
| risch | \(\frac {48 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{8} d}-\frac {8 i {\mathrm e}^{-3 i \left (d x +c \right )}}{a^{8} d}+\frac {8 i {\mathrm e}^{-5 i \left (d x +c \right )}}{5 a^{8} d}+\frac {i \left (15 \,{\mathrm e}^{3 i \left (d x +c \right )}+17 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{8} d}+\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{8} d}\) | \(143\) |
| derivativedivides | \(\frac {\frac {2 \left (\frac {1}{4}+4 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {63 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {2 \left (\frac {1}{4}-4 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {63 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {32 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {256}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {64}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {64}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{8} d}\) | \(184\) |
| default | \(\frac {\frac {2 \left (\frac {1}{4}+4 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {63 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {2 \left (\frac {1}{4}-4 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {63 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {32 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {256}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {64}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {64}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{8} d}\) | \(184\) |
48*I/a^8/d*exp(-I*(d*x+c))-8*I/a^8/d*exp(-3*I*(d*x+c))+8/5*I/a^8/d*exp(-5* I*(d*x+c))+I/d/a^8/(exp(2*I*(d*x+c))+1)^2*(15*exp(3*I*(d*x+c))+17*exp(I*(d *x+c)))-63/2/a^8/d*ln(exp(I*(d*x+c))+I)+63/2/a^8/d*ln(exp(I*(d*x+c))-I)
Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {315 \, {\left (e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, e^{\left (7 i \, d x + 7 i \, c\right )} + e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \, {\left (e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, e^{\left (7 i \, d x + 7 i \, c\right )} + e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 630 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 1050 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 336 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 48 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i}{10 \, {\left (a^{8} d e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, a^{8} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{8} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]
-1/10*(315*(e^(9*I*d*x + 9*I*c) + 2*e^(7*I*d*x + 7*I*c) + e^(5*I*d*x + 5*I *c))*log(e^(I*d*x + I*c) + I) - 315*(e^(9*I*d*x + 9*I*c) + 2*e^(7*I*d*x + 7*I*c) + e^(5*I*d*x + 5*I*c))*log(e^(I*d*x + I*c) - I) - 630*I*e^(8*I*d*x + 8*I*c) - 1050*I*e^(6*I*d*x + 6*I*c) - 336*I*e^(4*I*d*x + 4*I*c) + 48*I*e ^(2*I*d*x + 2*I*c) - 16*I)/(a^8*d*e^(9*I*d*x + 9*I*c) + 2*a^8*d*e^(7*I*d*x + 7*I*c) + a^8*d*e^(5*I*d*x + 5*I*c))
\[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{11}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]
Integral(sec(c + d*x)**11/(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
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 531 vs. \(2 (157) = 314\).
Time = 0.34 (sec) , antiderivative size = 531, normalized size of antiderivative = 2.90 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {630 \, {\left (\cos \left (9 \, d x + 9 \, c\right ) + 2 \, \cos \left (7 \, d x + 7 \, c\right ) + \cos \left (5 \, d x + 5 \, c\right ) + i \, \sin \left (9 \, d x + 9 \, c\right ) + 2 i \, \sin \left (7 \, d x + 7 \, c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 630 \, {\left (\cos \left (9 \, d x + 9 \, c\right ) + 2 \, \cos \left (7 \, d x + 7 \, c\right ) + \cos \left (5 \, d x + 5 \, c\right ) + i \, \sin \left (9 \, d x + 9 \, c\right ) + 2 i \, \sin \left (7 \, d x + 7 \, c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 315 \, {\left (i \, \cos \left (9 \, d x + 9 \, c\right ) + 2 i \, \cos \left (7 \, d x + 7 \, c\right ) + i \, \cos \left (5 \, d x + 5 \, c\right ) - \sin \left (9 \, d x + 9 \, c\right ) - 2 \, \sin \left (7 \, d x + 7 \, c\right ) - \sin \left (5 \, d x + 5 \, 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 ) + 315 \, {\left (-i \, \cos \left (9 \, d x + 9 \, c\right ) - 2 i \, \cos \left (7 \, d x + 7 \, c\right ) - i \, \cos \left (5 \, d x + 5 \, c\right ) + \sin \left (9 \, d x + 9 \, c\right ) + 2 \, \sin \left (7 \, d x + 7 \, c\right ) + \sin \left (5 \, d x + 5 \, 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 ) + 1260 \, \cos \left (8 \, d x + 8 \, c\right ) + 2100 \, \cos \left (6 \, d x + 6 \, c\right ) + 672 \, \cos \left (4 \, d x + 4 \, c\right ) - 96 \, \cos \left (2 \, d x + 2 \, c\right ) + 1260 i \, \sin \left (8 \, d x + 8 \, c\right ) + 2100 i \, \sin \left (6 \, d x + 6 \, c\right ) + 672 i \, \sin \left (4 \, d x + 4 \, c\right ) - 96 i \, \sin \left (2 \, d x + 2 \, c\right ) + 32}{-20 \, {\left (i \, a^{8} \cos \left (9 \, d x + 9 \, c\right ) + 2 i \, a^{8} \cos \left (7 \, d x + 7 \, c\right ) + i \, a^{8} \cos \left (5 \, d x + 5 \, c\right ) - a^{8} \sin \left (9 \, d x + 9 \, c\right ) - 2 \, a^{8} \sin \left (7 \, d x + 7 \, c\right ) - a^{8} \sin \left (5 \, d x + 5 \, c\right )\right )} d} \]
(630*(cos(9*d*x + 9*c) + 2*cos(7*d*x + 7*c) + cos(5*d*x + 5*c) + I*sin(9*d *x + 9*c) + 2*I*sin(7*d*x + 7*c) + I*sin(5*d*x + 5*c))*arctan2(cos(d*x + c ), sin(d*x + c) + 1) + 630*(cos(9*d*x + 9*c) + 2*cos(7*d*x + 7*c) + cos(5* d*x + 5*c) + I*sin(9*d*x + 9*c) + 2*I*sin(7*d*x + 7*c) + I*sin(5*d*x + 5*c ))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 315*(I*cos(9*d*x + 9*c) + 2* I*cos(7*d*x + 7*c) + I*cos(5*d*x + 5*c) - sin(9*d*x + 9*c) - 2*sin(7*d*x + 7*c) - sin(5*d*x + 5*c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 315*(-I*cos(9*d*x + 9*c) - 2*I*cos(7*d*x + 7*c) - I*cos(5*d*x + 5*c) + sin(9*d*x + 9*c) + 2*sin(7*d*x + 7*c) + sin(5*d*x + 5*c))*log(cos (d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 1260*cos(8*d*x + 8*c) + 2100*cos(6*d*x + 6*c) + 672*cos(4*d*x + 4*c) - 96*cos(2*d*x + 2*c) + 12 60*I*sin(8*d*x + 8*c) + 2100*I*sin(6*d*x + 6*c) + 672*I*sin(4*d*x + 4*c) - 96*I*sin(2*d*x + 2*c) + 32)/((-20*I*a^8*cos(9*d*x + 9*c) - 40*I*a^8*cos(7 *d*x + 7*c) - 20*I*a^8*cos(5*d*x + 5*c) + 20*a^8*sin(9*d*x + 9*c) + 40*a^8 *sin(7*d*x + 7*c) + 20*a^8*sin(5*d*x + 5*c))*d)
Time = 1.81 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {\frac {315 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {315 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {10 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{8}} - \frac {64 \, {\left (10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 45 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 55 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}}}{10 \, d} \]
-1/10*(315*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 315*log(tan(1/2*d*x + 1/2*c ) - 1)/a^8 - 10*(tan(1/2*d*x + 1/2*c)^3 - 16*I*tan(1/2*d*x + 1/2*c)^2 + ta n(1/2*d*x + 1/2*c) + 16*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^8) - 64*(10*t an(1/2*d*x + 1/2*c)^4 - 45*I*tan(1/2*d*x + 1/2*c)^3 - 85*tan(1/2*d*x + 1/2 *c)^2 + 55*I*tan(1/2*d*x + 1/2*c) + 13)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^5) )/d
Time = 8.67 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.55 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {63\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^8\,d}+\frac {\frac {1223\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^8}-\frac {1109\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^8}+\frac {309\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{a^8}-\frac {431\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,4407{}\mathrm {i}}{5\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,7351{}\mathrm {i}}{5\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,761{}\mathrm {i}}{a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,65{}\mathrm {i}}{a^8}+\frac {496{}\mathrm {i}}{5\,a^8}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,12{}\mathrm {i}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,26{}\mathrm {i}+26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,20{}\mathrm {i}-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \]
((1223*tan(c/2 + (d*x)/2)^3)/a^8 - (tan(c/2 + (d*x)/2)^2*4407i)/(5*a^8) + (tan(c/2 + (d*x)/2)^4*7351i)/(5*a^8) - (1109*tan(c/2 + (d*x)/2)^5)/a^8 - ( tan(c/2 + (d*x)/2)^6*761i)/a^8 + (309*tan(c/2 + (d*x)/2)^7)/a^8 + (tan(c/2 + (d*x)/2)^8*65i)/a^8 + 496i/(5*a^8) - (431*tan(c/2 + (d*x)/2))/a^8)/(d*( tan(c/2 + (d*x)/2)*5i - 12*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*20i + 26*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5*26i - 20*tan(c/2 + (d*x) /2)^6 - tan(c/2 + (d*x)/2)^7*12i + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x )/2)^9*1i + 1)) - (63*atanh(tan(c/2 + (d*x)/2)))/(a^8*d)