Integrand size = 24, antiderivative size = 119 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {7 \text {arctanh}(\sin (c+d x))}{8 a^3 d}-\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{12 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2} \] Output:
7/8*arctanh(sin(d*x+c))/a^3/d-7/15*I*sec(d*x+c)^5/a^3/d+7/8*sec(d*x+c)*tan (d*x+c)/a^3/d+7/12*sec(d*x+c)^3*tan(d*x+c)/a^3/d-2/3*I*sec(d*x+c)^7/a/d/(a +I*a*tan(d*x+c))^2
Time = 0.92 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^8(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (448+1680 i \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right ) \cos ^5(c+d x)+640 \cos (2 (c+d x))-150 i \sin (2 (c+d x))+105 i \sin (4 (c+d x))\right )}{960 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^3,x]
Output:
(Sec[c + d*x]^8*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(448 + (1680*I)*Ar cTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*Cos[c + d*x]^5 + 640*Cos[2*(c + d*x)] - (150*I)*Sin[2*(c + d*x)] + (105*I)*Sin[4*(c + d*x)]))/(960*a^3*d*(-I + T an[c + d*x])^3)
Time = 0.66 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 3981, 3042, 3982, 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.
| \(\displaystyle \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^9}{(a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {7 \int \frac {\sec ^7(c+d x)}{i \tan (c+d x) a+a}dx}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \int \frac {\sec (c+d x)^7}{i \tan (c+d x) a+a}dx}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3982 |
| \(\displaystyle \frac {7 \left (\frac {\int \sec ^5(c+d x)dx}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {7 \left (\frac {\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {7 \left (\frac {\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {7 \left (\frac {\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {i \sec ^5(c+d x)}{5 a d}\right )}{3 a^2}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\) |
Input:
Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^3,x]
Output:
(((-2*I)/3)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^2) + (7*(((-1/5*I) *Sec[c + d*x]^5)/(a*d) + ((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTan h[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)/a))/(3*a^2)
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[((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]
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 = 0.84 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(-\frac {i \left (105 \,{\mathrm e}^{9 i \left (d x +c \right )}+490 \,{\mathrm e}^{7 i \left (d x +c \right )}+896 \,{\mathrm e}^{5 i \left (d x +c \right )}+790 \,{\mathrm e}^{3 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a^{3} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 a^{3} d}\) | \(122\) |
| derivativedivides | \(\frac {-\frac {i}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {2 \left (\frac {1}{16}+\frac {13 i}{16}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 \left (-\frac {3}{8}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {2 \left (-\frac {5}{16}+\frac {11 i}{16}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {3}{4}+\frac {7 i}{24}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {i}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2 \left (\frac {5}{16}+\frac {11 i}{16}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\frac {3}{8}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2 \left (\frac {1}{16}-\frac {13 i}{16}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {2 \left (-\frac {3}{4}-\frac {7 i}{24}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}}{a^{3} d}\) | \(206\) |
| default | \(\frac {-\frac {i}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {2 \left (\frac {1}{16}+\frac {13 i}{16}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 \left (-\frac {3}{8}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {2 \left (-\frac {5}{16}+\frac {11 i}{16}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {3}{4}+\frac {7 i}{24}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {i}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2 \left (\frac {5}{16}+\frac {11 i}{16}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\frac {3}{8}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2 \left (\frac {1}{16}-\frac {13 i}{16}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {2 \left (-\frac {3}{4}-\frac {7 i}{24}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}}{a^{3} d}\) | \(206\) |
Input:
int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-1/60*I/d/a^3/(exp(2*I*(d*x+c))+1)^5*(105*exp(9*I*(d*x+c))+490*exp(7*I*(d* x+c))+896*exp(5*I*(d*x+c))+790*exp(3*I*(d*x+c))-105*exp(I*(d*x+c)))-7/8/a^ 3/d*ln(exp(I*(d*x+c))-I)+7/8/a^3/d*ln(exp(I*(d*x+c))+I)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (103) = 206\).
Time = 0.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.34 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {105 \, {\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 980 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 1792 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 1580 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, e^{\left (i \, d x + i \, c\right )}}{120 \, {\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/120*(105*(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d* x + I*c) + I) - 105*(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^ (6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1)*lo g(e^(I*d*x + I*c) - I) - 210*I*e^(9*I*d*x + 9*I*c) - 980*I*e^(7*I*d*x + 7* I*c) - 1792*I*e^(5*I*d*x + 5*I*c) - 1580*I*e^(3*I*d*x + 3*I*c) + 210*I*e^( I*d*x + I*c))/(a^3*d*e^(10*I*d*x + 10*I*c) + 5*a^3*d*e^(8*I*d*x + 8*I*c) + 10*a^3*d*e^(6*I*d*x + 6*I*c) + 10*a^3*d*e^(4*I*d*x + 4*I*c) + 5*a^3*d*e^( 2*I*d*x + 2*I*c) + a^3*d)
\[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{9}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**3,x)
Output:
I*Integral(sec(c + d*x)**9/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan( c + d*x) + I), x)/a**3
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (103) = 206\).
Time = 0.05 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.87 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {16 \, {\left (-\frac {15 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {320 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {390 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {400 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {960 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {390 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {360 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {15 i \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 136\right )}}{-120 i \, a^{3} + \frac {600 i \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1200 i \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1200 i \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {600 i \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {120 i \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} - \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{8 \, d} \] Input:
integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/8*(16*(-15*I*sin(d*x + c)/(cos(d*x + c) + 1) + 320*sin(d*x + c)^2/(cos(d *x + c) + 1)^2 + 390*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 400*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 960*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 390 *I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 360*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 15*I*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 136)/(-120*I*a^3 + 600 *I*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1200*I*a^3*sin(d*x + c)^4/(co s(d*x + c) + 1)^4 + 1200*I*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 600*I *a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 120*I*a^3*sin(d*x + c)^10/(cos( d*x + c) + 1)^10) + 7*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 - 7*log (sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d
Time = 0.36 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 400 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 136 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{3}}}{120 \, d} \] Input:
integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/120*(105*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 105*log(tan(1/2*d*x + 1/2*c ) - 1)/a^3 + 2*(15*tan(1/2*d*x + 1/2*c)^9 + 360*I*tan(1/2*d*x + 1/2*c)^8 - 390*tan(1/2*d*x + 1/2*c)^7 - 960*I*tan(1/2*d*x + 1/2*c)^6 + 400*I*tan(1/2 *d*x + 1/2*c)^4 + 390*tan(1/2*d*x + 1/2*c)^3 - 320*I*tan(1/2*d*x + 1/2*c)^ 2 - 15*tan(1/2*d*x + 1/2*c) + 136*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^3)) /d
Time = 3.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^3\,d}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,6{}\mathrm {i}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,16{}\mathrm {i}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,20{}\mathrm {i}}{3}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}}{3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {34}{15}{}\mathrm {i}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^5} \] Input:
int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^3),x)
Output:
(7*atanh(tan(c/2 + (d*x)/2)))/(4*a^3*d) + ((13*tan(c/2 + (d*x)/2)^3)/2 - ( tan(c/2 + (d*x)/2)^2*16i)/3 - tan(c/2 + (d*x)/2)/4 + (tan(c/2 + (d*x)/2)^4 *20i)/3 - tan(c/2 + (d*x)/2)^6*16i - (13*tan(c/2 + (d*x)/2)^7)/2 + tan(c/2 + (d*x)/2)^8*6i + tan(c/2 + (d*x)/2)^9/4 + 34i/15)/(a^3*d*(tan(c/2 + (d*x )/2)^2 - 1)^5)
Time = 0.17 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.34 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-105 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+210 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-105 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+105 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-210 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+105 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i -105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i +15 \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 \cos \left (d x +c \right ) i +160 \sin \left (d x +c \right )^{2} i -136 i}{120 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 105*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 210*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 105*cos(c + d*x)*log(t an((c + d*x)/2) - 1) + 105*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 210*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 105 *cos(c + d*x)*log(tan((c + d*x)/2) + 1) + 8*cos(c + d*x)*sin(c + d*x)**4*i - 105*cos(c + d*x)*sin(c + d*x)**3 - 16*cos(c + d*x)*sin(c + d*x)**2*i + 15*cos(c + d*x)*sin(c + d*x) + 8*cos(c + d*x)*i + 160*sin(c + d*x)**2*i - 136*i)/(120*cos(c + d*x)*a**3*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))