Integrand size = 24, antiderivative size = 126 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-4 a^3 x-\frac {4 a^3 \cot (c+d x)}{d}+\frac {2 i a^3 \cot ^2(c+d x)}{d}+\frac {4 a^3 \cot ^3(c+d x)}{3 d}-\frac {11 i a^3 \cot ^4(c+d x)}{20 d}+\frac {4 i a^3 \log (\sin (c+d x))}{d}-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d} \]
-4*a^3*x-4*a^3*cot(d*x+c)/d+2*I*a^3*cot(d*x+c)^2/d+4/3*a^3*cot(d*x+c)^3/d- 11/20*I*a^3*cot(d*x+c)^4/d+4*I*a^3*ln(sin(d*x+c))/d-1/5*cot(d*x+c)^5*(a^3+ I*a^3*tan(d*x+c))/d
Time = 0.45 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=a^3 \left (-\frac {4 \cot (c+d x)}{d}+\frac {2 i \cot ^2(c+d x)}{d}+\frac {4 \cot ^3(c+d x)}{3 d}-\frac {3 i \cot ^4(c+d x)}{4 d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {4 i \log (\tan (c+d x))}{d}-\frac {4 i \log (i+\tan (c+d x))}{d}\right ) \]
a^3*((-4*Cot[c + d*x])/d + ((2*I)*Cot[c + d*x]^2)/d + (4*Cot[c + d*x]^3)/( 3*d) - (((3*I)/4)*Cot[c + d*x]^4)/d - Cot[c + d*x]^5/(5*d) + ((4*I)*Log[Ta n[c + d*x]])/d - ((4*I)*Log[I + Tan[c + d*x]])/d)
Time = 0.94 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4036, 25, 3042, 4074, 27, 3042, 4012, 3042, 4012, 25, 3042, 4012, 3042, 4014, 3042, 25, 3956}
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.
| \(\displaystyle \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\tan (c+d x)^6}dx\) |
| \(\Big \downarrow \) 4036 |
| \(\displaystyle -\frac {1}{5} \int -\cot ^5(c+d x) (i \tan (c+d x) a+a) \left (11 i a^2-9 a^2 \tan (c+d x)\right )dx-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \int \cot ^5(c+d x) (i \tan (c+d x) a+a) \left (11 i a^2-9 a^2 \tan (c+d x)\right )dx-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(i \tan (c+d x) a+a) \left (11 i a^2-9 a^2 \tan (c+d x)\right )}{\tan (c+d x)^5}dx-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 4074 |
| \(\displaystyle \frac {1}{5} \left (\int -20 \cot ^4(c+d x) \left (i \tan (c+d x) a^3+a^3\right )dx-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-20 \int \cot ^4(c+d x) \left (i \tan (c+d x) a^3+a^3\right )dx-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-20 \int \frac {i \tan (c+d x) a^3+a^3}{\tan (c+d x)^4}dx-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\frac {a^3 \cot ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) \left (i a^3-a^3 \tan (c+d x)\right )dx\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\frac {a^3 \cot ^3(c+d x)}{3 d}+\int \frac {i a^3-a^3 \tan (c+d x)}{\tan (c+d x)^3}dx\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (\int -\cot ^2(c+d x) \left (i \tan (c+d x) a^3+a^3\right )dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\int \cot ^2(c+d x) \left (i \tan (c+d x) a^3+a^3\right )dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\int \frac {i \tan (c+d x) a^3+a^3}{\tan (c+d x)^2}dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\int \cot (c+d x) \left (i a^3-a^3 \tan (c+d x)\right )dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-\int \frac {i a^3-a^3 \tan (c+d x)}{\tan (c+d x)}dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-i a^3 \int \cot (c+d x)dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}+a^3 x\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (-i a^3 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}+a^3 x\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-20 \left (i a^3 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}+a^3 x\right )-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {1}{5} \left (-\frac {11 i a^3 \cot ^4(c+d x)}{4 d}-20 \left (-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {i a^3 \cot ^2(c+d x)}{2 d}+\frac {a^3 \cot (c+d x)}{d}-\frac {i a^3 \log (-\sin (c+d x))}{d}+a^3 x\right )\right )-\frac {\cot ^5(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
((((-11*I)/4)*a^3*Cot[c + d*x]^4)/d - 20*(a^3*x + (a^3*Cot[c + d*x])/d - ( (I/2)*a^3*Cot[c + d*x]^2)/d - (a^3*Cot[c + d*x]^3)/(3*d) - (I*a^3*Log[-Sin [c + d*x]])/d))/5 - (Cot[c + d*x]^5*(a^3 + I*a^3*Tan[c + d*x]))/(5*d)
3.1.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] )^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si mp[a/(d*(b*c + a*d)*(n + 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2 ))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^2, 0]
Time = 0.53 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-4 \cot \left (d x +c \right )-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-\frac {3 i \left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {4 \left (\cot ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\cot ^{2}\left (d x +c \right )\right )-2 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )+2 \pi -4 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
| default | \(\frac {a^{3} \left (-4 \cot \left (d x +c \right )-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-\frac {3 i \left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {4 \left (\cot ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\cot ^{2}\left (d x +c \right )\right )-2 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )+2 \pi -4 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
| parallelrisch | \(\frac {a^{3} \left (-12 \left (\cot ^{5}\left (d x +c \right )\right )-45 i \left (\cot ^{4}\left (d x +c \right )\right )+80 \left (\cot ^{3}\left (d x +c \right )\right )+120 i \left (\cot ^{2}\left (d x +c \right )\right )+240 i \ln \left (\tan \left (d x +c \right )\right )-120 i \ln \left (\sec ^{2}\left (d x +c \right )\right )-240 d x -240 \cot \left (d x +c \right )\right )}{60 d}\) | \(86\) |
| risch | \(\frac {8 a^{3} c}{d}-\frac {2 i a^{3} \left (240 \,{\mathrm e}^{8 i \left (d x +c \right )}-585 \,{\mathrm e}^{6 i \left (d x +c \right )}+695 \,{\mathrm e}^{4 i \left (d x +c \right )}-385 \,{\mathrm e}^{2 i \left (d x +c \right )}+83\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(100\) |
| norman | \(\frac {-\frac {a^{3}}{5 d}-4 a^{3} x \left (\tan ^{5}\left (d x +c \right )\right )+\frac {4 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{3 d}-\frac {4 a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{d}-\frac {3 i a^{3} \tan \left (d x +c \right )}{4 d}+\frac {2 i a^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{5}}+\frac {4 i a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(134\) |
1/d*a^3*(-4*cot(d*x+c)-1/5*cot(d*x+c)^5-3/4*I*cot(d*x+c)^4+4/3*cot(d*x+c)^ 3+2*I*cot(d*x+c)^2-2*I*ln(cot(d*x+c)^2+1)+2*Pi-4*arccot(cot(d*x+c)))
Time = 0.23 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.74 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (240 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 585 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 695 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 385 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 83 i \, a^{3} + 30 \, {\left (-i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 10 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
-2/15*(240*I*a^3*e^(8*I*d*x + 8*I*c) - 585*I*a^3*e^(6*I*d*x + 6*I*c) + 695 *I*a^3*e^(4*I*d*x + 4*I*c) - 385*I*a^3*e^(2*I*d*x + 2*I*c) + 83*I*a^3 + 30 *(-I*a^3*e^(10*I*d*x + 10*I*c) + 5*I*a^3*e^(8*I*d*x + 8*I*c) - 10*I*a^3*e^ (6*I*d*x + 6*I*c) + 10*I*a^3*e^(4*I*d*x + 4*I*c) - 5*I*a^3*e^(2*I*d*x + 2* I*c) + I*a^3)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(10*I*d*x + 10*I*c) - 5*d *e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)
Time = 0.33 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.73 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4 i a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 480 i a^{3} e^{8 i c} e^{8 i d x} + 1170 i a^{3} e^{6 i c} e^{6 i d x} - 1390 i a^{3} e^{4 i c} e^{4 i d x} + 770 i a^{3} e^{2 i c} e^{2 i d x} - 166 i a^{3}}{15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} - 15 d} \]
4*I*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-480*I*a**3*exp(8*I*c)*exp(8 *I*d*x) + 1170*I*a**3*exp(6*I*c)*exp(6*I*d*x) - 1390*I*a**3*exp(4*I*c)*exp (4*I*d*x) + 770*I*a**3*exp(2*I*c)*exp(2*I*d*x) - 166*I*a**3)/(15*d*exp(10* I*c)*exp(10*I*d*x) - 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6 *I*d*x) - 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) - 1 5*d)
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {240 \, {\left (d x + c\right )} a^{3} + 120 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 i \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {240 \, a^{3} \tan \left (d x + c\right )^{4} - 120 i \, a^{3} \tan \left (d x + c\right )^{3} - 80 \, a^{3} \tan \left (d x + c\right )^{2} + 45 i \, a^{3} \tan \left (d x + c\right ) + 12 \, a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]
-1/60*(240*(d*x + c)*a^3 + 120*I*a^3*log(tan(d*x + c)^2 + 1) - 240*I*a^3*l og(tan(d*x + c)) + (240*a^3*tan(d*x + c)^4 - 120*I*a^3*tan(d*x + c)^3 - 80 *a^3*tan(d*x + c)^2 + 45*I*a^3*tan(d*x + c) + 12*a^3)/tan(d*x + c)^5)/d
Time = 1.47 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.68 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 190 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 660 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7680 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 3840 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2460 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {-8768 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2460 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 660 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 190 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]
1/960*(6*a^3*tan(1/2*d*x + 1/2*c)^5 - 45*I*a^3*tan(1/2*d*x + 1/2*c)^4 - 19 0*a^3*tan(1/2*d*x + 1/2*c)^3 + 660*I*a^3*tan(1/2*d*x + 1/2*c)^2 - 7680*I*a ^3*log(tan(1/2*d*x + 1/2*c) + I) + 3840*I*a^3*log(tan(1/2*d*x + 1/2*c)) + 2460*a^3*tan(1/2*d*x + 1/2*c) + (-8768*I*a^3*tan(1/2*d*x + 1/2*c)^5 - 2460 *a^3*tan(1/2*d*x + 1/2*c)^4 + 660*I*a^3*tan(1/2*d*x + 1/2*c)^3 + 190*a^3*t an(1/2*d*x + 1/2*c)^2 - 45*I*a^3*tan(1/2*d*x + 1/2*c) - 6*a^3)/tan(1/2*d*x + 1/2*c)^5)/d
Time = 4.92 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {8\,a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {4\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4-a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,2{}\mathrm {i}-\frac {4\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3}+\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{4}+\frac {a^3}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]