Integrand size = 24, antiderivative size = 161 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {55 x}{8 a^3}+\frac {7 i \log (\cos (c+d x))}{a^3 d}-\frac {55 \tan (c+d x)}{8 a^3 d}+\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
55/8*x/a^3+7*I*ln(cos(d*x+c))/a^3/d-55/8*tan(d*x+c)/a^3/d+7/2*I*tan(d*x+c) ^2/a^3/d-1/6*tan(d*x+c)^5/d/(a+I*a*tan(d*x+c))^3+13/24*I*tan(d*x+c)^4/a/d/ (a+I*a*tan(d*x+c))^2+55/24*tan(d*x+c)^3/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.54 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {3 (56+111 \log (i-\tan (c+d x))+\log (i+\tan (c+d x)))+3 i (58+333 \log (i-\tan (c+d x))+3 \log (i+\tan (c+d x))) \tan (c+d x)+(318-999 \log (i-\tan (c+d x))-9 \log (i+\tan (c+d x))) \tan ^2(c+d x)-i (-428+333 \log (i-\tan (c+d x))+3 \log (i+\tan (c+d x))) \tan ^3(c+d x)-72 \tan ^4(c+d x)+24 i \tan ^5(c+d x)}{48 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]
Output:
(3*(56 + 111*Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]]) + (3*I)*(58 + 333*Log[I - Tan[c + d*x]] + 3*Log[I + Tan[c + d*x]])*Tan[c + d*x] + (318 - 999*Log[I - Tan[c + d*x]] - 9*Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 - I*( -428 + 333*Log[I - Tan[c + d*x]] + 3*Log[I + Tan[c + d*x]])*Tan[c + d*x]^3 - 72*Tan[c + d*x]^4 + (24*I)*Tan[c + d*x]^5)/(48*a^3*d*(-I + Tan[c + d*x] )^3)
Time = 1.03 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4041, 25, 3042, 4078, 27, 3042, 4078, 27, 3042, 4011, 3042, 4008, 3042, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^6}{(a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^4(c+d x) (5 a-8 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\tan ^4(c+d x) (5 a-8 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^4 (5 a-8 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {2 \tan ^3(c+d x) \left (29 \tan (c+d x) a^2+26 i a^2\right )}{i \tan (c+d x) a+a}dx}{4 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^3(c+d x) \left (29 \tan (c+d x) a^2+26 i a^2\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x)^3 \left (29 \tan (c+d x) a^2+26 i a^2\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {\int -3 \tan ^2(c+d x) \left (55 a^3-56 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \int \tan ^2(c+d x) \left (55 a^3-56 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \int \tan (c+d x)^2 \left (55 a^3-56 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \left (\int \tan (c+d x) \left (55 \tan (c+d x) a^3+56 i a^3\right )dx-\frac {28 i a^3 \tan ^2(c+d x)}{d}\right )}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \left (\int \tan (c+d x) \left (55 \tan (c+d x) a^3+56 i a^3\right )dx-\frac {28 i a^3 \tan ^2(c+d x)}{d}\right )}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4008 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \left (56 i a^3 \int \tan (c+d x)dx-\frac {28 i a^3 \tan ^2(c+d x)}{d}+\frac {55 a^3 \tan (c+d x)}{d}-55 a^3 x\right )}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \left (56 i a^3 \int \tan (c+d x)dx-\frac {28 i a^3 \tan ^2(c+d x)}{d}+\frac {55 a^3 \tan (c+d x)}{d}-55 a^3 x\right )}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {\frac {13 i a \tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {3 \left (-\frac {28 i a^3 \tan ^2(c+d x)}{d}+\frac {55 a^3 \tan (c+d x)}{d}-\frac {56 i a^3 \log (\cos (c+d x))}{d}-55 a^3 x\right )}{2 a^2}-\frac {55 a^2 \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}}{2 a^2}}{6 a^2}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
Input:
Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]
Output:
-1/6*Tan[c + d*x]^5/(d*(a + I*a*Tan[c + d*x])^3) + ((((13*I)/4)*a*Tan[c + d*x]^4)/(d*(a + I*a*Tan[c + d*x])^2) - ((-55*a^2*Tan[c + d*x]^3)/(2*d*(a + I*a*Tan[c + d*x])) + (3*(-55*a^3*x - ((56*I)*a^3*Log[Cos[c + d*x]])/d + ( 55*a^3*Tan[c + d*x])/d - ((28*I)*a^3*Tan[c + d*x]^2)/d))/(2*a^2))/(2*a^2)) /(6*a^2)
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_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[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] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[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_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.75 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(-\frac {3 \tan \left (d x +c \right )}{a^{3} d}+\frac {i \tan \left (d x +c \right )^{2}}{2 a^{3} d}-\frac {7 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}+\frac {55 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {11 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {1}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {49}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}\) | \(126\) |
| default | \(-\frac {3 \tan \left (d x +c \right )}{a^{3} d}+\frac {i \tan \left (d x +c \right )^{2}}{2 a^{3} d}-\frac {7 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}+\frac {55 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {11 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {1}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {49}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}\) | \(126\) |
| risch | \(\frac {111 x}{8 a^{3}}-\frac {39 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {9 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}+\frac {14 c}{a^{3} d}-\frac {2 i \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}+3\right )}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\) | \(127\) |
| norman | \(\frac {\frac {55 x}{8 a}-\frac {121 \tan \left (d x +c \right )^{5}}{8 a d}-\frac {3 \tan \left (d x +c \right )^{7}}{a d}+\frac {165 x \tan \left (d x +c \right )^{2}}{8 a}+\frac {165 x \tan \left (d x +c \right )^{4}}{8 a}+\frac {55 x \tan \left (d x +c \right )^{6}}{8 a}-\frac {77 i}{12 a d}-\frac {55 \tan \left (d x +c \right )}{8 a d}-\frac {55 \tan \left (d x +c \right )^{3}}{3 a d}+\frac {i \tan \left (d x +c \right )^{8}}{2 a d}-\frac {63 i \tan \left (d x +c \right )^{2}}{4 a d}-\frac {21 i \tan \left (d x +c \right )^{4}}{2 d a}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}-\frac {7 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}\) | \(209\) |
Input:
int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-3*tan(d*x+c)/a^3/d+1/2*I*tan(d*x+c)^2/a^3/d-7/2*I/d/a^3*ln(1+tan(d*x+c)^2 )+55/8/d/a^3*arctan(tan(d*x+c))-11/8*I/d/a^3/(-I+tan(d*x+c))^2+1/6/d/a^3/( -I+tan(d*x+c))^3-49/8/d/a^3/(-I+tan(d*x+c))
Time = 0.12 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {1332 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 6 \, {\left (444 \, d x - 103 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 9 \, {\left (148 \, d x - 113 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 672 \, {\left (-i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 2 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 182 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i}{96 \, {\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 2 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \] Input:
integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/96*(1332*d*x*e^(10*I*d*x + 10*I*c) + 6*(444*d*x - 103*I)*e^(8*I*d*x + 8* I*c) + 9*(148*d*x - 113*I)*e^(6*I*d*x + 6*I*c) - 672*(-I*e^(10*I*d*x + 10* I*c) - 2*I*e^(8*I*d*x + 8*I*c) - I*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2 *I*c) + 1) - 182*I*e^(4*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2*I*c) - 2*I)/( a^3*d*e^(10*I*d*x + 10*I*c) + 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d *x + 6*I*c))
Time = 0.38 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.60 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {- 4 i e^{2 i c} e^{2 i d x} - 6 i}{a^{3} d e^{4 i c} e^{4 i d x} + 2 a^{3} d e^{2 i c} e^{2 i d x} + a^{3} d} + \begin {cases} \frac {\left (- 59904 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 6912 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (111 e^{6 i c} - 39 e^{4 i c} + 9 e^{2 i c} - 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {111}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {111 x}{8 a^{3}} + \frac {7 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \] Input:
integrate(tan(d*x+c)**6/(a+I*a*tan(d*x+c))**3,x)
Output:
(-4*I*exp(2*I*c)*exp(2*I*d*x) - 6*I)/(a**3*d*exp(4*I*c)*exp(4*I*d*x) + 2*a **3*d*exp(2*I*c)*exp(2*I*d*x) + a**3*d) + Piecewise(((-59904*I*a**6*d**2*e xp(10*I*c)*exp(-2*I*d*x) + 6912*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 512 *I*a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne( a**9*d**3*exp(12*I*c), 0)), (x*((111*exp(6*I*c) - 39*exp(4*I*c) + 9*exp(2* I*c) - 1)*exp(-6*I*c)/(8*a**3) - 111/(8*a**3)), True)) + 111*x/(8*a**3) + 7*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)
Exception generated. \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {111 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} - \frac {i \, {\left (-147 i \, \tan \left (d x + c\right )^{2} - 261 \, \tan \left (d x + c\right ) + 118 i\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} + \frac {i \, {\left (a^{3} d \tan \left (d x + c\right )^{2} + 6 i \, a^{3} d \tan \left (d x + c\right )\right )}}{2 \, a^{6} d^{2}} \] Input:
integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/16*I*log(tan(d*x + c) + I)/(a^3*d) - 111/16*I*log(tan(d*x + c) - I)/(a^ 3*d) - 1/24*I*(-147*I*tan(d*x + c)^2 - 261*tan(d*x + c) + 118*I)/(a^3*d*(t an(d*x + c) - I)^3) + 1/2*I*(a^3*d*tan(d*x + c)^2 + 6*I*a^3*d*tan(d*x + c) )/(a^6*d^2)
Time = 1.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {87\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^3}-\frac {59{}\mathrm {i}}{12\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,49{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,111{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}-\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{a^3\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a^3\,d} \] Input:
int(tan(c + d*x)^6/(a + a*tan(c + d*x)*1i)^3,x)
Output:
((87*tan(c + d*x))/(8*a^3) - 59i/(12*a^3) + (tan(c + d*x)^2*49i)/(8*a^3))/ (d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) - (log(ta n(c + d*x) - 1i)*111i)/(16*a^3*d) - (log(tan(c + d*x) + 1i)*1i)/(16*a^3*d) - (3*tan(c + d*x))/(a^3*d) + (tan(c + d*x)^2*1i)/(2*a^3*d)
\[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {30 \left (\int \frac {\tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3}-3 \tan \left (d x +c \right )^{2} i -3 \tan \left (d x +c \right )+i}d x \right ) d i +48 \left (\int \frac {\tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3}-3 \tan \left (d x +c \right )^{2} i -3 \tan \left (d x +c \right )+i}d x \right ) d -20 \left (\int \frac {1}{\tan \left (d x +c \right )^{3}-3 \tan \left (d x +c \right )^{2} i -3 \tan \left (d x +c \right )+i}d x \right ) d i -7 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) i +\tan \left (d x +c \right )^{2} i -6 \tan \left (d x +c \right )+26 d x}{2 a^{3} d} \] Input:
int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x)
Output:
(30*int(tan(c + d*x)**2/(tan(c + d*x)**3 - 3*tan(c + d*x)**2*i - 3*tan(c + d*x) + i),x)*d*i + 48*int(tan(c + d*x)/(tan(c + d*x)**3 - 3*tan(c + d*x)* *2*i - 3*tan(c + d*x) + i),x)*d - 20*int(1/(tan(c + d*x)**3 - 3*tan(c + d* x)**2*i - 3*tan(c + d*x) + i),x)*d*i - 7*log(tan(c + d*x)**2 + 1)*i + tan( c + d*x)**2*i - 6*tan(c + d*x) + 26*d*x)/(2*a**3*d)