\(\int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [56]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {25 x}{4 a^2}-\frac {6 i \log (\cos (c+d x))}{a^2 d}+\frac {25 \tan (c+d x)}{4 a^2 d}-\frac {3 i \tan ^2(c+d x)}{a^2 d}-\frac {25 \tan ^3(c+d x)}{12 a^2 d}+\frac {3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:

-25/4*x/a^2-6*I*ln(cos(d*x+c))/a^2/d+25/4*tan(d*x+c)/a^2/d-3*I*tan(d*x+c)^ 
2/a^2/d-25/12*tan(d*x+c)^3/a^2/d+3/2*I*tan(d*x+c)^4/a^2/d/(1+I*tan(d*x+c)) 
-1/4*tan(d*x+c)^5/d/(a+I*a*tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {-3 i (50+49 \log (i-\tan (c+d x))-\log (i+\tan (c+d x)))+6 (25+49 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan (c+d x)+3 i (-26+49 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan ^2(c+d x)+56 \tan ^3(c+d x)-8 i \tan ^4(c+d x)-8 \tan ^5(c+d x)}{24 a^2 d (-i+\tan (c+d x))^2} \] Input:

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

((-3*I)*(50 + 49*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]]) + 6*(25 + 
49*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x] + (3*I)*(-2 
6 + 49*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + 56* 
Tan[c + d*x]^3 - (8*I)*Tan[c + d*x]^4 - 8*Tan[c + d*x]^5)/(24*a^2*d*(-I + 
Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 4041, 25, 3042, 4078, 27, 3042, 4011, 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.

\(\displaystyle \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^6}{(a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan ^4(c+d x) (5 a-7 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\tan ^4(c+d x) (5 a-7 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^4 (5 a-7 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int 2 \tan ^3(c+d x) \left (25 \tan (c+d x) a^2+24 i a^2\right )dx}{2 a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan ^3(c+d x) \left (25 \tan (c+d x) a^2+24 i a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x)^3 \left (25 \tan (c+d x) a^2+24 i a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) \left (24 i a^2 \tan (c+d x)-25 a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\int \tan (c+d x)^2 \left (24 i a^2 \tan (c+d x)-25 a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x) \left (-25 \tan (c+d x) a^2-24 i a^2\right )dx+\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\frac {12 i a^2 \tan ^2(c+d x)}{d}}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x) \left (-25 \tan (c+d x) a^2-24 i a^2\right )dx+\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\frac {12 i a^2 \tan ^2(c+d x)}{d}}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4008

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {-24 i a^2 \int \tan (c+d x)dx+\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\frac {12 i a^2 \tan ^2(c+d x)}{d}-\frac {25 a^2 \tan (c+d x)}{d}+25 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {-24 i a^2 \int \tan (c+d x)dx+\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\frac {12 i a^2 \tan ^2(c+d x)}{d}-\frac {25 a^2 \tan (c+d x)}{d}+25 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {6 i \tan ^4(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {25 a^2 \tan ^3(c+d x)}{3 d}+\frac {12 i a^2 \tan ^2(c+d x)}{d}-\frac {25 a^2 \tan (c+d x)}{d}+\frac {24 i a^2 \log (\cos (c+d x))}{d}+25 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/4*Tan[c + d*x]^5/(d*(a + I*a*Tan[c + d*x])^2) + (((6*I)*Tan[c + d*x]^4) 
/(d*(1 + I*Tan[c + d*x])) - (25*a^2*x + ((24*I)*a^2*Log[Cos[c + d*x]])/d - 
 (25*a^2*Tan[c + d*x])/d + ((12*I)*a^2*Tan[c + d*x]^2)/d + (25*a^2*Tan[c + 
 d*x]^3)/(3*d))/a^2)/(4*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4008

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]
 

rule 4011

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]
 

rule 4041

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])
 

rule 4078

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]
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {49 x}{4 a^{2}}+\frac {5 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {12 c}{a^{2} d}+\frac {2 i \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{2 i \left (d x +c \right )}+13\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {6 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(120\)
derivativedivides	\(\frac {4 \tan \left (d x +c \right )}{a^{2} d}-\frac {\tan \left (d x +c \right )^{3}}{3 a^{2} d}-\frac {i \tan \left (d x +c \right )^{2}}{d \,a^{2}}+\frac {3 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}-\frac {25 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {11}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}\)	\(123\)
default	\(\frac {4 \tan \left (d x +c \right )}{a^{2} d}-\frac {\tan \left (d x +c \right )^{3}}{3 a^{2} d}-\frac {i \tan \left (d x +c \right )^{2}}{d \,a^{2}}+\frac {3 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}-\frac {25 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {11}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}\)	\(123\)
norman	\(\frac {-\frac {25 x}{4 a}+\frac {10 \tan \left (d x +c \right )^{5}}{3 a d}-\frac {\tan \left (d x +c \right )^{7}}{3 a d}-\frac {25 x \tan \left (d x +c \right )^{2}}{2 a}-\frac {25 x \tan \left (d x +c \right )^{4}}{4 a}+\frac {9 i}{2 a d}+\frac {25 \tan \left (d x +c \right )}{4 a d}+\frac {125 \tan \left (d x +c \right )^{3}}{12 a d}-\frac {i \tan \left (d x +c \right )^{6}}{a d}+\frac {6 i \tan \left (d x +c \right )^{2}}{a d}}{a \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {3 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}\)	\(178\)

Input:

int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-49/4*x/a^2+5/4*I/a^2/d*exp(-2*I*(d*x+c))-1/16*I/a^2/d*exp(-4*I*(d*x+c))-1 
2/a^2/d*c+2/3*I*(9*exp(4*I*(d*x+c))+18*exp(2*I*(d*x+c))+13)/d/a^2/(exp(2*I 
*(d*x+c))+1)^3-6*I/a^2/d*ln(exp(2*I*(d*x+c))+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.39 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {588 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 12 \, {\left (147 \, d x - 29 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, {\left (588 \, d x - 251 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (588 \, d x - 587 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 288 \, {\left (i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 3 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 51 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{48 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 3 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/48*(588*d*x*e^(10*I*d*x + 10*I*c) + 12*(147*d*x - 29*I)*e^(8*I*d*x + 8* 
I*c) + 3*(588*d*x - 251*I)*e^(6*I*d*x + 6*I*c) + (588*d*x - 587*I)*e^(4*I* 
d*x + 4*I*c) + 288*(I*e^(10*I*d*x + 10*I*c) + 3*I*e^(8*I*d*x + 8*I*c) + 3* 
I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1 
) - 51*I*e^(2*I*d*x + 2*I*c) + 3*I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 3*a^2*d 
*e^(8*I*d*x + 8*I*c) + 3*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4* 
I*c))
 

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.85 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {18 i e^{4 i c} e^{4 i d x} + 36 i e^{2 i c} e^{2 i d x} + 26 i}{3 a^{2} d e^{6 i c} e^{6 i d x} + 9 a^{2} d e^{4 i c} e^{4 i d x} + 9 a^{2} d e^{2 i c} e^{2 i d x} + 3 a^{2} d} + \begin {cases} \frac {\left (80 i a^{2} d e^{4 i c} e^{- 2 i d x} - 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 49 e^{4 i c} + 10 e^{2 i c} - 1\right ) e^{- 4 i c}}{4 a^{2}} + \frac {49}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {49 x}{4 a^{2}} - \frac {6 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \] Input:

integrate(tan(d*x+c)**6/(a+I*a*tan(d*x+c))**2,x)
 

Output:

(18*I*exp(4*I*c)*exp(4*I*d*x) + 36*I*exp(2*I*c)*exp(2*I*d*x) + 26*I)/(3*a* 
*2*d*exp(6*I*c)*exp(6*I*d*x) + 9*a**2*d*exp(4*I*c)*exp(4*I*d*x) + 9*a**2*d 
*exp(2*I*c)*exp(2*I*d*x) + 3*a**2*d) + Piecewise(((80*I*a**2*d*exp(4*I*c)* 
exp(-2*I*d*x) - 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4* 
d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((-49*exp(4*I*c) + 10*exp(2*I*c) - 
 1)*exp(-4*I*c)/(4*a**2) + 49/(4*a**2)), True)) - 49*x/(4*a**2) - 6*I*log( 
exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{8 \, a^{2} d} + \frac {49 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{8 \, a^{2} d} + \frac {11 \, \tan \left (d x + c\right ) - 10 i}{4 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} - \frac {a^{4} d^{2} \tan \left (d x + c\right )^{3} + 3 i \, a^{4} d^{2} \tan \left (d x + c\right )^{2} - 12 \, a^{4} d^{2} \tan \left (d x + c\right )}{3 \, a^{6} d^{3}} \] Input:

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/8*I*log(tan(d*x + c) + I)/(a^2*d) + 49/8*I*log(tan(d*x + c) - I)/(a^2*d 
) + 1/4*(11*tan(d*x + c) - 10*I)/(a^2*d*(tan(d*x + c) - I)^2) - 1/3*(a^4*d 
^2*tan(d*x + c)^3 + 3*I*a^4*d^2*tan(d*x + c)^2 - 12*a^4*d^2*tan(d*x + c))/ 
(a^6*d^3)
 

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,49{}\mathrm {i}}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}+\frac {4\,\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{a^2\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,a^2\,d}+\frac {\frac {5}{2\,a^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,11{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \] Input:

int(tan(c + d*x)^6/(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

(log(tan(c + d*x) - 1i)*49i)/(8*a^2*d) - (log(tan(c + d*x) + 1i)*1i)/(8*a^ 
2*d) + (4*tan(c + d*x))/(a^2*d) - (tan(c + d*x)^2*1i)/(a^2*d) - tan(c + d* 
x)^3/(3*a^2*d) + ((tan(c + d*x)*11i)/(4*a^2) + 5/(2*a^2))/(d*(2*tan(c + d* 
x) + tan(c + d*x)^2*1i - 1i))
 

Reduce [F]

\[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{6}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - int(tan(c + d*x)**6/(tan(c + d*x)**2 - 2*tan(c + d*x)*i - 1),x))/a**2