\(\int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx\) [1082]

Optimal result

Integrand size = 28, antiderivative size = 140 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\frac {(c-i d)^3 x}{8 a^3}+\frac {(c+i d) (c-3 i d) (i c+d)}{8 a^3 f (1+i \tan (e+f x))}+\frac {(c+i d)^2 (i c+d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3} \] Output:

1/8*(c-I*d)^3*x/a^3+1/8*(c+I*d)*(c-3*I*d)*(I*c+d)/a^3/f/(1+I*tan(f*x+e))+1 
/8*(c+I*d)^2*(I*c+d)/a/f/(a+I*a*tan(f*x+e))^2+1/6*I*(c+d*tan(f*x+e))^3/f/( 
a+I*a*tan(f*x+e))^3
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.93 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\frac {\sec ^3(e+f x) \left (-9 \left (3 c^3-i c^2 d+3 c d^2-i d^3\right ) \cos (e+f x)-13 c^3 \cos (3 (e+f x))+15 i c^2 d \cos (3 (e+f x))+15 c d^2 \cos (3 (e+f x))+3 i d^3 \cos (3 (e+f x))-9 i c^3 \sin (e+f x)-27 c^2 d \sin (e+f x)-9 i c d^2 \sin (e+f x)-27 d^3 \sin (e+f x)-12 (i c+d)^3 \arctan (\tan (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-9 i c^3 \sin (3 (e+f x))-27 c^2 d \sin (3 (e+f x))+3 i c d^2 \sin (3 (e+f x))+d^3 \sin (3 (e+f x))\right )}{96 a^3 f (-i+\tan (e+f x))^3} \] Input:

Integrate[(c + d*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

(Sec[e + f*x]^3*(-9*(3*c^3 - I*c^2*d + 3*c*d^2 - I*d^3)*Cos[e + f*x] - 13* 
c^3*Cos[3*(e + f*x)] + (15*I)*c^2*d*Cos[3*(e + f*x)] + 15*c*d^2*Cos[3*(e + 
 f*x)] + (3*I)*d^3*Cos[3*(e + f*x)] - (9*I)*c^3*Sin[e + f*x] - 27*c^2*d*Si 
n[e + f*x] - (9*I)*c*d^2*Sin[e + f*x] - 27*d^3*Sin[e + f*x] - 12*(I*c + d) 
^3*ArcTan[Tan[e + f*x]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) - (9*I)*c^ 
3*Sin[3*(e + f*x)] - 27*c^2*d*Sin[3*(e + f*x)] + (3*I)*c*d^2*Sin[3*(e + f* 
x)] + d^3*Sin[3*(e + f*x)]))/(96*a^3*f*(-I + Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4029, 3042, 4023, 3042, 4009, 24}

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.

\(\displaystyle \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4029

\(\displaystyle \frac {(c-i d) \int \frac {(c+d \tan (e+f x))^2}{(i \tan (e+f x) a+a)^2}dx}{2 a}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c-i d) \int \frac {(c+d \tan (e+f x))^2}{(i \tan (e+f x) a+a)^2}dx}{2 a}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4023

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

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

\(\Big \downarrow \) 24

\(\displaystyle \frac {(c-i d) \left (\frac {\frac {(c+i d) (3 d+i c)}{2 f (1+i \tan (e+f x))}+\frac {1}{2} x (c-i d)^2}{2 a^2}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}\)

Input:

Int[(c + d*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

((I/6)*(c + d*Tan[e + f*x])^3)/(f*(a + I*a*Tan[e + f*x])^3) + ((c - I*d)*( 
(((c - I*d)^2*x)/2 + ((c + I*d)*(I*c + 3*d))/(2*f*(1 + I*Tan[e + f*x])))/( 
2*a^2) + ((I/4)*(c + I*d)^2)/(f*(a + I*a*Tan[e + f*x])^2)))/(2*a)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 4009

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/(2*a 
*f*m)), x] + Simp[(b*c + a*d)/(2*a*b)   Int[(a + b*Tan[e + f*x])^(m + 1), x 
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 
, 0] && LtQ[m, 0]
 

rule 4023

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ 
m/(2*a^3*f*m)), x] + Simp[1/(2*a^2)   Int[(a + b*Tan[e + f*x])^(m + 1)*Simp 
[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, 
 c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
 

rule 4029

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^n/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (122 ) = 244\).

Time = 0.22 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.10

method	result	size
risch	\(\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} c^{3}}{32 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} c \,d^{2}}{32 a^{3} f}+\frac {x \,c^{3}}{8 a^{3}}-\frac {3 x c \,d^{2}}{8 a^{3}}+\frac {3 \,{\mathrm e}^{-2 i \left (f x +e \right )} c^{2} d}{16 a^{3} f}+\frac {3 \,{\mathrm e}^{-2 i \left (f x +e \right )} d^{3}}{16 a^{3} f}+\frac {i x \,d^{3}}{8 a^{3}}-\frac {3 i x \,c^{2} d}{8 a^{3}}-\frac {3 \,{\mathrm e}^{-4 i \left (f x +e \right )} c^{2} d}{32 a^{3} f}-\frac {3 \,{\mathrm e}^{-4 i \left (f x +e \right )} d^{3}}{32 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} c^{3}}{48 a^{3} f}-\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} c \,d^{2}}{16 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} c^{2} d}{16 a^{3} f}+\frac {{\mathrm e}^{-6 i \left (f x +e \right )} d^{3}}{48 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c^{3}}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c \,d^{2}}{16 a^{3} f}\)	\(294\)
derivativedivides	\(-\frac {c^{3}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {c^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {7 i d^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 i c^{2} d \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2} d}{2 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 d^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {3 i c^{2} d}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 c^{2} d}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}+\frac {c \,d^{2}}{2 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {9 i c \,d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {i d^{3}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {3 c \,d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i d^{3} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {3 c \,d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}\)	\(354\)
default	\(-\frac {c^{3}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {c^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {7 i d^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 i c^{2} d \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2} d}{2 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 d^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {3 i c^{2} d}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 c^{2} d}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}+\frac {c \,d^{2}}{2 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {9 i c \,d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {i d^{3}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {3 c \,d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i d^{3} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {3 c \,d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}\)	\(354\)

Input:

int((c+d*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

3/32*I/a^3/f*exp(-4*I*(f*x+e))*c^3+3/32*I/a^3/f*exp(-4*I*(f*x+e))*c*d^2+1/ 
8*x/a^3*c^3-3/8*x/a^3*c*d^2+3/16/a^3/f*exp(-2*I*(f*x+e))*c^2*d+3/16/a^3/f* 
exp(-2*I*(f*x+e))*d^3+1/8*I*x/a^3*d^3-3/8*I*x/a^3*c^2*d-3/32/a^3/f*exp(-4* 
I*(f*x+e))*c^2*d-3/32/a^3/f*exp(-4*I*(f*x+e))*d^3+1/48*I/a^3/f*exp(-6*I*(f 
*x+e))*c^3-1/16*I/a^3/f*exp(-6*I*(f*x+e))*c*d^2-1/16/a^3/f*exp(-6*I*(f*x+e 
))*c^2*d+1/48/a^3/f*exp(-6*I*(f*x+e))*d^3+3/16*I/a^3/f*exp(-2*I*(f*x+e))*c 
^3+3/16*I/a^3/f*exp(-2*I*(f*x+e))*c*d^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, {\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{3} - 6 \, c^{2} d - 6 i \, c d^{2} + 2 \, d^{3} - 18 \, {\left (-i \, c^{3} - c^{2} d - i \, c d^{2} - d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 9 \, {\left (-i \, c^{3} + c^{2} d - i \, c d^{2} + d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \] Input:

integrate((c+d*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

1/96*(12*(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*f*x*e^(6*I*f*x + 6*I*e) + 2*I 
*c^3 - 6*c^2*d - 6*I*c*d^2 + 2*d^3 - 18*(-I*c^3 - c^2*d - I*c*d^2 - d^3)*e 
^(4*I*f*x + 4*I*e) - 9*(-I*c^3 + c^2*d - I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e 
))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
 

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 552, normalized size of antiderivative = 3.94 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (512 i a^{6} c^{3} f^{2} e^{6 i e} - 1536 a^{6} c^{2} d f^{2} e^{6 i e} - 1536 i a^{6} c d^{2} f^{2} e^{6 i e} + 512 a^{6} d^{3} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{3} f^{2} e^{8 i e} - 2304 a^{6} c^{2} d f^{2} e^{8 i e} + 2304 i a^{6} c d^{2} f^{2} e^{8 i e} - 2304 a^{6} d^{3} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{3} f^{2} e^{10 i e} + 4608 a^{6} c^{2} d f^{2} e^{10 i e} + 4608 i a^{6} c d^{2} f^{2} e^{10 i e} + 4608 a^{6} d^{3} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {c^{3} - 3 i c^{2} d - 3 c d^{2} + i d^{3}}{8 a^{3}} + \frac {\left (c^{3} e^{6 i e} + 3 c^{3} e^{4 i e} + 3 c^{3} e^{2 i e} + c^{3} - 3 i c^{2} d e^{6 i e} - 3 i c^{2} d e^{4 i e} + 3 i c^{2} d e^{2 i e} + 3 i c^{2} d - 3 c d^{2} e^{6 i e} + 3 c d^{2} e^{4 i e} + 3 c d^{2} e^{2 i e} - 3 c d^{2} + i d^{3} e^{6 i e} - 3 i d^{3} e^{4 i e} + 3 i d^{3} e^{2 i e} - i d^{3}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c^{3} - 3 i c^{2} d - 3 c d^{2} + i d^{3}\right )}{8 a^{3}} \] Input:

integrate((c+d*tan(f*x+e))**3/(a+I*a*tan(f*x+e))**3,x)
 

Output:

Piecewise((((512*I*a**6*c**3*f**2*exp(6*I*e) - 1536*a**6*c**2*d*f**2*exp(6 
*I*e) - 1536*I*a**6*c*d**2*f**2*exp(6*I*e) + 512*a**6*d**3*f**2*exp(6*I*e) 
)*exp(-6*I*f*x) + (2304*I*a**6*c**3*f**2*exp(8*I*e) - 2304*a**6*c**2*d*f** 
2*exp(8*I*e) + 2304*I*a**6*c*d**2*f**2*exp(8*I*e) - 2304*a**6*d**3*f**2*ex 
p(8*I*e))*exp(-4*I*f*x) + (4608*I*a**6*c**3*f**2*exp(10*I*e) + 4608*a**6*c 
**2*d*f**2*exp(10*I*e) + 4608*I*a**6*c*d**2*f**2*exp(10*I*e) + 4608*a**6*d 
**3*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(24576*a**9*f**3), Ne(a* 
*9*f**3*exp(12*I*e), 0)), (x*(-(c**3 - 3*I*c**2*d - 3*c*d**2 + I*d**3)/(8* 
a**3) + (c**3*exp(6*I*e) + 3*c**3*exp(4*I*e) + 3*c**3*exp(2*I*e) + c**3 - 
3*I*c**2*d*exp(6*I*e) - 3*I*c**2*d*exp(4*I*e) + 3*I*c**2*d*exp(2*I*e) + 3* 
I*c**2*d - 3*c*d**2*exp(6*I*e) + 3*c*d**2*exp(4*I*e) + 3*c*d**2*exp(2*I*e) 
 - 3*c*d**2 + I*d**3*exp(6*I*e) - 3*I*d**3*exp(4*I*e) + 3*I*d**3*exp(2*I*e 
) - I*d**3)*exp(-6*I*e)/(8*a**3)), True)) + x*(c**3 - 3*I*c**2*d - 3*c*d** 
2 + I*d**3)/(8*a**3)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c+d*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{16 \, a^{3} f} - \frac {{\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{16 \, a^{3} f} - \frac {10 \, c^{3} - 6 i \, c^{2} d + 6 \, c d^{2} - 10 i \, d^{3} - 3 \, {\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} - 7 i \, d^{3}\right )} \tan \left (f x + e\right )^{2} + 9 \, {\left (i \, c^{3} + 3 \, c^{2} d + i \, c d^{2} + 3 \, d^{3}\right )} \tan \left (f x + e\right )}{24 \, a^{3} f {\left (\tan \left (f x + e\right ) - i\right )}^{3}} \] Input:

integrate((c+d*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/16*(-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*log(tan(f*x + e) + I)/(a^3*f) - 
 1/16*(I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*log(tan(f*x + e) - I)/(a^3*f) - 
1/24*(10*c^3 - 6*I*c^2*d + 6*c*d^2 - 10*I*d^3 - 3*(c^3 - 3*I*c^2*d - 3*c*d 
^2 - 7*I*d^3)*tan(f*x + e)^2 + 9*(I*c^3 + 3*c^2*d + I*c*d^2 + 3*d^3)*tan(f 
*x + e))/(a^3*f*(tan(f*x + e) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 2.47 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {5\,d^3}{12\,a^3}-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^3}{8\,a^3}-\frac {d^3\,9{}\mathrm {i}}{8\,a^3}+\frac {3\,c\,d^2}{8\,a^3}-\frac {c^2\,d\,9{}\mathrm {i}}{8\,a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {7\,d^3}{8\,a^3}+\frac {3\,c^2\,d}{8\,a^3}+\frac {c^3\,1{}\mathrm {i}}{8\,a^3}-\frac {c\,d^2\,3{}\mathrm {i}}{8\,a^3}\right )+\frac {c^2\,d}{4\,a^3}+\frac {c^3\,5{}\mathrm {i}}{12\,a^3}+\frac {c\,d^2\,1{}\mathrm {i}}{4\,a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{8\,a^3} \] Input:

int((c + d*tan(e + f*x))^3/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

((c^3*5i)/(12*a^3) - tan(e + f*x)*((3*c^3)/(8*a^3) - (d^3*9i)/(8*a^3) + (3 
*c*d^2)/(8*a^3) - (c^2*d*9i)/(8*a^3)) + (5*d^3)/(12*a^3) - tan(e + f*x)^2* 
((c^3*1i)/(8*a^3) + (7*d^3)/(8*a^3) - (c*d^2*3i)/(8*a^3) + (3*c^2*d)/(8*a^ 
3)) + (c*d^2*1i)/(4*a^3) + (c^2*d)/(4*a^3))/(f*(tan(e + f*x)*3i - 3*tan(e 
+ f*x)^2 - tan(e + f*x)^3*1i + 1)) + (x*(c*1i + d)^3*1i)/(8*a^3)
 

Reduce [F]

\[ \int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx=\frac {-\left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) d^{3}-3 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) c \,d^{2}-3 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) c^{2} d -\left (\int \frac {1}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) c^{3}}{a^{3}} \] Input:

int((c+d*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x)
 

Output:

( - int(tan(e + f*x)**3/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + 
 f*x)*i - 1),x)*d**3 - 3*int(tan(e + f*x)**2/(tan(e + f*x)**3*i + 3*tan(e 
+ f*x)**2 - 3*tan(e + f*x)*i - 1),x)*c*d**2 - 3*int(tan(e + f*x)/(tan(e + 
f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*c**2*d - int(1/(t 
an(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*c**3)/a**3