\(\int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\) [1076]

Optimal result

Integrand size = 28, antiderivative size = 129 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {(c-i d)^2 x}{8 a^3}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \left (\frac {3 \sec ^2(e+f x) \left (2 c^2+2 d^2+2 c^2 \cos (2 (e+f x))+i c^2 \sin (2 (e+f x))+2 c d \sin (2 (e+f x))+i d^2 \sin (2 (e+f x))+2 (c-i d)^2 \arctan (\tan (e+f x)) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))\right )}{a^3 (-i+\tan (e+f x))^2}-\frac {8 (c+d \tan (e+f x))^3}{(c+i d) (a+i a \tan (e+f x))^3}\right )}{48 f} \] Input:

Integrate[(c + d*Tan[e + f*x])^2/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

((-1/48*I)*((3*Sec[e + f*x]^2*(2*c^2 + 2*d^2 + 2*c^2*Cos[2*(e + f*x)] + I* 
c^2*Sin[2*(e + f*x)] + 2*c*d*Sin[2*(e + f*x)] + I*d^2*Sin[2*(e + f*x)] + 2 
*(c - I*d)^2*ArcTan[Tan[e + f*x]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x) 
])))/(a^3*(-I + Tan[e + f*x])^2) - (8*(c + d*Tan[e + f*x])^3)/((c + I*d)*( 
a + I*a*Tan[e + f*x])^3)))/f
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4023, 3042, 4009, 3042, 3960, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4023

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3960

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

\(\Big \downarrow \) 24

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

Input:

Int[(c + d*Tan[e + f*x])^2/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

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

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 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a)   Int[(a + b*Tan[c + d*x])^ 
(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
 

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]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.64

method	result	size
risch	\(-\frac {i x c d}{4 a^{3}}+\frac {x \,c^{2}}{8 a^{3}}-\frac {x \,d^{2}}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} c d}{8 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c^{2}}{16 a^{3} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} d^{2}}{16 a^{3} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} c d}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} c^{2}}{32 a^{3} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} d^{2}}{32 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} c d}{24 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} c^{2}}{48 a^{3} f}-\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} d^{2}}{48 a^{3} f}\)	\(212\)
derivativedivides	\(-\frac {i c d}{4 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {c d}{4 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {c^{2}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}\)	\(255\)
default	\(-\frac {i c d}{4 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {c d}{4 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {c^{2}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}\)	\(255\)

Input:

int((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/4*I*x/a^3*c*d+1/8*x/a^3*c^2-1/8*x/a^3*d^2+1/8/a^3/f*exp(-2*I*(f*x+e))*c 
*d+3/16*I/a^3/f*exp(-2*I*(f*x+e))*c^2+1/16*I/a^3/f*exp(-2*I*(f*x+e))*d^2-1 
/16/a^3/f*exp(-4*I*(f*x+e))*c*d+3/32*I/a^3/f*exp(-4*I*(f*x+e))*c^2+1/32*I/ 
a^3/f*exp(-4*I*(f*x+e))*d^2-1/24/a^3/f*exp(-6*I*(f*x+e))*c*d+1/48*I/a^3/f* 
exp(-6*I*(f*x+e))*c^2-1/48*I/a^3/f*exp(-6*I*(f*x+e))*d^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, {\left (c^{2} - 2 i \, c d - d^{2}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{2} - 4 \, c d - 2 i \, d^{2} - 6 \, {\left (-3 i \, c^{2} - 2 \, c d - i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, c^{2} + 2 \, c d - i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \] Input:

integrate((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

1/96*(12*(c^2 - 2*I*c*d - d^2)*f*x*e^(6*I*f*x + 6*I*e) + 2*I*c^2 - 4*c*d - 
 2*I*d^2 - 6*(-3*I*c^2 - 2*c*d - I*d^2)*e^(4*I*f*x + 4*I*e) - 3*(-3*I*c^2 
+ 2*c*d - I*d^2)*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 = 0.34 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.11 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (512 i a^{6} c^{2} f^{2} e^{6 i e} - 1024 a^{6} c d f^{2} e^{6 i e} - 512 i a^{6} d^{2} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{2} f^{2} e^{8 i e} - 1536 a^{6} c d f^{2} e^{8 i e} + 768 i a^{6} d^{2} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{2} f^{2} e^{10 i e} + 3072 a^{6} c d f^{2} e^{10 i e} + 1536 i a^{6} d^{2} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {c^{2} - 2 i c d - d^{2}}{8 a^{3}} + \frac {\left (c^{2} e^{6 i e} + 3 c^{2} e^{4 i e} + 3 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{6 i e} - 2 i c d e^{4 i e} + 2 i c d e^{2 i e} + 2 i c d - d^{2} e^{6 i e} + d^{2} e^{4 i e} + d^{2} e^{2 i e} - d^{2}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c^{2} - 2 i c d - d^{2}\right )}{8 a^{3}} \] Input:

integrate((c+d*tan(f*x+e))**2/(a+I*a*tan(f*x+e))**3,x)
 

Output:

Piecewise((((512*I*a**6*c**2*f**2*exp(6*I*e) - 1024*a**6*c*d*f**2*exp(6*I* 
e) - 512*I*a**6*d**2*f**2*exp(6*I*e))*exp(-6*I*f*x) + (2304*I*a**6*c**2*f* 
*2*exp(8*I*e) - 1536*a**6*c*d*f**2*exp(8*I*e) + 768*I*a**6*d**2*f**2*exp(8 
*I*e))*exp(-4*I*f*x) + (4608*I*a**6*c**2*f**2*exp(10*I*e) + 3072*a**6*c*d* 
f**2*exp(10*I*e) + 1536*I*a**6*d**2*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**2 - 2*I 
*c*d - d**2)/(8*a**3) + (c**2*exp(6*I*e) + 3*c**2*exp(4*I*e) + 3*c**2*exp( 
2*I*e) + c**2 - 2*I*c*d*exp(6*I*e) - 2*I*c*d*exp(4*I*e) + 2*I*c*d*exp(2*I* 
e) + 2*I*c*d - d**2*exp(6*I*e) + d**2*exp(4*I*e) + d**2*exp(2*I*e) - d**2) 
*exp(-6*I*e)/(8*a**3)), True)) + x*(c**2 - 2*I*c*d - d**2)/(8*a**3)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c+d*tan(f*x+e))^2/(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 = 144, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{16 \, a^{3} f} - \frac {{\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{16 \, a^{3} f} + \frac {3 \, {\left (c^{2} - 2 i \, c d - d^{2}\right )} \tan \left (f x + e\right )^{2} - 10 \, c^{2} + 4 i \, c d - 2 \, d^{2} - 3 \, {\left (3 i \, c^{2} + 6 \, c d + i \, d^{2}\right )} \tan \left (f x + e\right )}{24 \, a^{3} f {\left (\tan \left (f x + e\right ) - i\right )}^{3}} \] Input:

integrate((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/16*(-I*c^2 - 2*c*d + I*d^2)*log(tan(f*x + e) + I)/(a^3*f) - 1/16*(I*c^2 
 + 2*c*d - I*d^2)*log(tan(f*x + e) - I)/(a^3*f) + 1/24*(3*(c^2 - 2*I*c*d - 
 d^2)*tan(f*x + e)^2 - 10*c^2 + 4*I*c*d - 2*d^2 - 3*(3*I*c^2 + 6*c*d + I*d 
^2)*tan(f*x + e))/(a^3*f*(tan(f*x + e) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 2.34 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {c\,d}{6\,a^3}-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^2}{8\,a^3}+\frac {d^2}{8\,a^3}-\frac {c\,d\,3{}\mathrm {i}}{4\,a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {c\,d}{4\,a^3}+\frac {c^2\,1{}\mathrm {i}}{8\,a^3}-\frac {d^2\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {c^2\,5{}\mathrm {i}}{12\,a^3}+\frac {d^2\,1{}\mathrm {i}}{12\,a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^2}{8\,a^3} \] Input:

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

Output:

((c^2*5i)/(12*a^3) - tan(e + f*x)*((3*c^2)/(8*a^3) + d^2/(8*a^3) - (c*d*3i 
)/(4*a^3)) - tan(e + f*x)^2*((c^2*1i)/(8*a^3) - (d^2*1i)/(8*a^3) + (c*d)/( 
4*a^3)) + (d^2*1i)/(12*a^3) + (c*d)/(6*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)^2)/(8*a^3)
 

Reduce [F]

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

int((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x)
 

Output:

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