\(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx\) [937]

Optimal result

Integrand size = 31, antiderivative size = 114 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\frac {5 x}{16 a^3 c^2}+\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f} \] Output:

5/16*x/a^3/c^2+1/6*I*cos(f*x+e)^6/a^3/c^2/f+5/16*cos(f*x+e)*sin(f*x+e)/a^3 
/c^2/f+5/24*cos(f*x+e)^3*sin(f*x+e)/a^3/c^2/f+1/6*cos(f*x+e)^5*sin(f*x+e)/ 
a^3/c^2/f
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=-\frac {\sec ^5(e+f x) (-80 \cos (e+f x)+15 \cos (3 (e+f x))+\cos (5 (e+f x))+120 i \arctan (\tan (e+f x)) (\cos (e+f x)+i \sin (e+f x))+40 i \sin (e+f x)+45 i \sin (3 (e+f x))+5 i \sin (5 (e+f x)))}{384 a^3 c^2 f (-i+\tan (e+f x))^3 (i+\tan (e+f x))^2} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]
 

Output:

-1/384*(Sec[e + f*x]^5*(-80*Cos[e + f*x] + 15*Cos[3*(e + f*x)] + Cos[5*(e 
+ f*x)] + (120*I)*ArcTan[Tan[e + f*x]]*(Cos[e + f*x] + I*Sin[e + f*x]) + ( 
40*I)*Sin[e + f*x] + (45*I)*Sin[3*(e + f*x)] + (5*I)*Sin[5*(e + f*x)]))/(a 
^3*c^2*f*(-I + Tan[e + f*x])^3*(I + Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4005, 3042, 3967, 3042, 3115, 3042, 3115, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

\(\displaystyle \frac {c \int \cos ^6(e+f x)dx+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \int \sin \left (e+f x+\frac {\pi }{2}\right )^6dx+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {c \left (\frac {5}{6} \int \cos ^4(e+f x)dx+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \left (\frac {5}{6} \int \sin \left (e+f x+\frac {\pi }{2}\right )^4dx+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {c \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(e+f x)dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {c \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}\right )+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {c \left (\frac {\sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac {5}{6} \left (\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3}{4} \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )\right )\right )+\frac {i c \cos ^6(e+f x)}{6 f}}{a^3 c^3}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]
 

Output:

(((I/6)*c*Cos[e + f*x]^6)/f + c*((Cos[e + f*x]^5*Sin[e + f*x])/(6*f) + (5* 
((Cos[e + f*x]^3*Sin[e + f*x])/(4*f) + (3*(x/2 + (Cos[e + f*x]*Sin[e + f*x 
])/(2*f)))/4))/6))/(a^3*c^3)
 

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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 4005

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Sec[e + f*x]^(2*m)*(c + 
 d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[ 
m, 0] || GtQ[m, n]))
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {5 x}{16 a^{3} c^{2}}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )}}{192 a^{3} c^{2} f}+\frac {i \cos \left (4 f x +4 e \right )}{32 a^{3} c^{2} f}+\frac {3 \sin \left (4 f x +4 e \right )}{64 a^{3} c^{2} f}+\frac {5 i \cos \left (2 f x +2 e \right )}{64 a^{3} c^{2} f}+\frac {15 \sin \left (2 f x +2 e \right )}{64 a^{3} c^{2} f}\)	\(114\)
derivativedivides	\(\frac {i}{32 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{3} c^{2}}+\frac {1}{8 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )}-\frac {3 i}{32 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{24 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )}\)	\(132\)
default	\(\frac {i}{32 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{3} c^{2}}+\frac {1}{8 f \,a^{3} c^{2} \left (i+\tan \left (f x +e \right )\right )}-\frac {3 i}{32 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{24 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f \,a^{3} c^{2} \left (-i+\tan \left (f x +e \right )\right )}\)	\(132\)
norman	\(\frac {\frac {5 x}{16 a c}+\frac {i}{6 a c f}+\frac {11 \tan \left (f x +e \right )}{16 a c f}+\frac {5 \tan \left (f x +e \right )^{3}}{6 a c f}+\frac {5 \tan \left (f x +e \right )^{5}}{16 a c f}+\frac {15 x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {15 x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {5 x \tan \left (f x +e \right )^{6}}{16 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a^{2} c}\)	\(148\)

Input:

int(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

5/16*x/a^3/c^2+1/192*I/a^3/c^2/f*exp(-6*I*(f*x+e))+1/32*I/a^3/c^2/f*cos(4* 
f*x+4*e)+3/64/a^3/c^2/f*sin(4*f*x+4*e)+5/64*I/a^3/c^2/f*cos(2*f*x+2*e)+15/ 
64/a^3/c^2/f*sin(2*f*x+2*e)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\frac {{\left (120 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 60 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{2} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas 
")
 

Output:

1/384*(120*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(10*I*f*x + 10*I*e) - 30*I*e^(8 
*I*f*x + 8*I*e) + 60*I*e^(4*I*f*x + 4*I*e) + 15*I*e^(2*I*f*x + 2*I*e) + 2* 
I)*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.26 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (- 50331648 i a^{12} c^{8} f^{4} e^{16 i e} e^{4 i f x} - 503316480 i a^{12} c^{8} f^{4} e^{14 i e} e^{2 i f x} + 1006632960 i a^{12} c^{8} f^{4} e^{10 i e} e^{- 2 i f x} + 251658240 i a^{12} c^{8} f^{4} e^{8 i e} e^{- 4 i f x} + 33554432 i a^{12} c^{8} f^{4} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{6442450944 a^{15} c^{10} f^{5}} & \text {for}\: a^{15} c^{10} f^{5} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 6 i e}}{32 a^{3} c^{2}} - \frac {5}{16 a^{3} c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{3} c^{2}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)
 

Output:

Piecewise(((-50331648*I*a**12*c**8*f**4*exp(16*I*e)*exp(4*I*f*x) - 5033164 
80*I*a**12*c**8*f**4*exp(14*I*e)*exp(2*I*f*x) + 1006632960*I*a**12*c**8*f* 
*4*exp(10*I*e)*exp(-2*I*f*x) + 251658240*I*a**12*c**8*f**4*exp(8*I*e)*exp( 
-4*I*f*x) + 33554432*I*a**12*c**8*f**4*exp(6*I*e)*exp(-6*I*f*x))*exp(-12*I 
*e)/(6442450944*a**15*c**10*f**5), Ne(a**15*c**10*f**5*exp(12*I*e), 0)), ( 
x*((exp(10*I*e) + 5*exp(8*I*e) + 10*exp(6*I*e) + 10*exp(4*I*e) + 5*exp(2*I 
*e) + 1)*exp(-6*I*e)/(32*a**3*c**2) - 5/(16*a**3*c**2)), True)) + 5*x/(16* 
a**3*c**2)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\frac {5 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{32 \, a^{3} c^{2} f} - \frac {5 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{32 \, a^{3} c^{2} f} + \frac {i \, {\left (-15 i \, \tan \left (f x + e\right )^{4} - 15 \, \tan \left (f x + e\right )^{3} - 25 i \, \tan \left (f x + e\right )^{2} - 25 \, \tan \left (f x + e\right ) - 8 i\right )}}{48 \, a^{3} c^{2} f {\left (\tan \left (f x + e\right ) + i\right )}^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{3}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

5/32*I*log(tan(f*x + e) + I)/(a^3*c^2*f) - 5/32*I*log(tan(f*x + e) - I)/(a 
^3*c^2*f) + 1/48*I*(-15*I*tan(f*x + e)^4 - 15*tan(f*x + e)^3 - 25*I*tan(f* 
x + e)^2 - 25*tan(f*x + e) - 8*I)/(a^3*c^2*f*(tan(f*x + e) + I)^2*(tan(f*x 
 + e) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 2.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\frac {5\,x}{16\,a^3\,c^2}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}}{16}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,25{}\mathrm {i}}{48}+\frac {25\,\mathrm {tan}\left (e+f\,x\right )}{48}+\frac {1}{6}{}\mathrm {i}}{a^3\,c^2\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^2} \] Input:

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

Output:

(5*x)/(16*a^3*c^2) - ((25*tan(e + f*x))/48 + (tan(e + f*x)^2*25i)/48 + (5* 
tan(e + f*x)^3)/16 + (tan(e + f*x)^4*5i)/16 + 1i/6)/(a^3*c^2*f*(tan(e + f* 
x)*1i + 1)^3*(tan(e + f*x) + 1i)^2)
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx=\frac {\int \frac {1}{\tan \left (f x +e \right )^{5} i +\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right ) i +1}d x}{a^{3} c^{2}} \] Input:

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

Output:

int(1/(tan(e + f*x)**5*i + tan(e + f*x)**4 + 2*tan(e + f*x)**3*i + 2*tan(e 
 + f*x)**2 + tan(e + f*x)*i + 1),x)/(a**3*c**2)