\(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx\) [954]

Optimal result

Integrand size = 31, antiderivative size = 193 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {15 x}{64 a^2 c^4}-\frac {i}{32 a^2 f (c-i c \tan (e+f x))^4}-\frac {i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )} \] Output:

15/64*x/a^2/c^4-1/32*I/a^2/f/(c-I*c*tan(f*x+e))^4-1/16*I/a^2/c/f/(c-I*c*ta 
n(f*x+e))^3-3/32*I/a^2/f/(c^2-I*c^2*tan(f*x+e))^2+1/64*I/a^2/f/(c^2+I*c^2* 
tan(f*x+e))^2-5/32*I/a^2/f/(c^4-I*c^4*tan(f*x+e))+5/64*I/a^2/f/(c^4+I*c^4* 
tan(f*x+e))
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=-\frac {\sec ^6(e+f x) (-80 i-65 i \cos (2 (e+f x))+16 i \cos (4 (e+f x))+i \cos (6 (e+f x))+120 \arctan (\tan (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))-5 \sin (2 (e+f x))+32 \sin (4 (e+f x))+3 \sin (6 (e+f x)))}{512 a^2 c^4 f (-i+\tan (e+f x))^2 (i+\tan (e+f x))^4} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4),x]
 

Output:

-1/512*(Sec[e + f*x]^6*(-80*I - (65*I)*Cos[2*(e + f*x)] + (16*I)*Cos[4*(e 
+ f*x)] + I*Cos[6*(e + f*x)] + 120*ArcTan[Tan[e + f*x]]*(Cos[2*(e + f*x)] 
- I*Sin[2*(e + f*x)]) - 5*Sin[2*(e + f*x)] + 32*Sin[4*(e + f*x)] + 3*Sin[6 
*(e + f*x)]))/(a^2*c^4*f*(-I + Tan[e + f*x])^2*(I + Tan[e + f*x])^4)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4005, 3042, 3968, 54, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i c^3 \int \frac {1}{(c-i c \tan (e+f x))^5 (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {i c^3 \int \left (\frac {5}{32 c^6 (c-i c \tan (e+f x))^2}+\frac {5}{64 c^6 (i \tan (e+f x) c+c)^2}+\frac {3}{16 c^5 (c-i c \tan (e+f x))^3}+\frac {1}{32 c^5 (i \tan (e+f x) c+c)^3}+\frac {3}{16 c^4 (c-i c \tan (e+f x))^4}+\frac {1}{8 c^3 (c-i c \tan (e+f x))^5}+\frac {15}{64 c^6 \left (\tan ^2(e+f x) c^2+c^2\right )}\right )d(-i c \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i c^3 \left (-\frac {15 i \arctan (\tan (e+f x))}{64 c^7}-\frac {5}{32 c^6 (c-i c \tan (e+f x))}+\frac {5}{64 c^6 (c+i c \tan (e+f x))}-\frac {3}{32 c^5 (c-i c \tan (e+f x))^2}+\frac {1}{64 c^5 (c+i c \tan (e+f x))^2}-\frac {1}{16 c^4 (c-i c \tan (e+f x))^3}-\frac {1}{32 c^3 (c-i c \tan (e+f x))^4}\right )}{a^2 f}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4),x]
 

Output:

(I*c^3*((((-15*I)/64)*ArcTan[Tan[e + f*x]])/c^7 - 1/(32*c^3*(c - I*c*Tan[e 
 + f*x])^4) - 1/(16*c^4*(c - I*c*Tan[e + f*x])^3) - 3/(32*c^5*(c - I*c*Tan 
[e + f*x])^2) - 5/(32*c^6*(c - I*c*Tan[e + f*x])) + 1/(64*c^5*(c + I*c*Tan 
[e + f*x])^2) + 5/(64*c^6*(c + I*c*Tan[e + f*x]))))/(a^2*f)
 

Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(a^(m - 2)*b*f)   Subst[Int[(a - x)^(m/2 - 1)*(a + x 
)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && 
 EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
 

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 = 0.31 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {15 x}{64 a^{2} c^{4}}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )}}{512 a^{2} c^{4} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )}}{64 a^{2} c^{4} f}-\frac {7 i \cos \left (4 f x +4 e \right )}{128 a^{2} c^{4} f}+\frac {\sin \left (4 f x +4 e \right )}{16 a^{2} c^{4} f}-\frac {7 i \cos \left (2 f x +2 e \right )}{64 a^{2} c^{4} f}+\frac {13 \sin \left (2 f x +2 e \right )}{64 a^{2} c^{4} f}\)	\(135\)
derivativedivides	\(\frac {15 \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}-\frac {i}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {3 i}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {1}{16 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {5}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}\)	\(155\)
default	\(\frac {15 \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}-\frac {i}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {3 i}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {1}{16 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {5}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}\)	\(155\)
norman	\(\frac {\frac {15 x}{64 a c}+\frac {49 \tan \left (f x +e \right )}{64 a c f}+\frac {73 \tan \left (f x +e \right )^{3}}{64 a c f}+\frac {55 \tan \left (f x +e \right )^{5}}{64 a c f}+\frac {15 \tan \left (f x +e \right )^{7}}{64 a c f}+\frac {15 x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {45 x \tan \left (f x +e \right )^{4}}{32 a c}+\frac {15 x \tan \left (f x +e \right )^{6}}{16 a c}+\frac {15 x \tan \left (f x +e \right )^{8}}{64 a c}-\frac {i}{4 a c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{4} a \,c^{3}}\)	\(184\)

Input:

int(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)
 

Output:

15/64*x/a^2/c^4-1/512*I/a^2/c^4/f*exp(8*I*(f*x+e))-1/64*I/a^2/c^4/f*exp(6* 
I*(f*x+e))-7/128*I/a^2/c^4/f*cos(4*f*x+4*e)+1/16/a^2/c^4/f*sin(4*f*x+4*e)- 
7/64*I/a^2/c^4/f*cos(2*f*x+2*e)+13/64/a^2/c^4/f*sin(2*f*x+2*e)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {{\left (120 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 80 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 24 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{512 \, a^{2} c^{4} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas 
")
 

Output:

1/512*(120*f*x*e^(4*I*f*x + 4*I*e) - I*e^(12*I*f*x + 12*I*e) - 8*I*e^(10*I 
*f*x + 10*I*e) - 30*I*e^(8*I*f*x + 8*I*e) - 80*I*e^(6*I*f*x + 6*I*e) + 24* 
I*e^(2*I*f*x + 2*I*e) + 2*I)*e^(-4*I*f*x - 4*I*e)/(a^2*c^4*f)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\begin {cases} \frac {\left (- 8589934592 i a^{10} c^{20} f^{5} e^{14 i e} e^{8 i f x} - 68719476736 i a^{10} c^{20} f^{5} e^{12 i e} e^{6 i f x} - 257698037760 i a^{10} c^{20} f^{5} e^{10 i e} e^{4 i f x} - 687194767360 i a^{10} c^{20} f^{5} e^{8 i e} e^{2 i f x} + 206158430208 i a^{10} c^{20} f^{5} e^{4 i e} e^{- 2 i f x} + 17179869184 i a^{10} c^{20} f^{5} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{4398046511104 a^{12} c^{24} f^{6}} & \text {for}\: a^{12} c^{24} f^{6} e^{6 i e} \neq 0 \\x \left (\frac {\left (e^{12 i e} + 6 e^{10 i e} + 15 e^{8 i e} + 20 e^{6 i e} + 15 e^{4 i e} + 6 e^{2 i e} + 1\right ) e^{- 4 i e}}{64 a^{2} c^{4}} - \frac {15}{64 a^{2} c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {15 x}{64 a^{2} c^{4}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e))**4,x)
 

Output:

Piecewise(((-8589934592*I*a**10*c**20*f**5*exp(14*I*e)*exp(8*I*f*x) - 6871 
9476736*I*a**10*c**20*f**5*exp(12*I*e)*exp(6*I*f*x) - 257698037760*I*a**10 
*c**20*f**5*exp(10*I*e)*exp(4*I*f*x) - 687194767360*I*a**10*c**20*f**5*exp 
(8*I*e)*exp(2*I*f*x) + 206158430208*I*a**10*c**20*f**5*exp(4*I*e)*exp(-2*I 
*f*x) + 17179869184*I*a**10*c**20*f**5*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I* 
e)/(4398046511104*a**12*c**24*f**6), Ne(a**12*c**24*f**6*exp(6*I*e), 0)), 
(x*((exp(12*I*e) + 6*exp(10*I*e) + 15*exp(8*I*e) + 20*exp(6*I*e) + 15*exp( 
4*I*e) + 6*exp(2*I*e) + 1)*exp(-4*I*e)/(64*a**2*c**4) - 15/(64*a**2*c**4)) 
, True)) + 15*x/(64*a**2*c**4)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {15 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{128 \, a^{2} c^{4} f} - \frac {15 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{128 \, a^{2} c^{4} f} + \frac {15 \, \tan \left (f x + e\right )^{5} + 30 i \, \tan \left (f x + e\right )^{4} + 10 \, \tan \left (f x + e\right )^{3} + 50 i \, \tan \left (f x + e\right )^{2} - 17 \, \tan \left (f x + e\right ) + 16 i}{64 \, a^{2} c^{4} f {\left (\tan \left (f x + e\right ) + i\right )}^{4} {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")
 

Output:

15/128*I*log(tan(f*x + e) + I)/(a^2*c^4*f) - 15/128*I*log(tan(f*x + e) - I 
)/(a^2*c^4*f) + 1/64*(15*tan(f*x + e)^5 + 30*I*tan(f*x + e)^4 + 10*tan(f*x 
 + e)^3 + 50*I*tan(f*x + e)^2 - 17*tan(f*x + e) + 16*I)/(a^2*c^4*f*(tan(f* 
x + e) + I)^4*(tan(f*x + e) - I)^2)
 

Mupad [B] (verification not implemented)

Time = 3.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.51 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {15\,x}{64\,a^2\,c^4}-\frac {\frac {15\,{\mathrm {tan}\left (e+f\,x\right )}^5}{64}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,15{}\mathrm {i}}{32}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{32}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,25{}\mathrm {i}}{32}-\frac {17\,\mathrm {tan}\left (e+f\,x\right )}{64}+\frac {1}{4}{}\mathrm {i}}{a^2\,c^4\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \] Input:

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

Output:

(15*x)/(64*a^2*c^4) - ((tan(e + f*x)^2*25i)/32 - (17*tan(e + f*x))/64 + (5 
*tan(e + f*x)^3)/32 + (tan(e + f*x)^4*15i)/32 + (15*tan(e + f*x)^5)/64 + 1 
i/4)/(a^2*c^4*f*(tan(e + f*x)*1i + 1)^2*(tan(e + f*x) + 1i)^4)
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=-\frac {\int \frac {1}{\tan \left (f x +e \right )^{6}+2 \tan \left (f x +e \right )^{5} i +\tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i -\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x}{a^{2} c^{4}} \] Input:

int(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x)
 

Output:

( - int(1/(tan(e + f*x)**6 + 2*tan(e + f*x)**5*i + tan(e + f*x)**4 + 4*tan 
(e + f*x)**3*i - tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x))/(a**2*c**4)