3.8.25 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx\) [725]

Optimal. Leaf size=221 \[ \frac {5 (3 A+i B) x}{64 a^2 c^4}-\frac {i A-B}{64 a^2 c^4 f (i-\tan (e+f x))^2}-\frac {5 A+3 i B}{64 a^2 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{32 a^2 c^4 f (i+\tan (e+f x))^4}-\frac {3 A-i B}{48 a^2 c^4 f (i+\tan (e+f x))^3}+\frac {3 i A}{32 a^2 c^4 f (i+\tan (e+f x))^2}+\frac {5 A+i B}{32 a^2 c^4 f (i+\tan (e+f x))} \]

[Out]

5/64*(3*A+I*B)*x/a^2/c^4+1/64*(-I*A+B)/a^2/c^4/f/(I-tan(f*x+e))^2+1/64*(-5*A-3*I*B)/a^2/c^4/f/(I-tan(f*x+e))+1
/32*(-I*A-B)/a^2/c^4/f/(I+tan(f*x+e))^4+1/48*(-3*A+I*B)/a^2/c^4/f/(I+tan(f*x+e))^3+3/32*I*A/a^2/c^4/f/(I+tan(f
*x+e))^2+1/32*(5*A+I*B)/a^2/c^4/f/(I+tan(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3669, 78, 209} \begin {gather*} -\frac {5 A+3 i B}{64 a^2 c^4 f (-\tan (e+f x)+i)}+\frac {5 A+i B}{32 a^2 c^4 f (\tan (e+f x)+i)}-\frac {-B+i A}{64 a^2 c^4 f (-\tan (e+f x)+i)^2}-\frac {3 A-i B}{48 a^2 c^4 f (\tan (e+f x)+i)^3}-\frac {B+i A}{32 a^2 c^4 f (\tan (e+f x)+i)^4}+\frac {5 x (3 A+i B)}{64 a^2 c^4}+\frac {3 i A}{32 a^2 c^4 f (\tan (e+f x)+i)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(3*A + I*B)*x)/(64*a^2*c^4) - (I*A - B)/(64*a^2*c^4*f*(I - Tan[e + f*x])^2) - (5*A + (3*I)*B)/(64*a^2*c^4*f
*(I - Tan[e + f*x])) - (I*A + B)/(32*a^2*c^4*f*(I + Tan[e + f*x])^4) - (3*A - I*B)/(48*a^2*c^4*f*(I + Tan[e +
f*x])^3) + (((3*I)/32)*A)/(a^2*c^4*f*(I + Tan[e + f*x])^2) + (5*A + I*B)/(32*a^2*c^4*f*(I + Tan[e + f*x]))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3669

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^5} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {i (A+i B)}{32 a^3 c^5 (-i+x)^3}+\frac {-5 A-3 i B}{64 a^3 c^5 (-i+x)^2}+\frac {i A+B}{8 a^3 c^5 (i+x)^5}+\frac {3 A-i B}{16 a^3 c^5 (i+x)^4}-\frac {3 i A}{16 a^3 c^5 (i+x)^3}+\frac {-5 A-i B}{32 a^3 c^5 (i+x)^2}+\frac {5 (3 A+i B)}{64 a^3 c^5 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A-B}{64 a^2 c^4 f (i-\tan (e+f x))^2}-\frac {5 A+3 i B}{64 a^2 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{32 a^2 c^4 f (i+\tan (e+f x))^4}-\frac {3 A-i B}{48 a^2 c^4 f (i+\tan (e+f x))^3}+\frac {3 i A}{32 a^2 c^4 f (i+\tan (e+f x))^2}+\frac {5 A+i B}{32 a^2 c^4 f (i+\tan (e+f x))}+\frac {(5 (3 A+i B)) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{64 a^2 c^4 f}\\ &=\frac {5 (3 A+i B) x}{64 a^2 c^4}-\frac {i A-B}{64 a^2 c^4 f (i-\tan (e+f x))^2}-\frac {5 A+3 i B}{64 a^2 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{32 a^2 c^4 f (i+\tan (e+f x))^4}-\frac {3 A-i B}{48 a^2 c^4 f (i+\tan (e+f x))^3}+\frac {3 i A}{32 a^2 c^4 f (i+\tan (e+f x))^2}+\frac {5 A+i B}{32 a^2 c^4 f (i+\tan (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.19, size = 232, normalized size = 1.05 \begin {gather*} \frac {\sec ^2(e+f x) (-i \cos (4 (e+f x))+\sin (4 (e+f x))) (-240 A+30 (A (-3-12 i f x)+B (i+4 f x)) \cos (2 (e+f x))+16 (3 A+4 i B) \cos (4 (e+f x))+3 A \cos (6 (e+f x))+9 i B \cos (6 (e+f x))-90 i A \sin (2 (e+f x))-30 B \sin (2 (e+f x))-360 A f x \sin (2 (e+f x))-120 i B f x \sin (2 (e+f x))-96 i A \sin (4 (e+f x))+32 B \sin (4 (e+f x))-9 i A \sin (6 (e+f x))+3 B \sin (6 (e+f x)))}{1536 a^2 c^4 f (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*((-I)*Cos[4*(e + f*x)] + Sin[4*(e + f*x)])*(-240*A + 30*(A*(-3 - (12*I)*f*x) + B*(I + 4*f*x))*
Cos[2*(e + f*x)] + 16*(3*A + (4*I)*B)*Cos[4*(e + f*x)] + 3*A*Cos[6*(e + f*x)] + (9*I)*B*Cos[6*(e + f*x)] - (90
*I)*A*Sin[2*(e + f*x)] - 30*B*Sin[2*(e + f*x)] - 360*A*f*x*Sin[2*(e + f*x)] - (120*I)*B*f*x*Sin[2*(e + f*x)] -
(96*I)*A*Sin[4*(e + f*x)] + 32*B*Sin[4*(e + f*x)] - (9*I)*A*Sin[6*(e + f*x)] + 3*B*Sin[6*(e + f*x)]))/(1536*a
^2*c^4*f*(-I + Tan[e + f*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.36, size = 170, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {\frac {3 i A}{32 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {3 A}{16}-\frac {i B}{16}}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}+\left (-\frac {5 B}{128}+\frac {15 i A}{128}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {\frac {i A}{8}+\frac {B}{8}}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {-\frac {5 A}{32}-\frac {i B}{32}}{i+\tan \left (f x +e \right )}+\left (\frac {5 B}{128}-\frac {15 i A}{128}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {\frac {i A}{32}-\frac {B}{32}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {5 A}{64}-\frac {3 i B}{64}}{-i+\tan \left (f x +e \right )}}{f \,a^{2} c^{4}}\)	\(170\)
default	\(\frac {\frac {3 i A}{32 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {3 A}{16}-\frac {i B}{16}}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}+\left (-\frac {5 B}{128}+\frac {15 i A}{128}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {\frac {i A}{8}+\frac {B}{8}}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {-\frac {5 A}{32}-\frac {i B}{32}}{i+\tan \left (f x +e \right )}+\left (\frac {5 B}{128}-\frac {15 i A}{128}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {\frac {i A}{32}-\frac {B}{32}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {5 A}{64}-\frac {3 i B}{64}}{-i+\tan \left (f x +e \right )}}{f \,a^{2} c^{4}}\)	\(170\)
norman	\(\frac {\frac {5 \left (i B +3 A \right ) x}{64 a c}-\frac {3 i A +B}{12 a c f}+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{6 a c f}+\frac {73 \left (i B +3 A \right ) \left (\tan ^{3}\left (f x +e \right )\right )}{192 a c f}+\frac {55 \left (i B +3 A \right ) \left (\tan ^{5}\left (f x +e \right )\right )}{192 a c f}+\frac {5 \left (i B +3 A \right ) \left (\tan ^{7}\left (f x +e \right )\right )}{64 a c f}+\frac {5 \left (i B +3 A \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{16 a c}+\frac {15 \left (i B +3 A \right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{32 a c}+\frac {5 \left (i B +3 A \right ) x \left (\tan ^{6}\left (f x +e \right )\right )}{16 a c}+\frac {5 \left (i B +3 A \right ) x \left (\tan ^{8}\left (f x +e \right )\right )}{64 a c}+\frac {\left (-5 i B +49 A \right ) \tan \left (f x +e \right )}{64 a c f}}{a \,c^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{4}}\)	\(281\)
risch	\(\frac {5 i x B}{64 a^{2} c^{4}}+\frac {15 x A}{64 a^{2} c^{4}}-\frac {{\mathrm e}^{8 i \left (f x +e \right )} B}{512 a^{2} c^{4} f}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )} A}{512 a^{2} c^{4} f}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{96 a^{2} c^{4} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A}{64 a^{2} c^{4} f}-\frac {3 \cos \left (4 f x +4 e \right ) B}{128 a^{2} c^{4} f}-\frac {7 i \cos \left (4 f x +4 e \right ) A}{128 a^{2} c^{4} f}-\frac {i \sin \left (4 f x +4 e \right ) B}{64 a^{2} c^{4} f}+\frac {\sin \left (4 f x +4 e \right ) A}{16 a^{2} c^{4} f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a^{2} c^{4} f}-\frac {7 i \cos \left (2 f x +2 e \right ) A}{64 a^{2} c^{4} f}+\frac {i \sin \left (2 f x +2 e \right ) B}{32 a^{2} c^{4} f}+\frac {13 \sin \left (2 f x +2 e \right ) A}{64 a^{2} c^{4} f}\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/a^2/c^4*(3/32*I*A/(I+tan(f*x+e))^2-1/3*(3/16*A-1/16*I*B)/(I+tan(f*x+e))^3+(-5/128*B+15/128*I*A)*ln(I+tan(f
*x+e))-1/4*(1/8*I*A+1/8*B)/(I+tan(f*x+e))^4-(-5/32*A-1/32*I*B)/(I+tan(f*x+e))+(5/128*B-15/128*I*A)*ln(-I+tan(f
*x+e))-1/2*(1/32*I*A-1/32*B)/(-I+tan(f*x+e))^2-(-5/64*A-3/64*I*B)/(-I+tan(f*x+e)))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

Fricas [A]
time = 0.83, size = 134, normalized size = 0.61 \begin {gather*} \frac {{\left (120 \, {\left (3 \, A + i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} - 8 \, {\left (3 i \, A + 2 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 30 \, {\left (3 i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 240 i \, A e^{\left (6 i \, f x + 6 i \, e\right )} - 24 \, {\left (-3 i \, A + 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, A - 6 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{1536 \, a^{2} c^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(120*(3*A + I*B)*f*x*e^(4*I*f*x + 4*I*e) - 3*(I*A + B)*e^(12*I*f*x + 12*I*e) - 8*(3*I*A + 2*B)*e^(10*I*
f*x + 10*I*e) - 30*(3*I*A + B)*e^(8*I*f*x + 8*I*e) - 240*I*A*e^(6*I*f*x + 6*I*e) - 24*(-3*I*A + 2*B)*e^(2*I*f*
x + 2*I*e) + 6*I*A - 6*B)*e^(-4*I*f*x - 4*I*e)/(a^2*c^4*f)

________________________________________________________________________________________

Sympy [A]
time = 0.62, size = 498, normalized size = 2.25 \begin {gather*} \begin {cases} \frac {\left (- 2061584302080 i A a^{10} c^{20} f^{5} e^{8 i e} e^{2 i f x} + \left (51539607552 i A a^{10} c^{20} f^{5} e^{2 i e} - 51539607552 B a^{10} c^{20} f^{5} e^{2 i e}\right ) e^{- 4 i f x} + \left (618475290624 i A a^{10} c^{20} f^{5} e^{4 i e} - 412316860416 B a^{10} c^{20} f^{5} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 773094113280 i A a^{10} c^{20} f^{5} e^{10 i e} - 257698037760 B a^{10} c^{20} f^{5} e^{10 i e}\right ) e^{4 i f x} + \left (- 206158430208 i A a^{10} c^{20} f^{5} e^{12 i e} - 137438953472 B a^{10} c^{20} f^{5} e^{12 i e}\right ) e^{6 i f x} + \left (- 25769803776 i A a^{10} c^{20} f^{5} e^{14 i e} - 25769803776 B a^{10} c^{20} f^{5} e^{14 i e}\right ) e^{8 i f x}\right ) e^{- 6 i e}}{13194139533312 a^{12} c^{24} f^{6}} & \text {for}\: a^{12} c^{24} f^{6} e^{6 i e} \neq 0 \\x \left (- \frac {15 A + 5 i B}{64 a^{2} c^{4}} + \frac {\left (A e^{12 i e} + 6 A e^{10 i e} + 15 A e^{8 i e} + 20 A e^{6 i e} + 15 A e^{4 i e} + 6 A e^{2 i e} + A - i B e^{12 i e} - 4 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{4 i e} + 4 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{64 a^{2} c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (15 A + 5 i B\right )}{64 a^{2} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2061584302080*I*A*a**10*c**20*f**5*exp(8*I*e)*exp(2*I*f*x) + (51539607552*I*A*a**10*c**20*f**5*ex
p(2*I*e) - 51539607552*B*a**10*c**20*f**5*exp(2*I*e))*exp(-4*I*f*x) + (618475290624*I*A*a**10*c**20*f**5*exp(4
*I*e) - 412316860416*B*a**10*c**20*f**5*exp(4*I*e))*exp(-2*I*f*x) + (-773094113280*I*A*a**10*c**20*f**5*exp(10
*I*e) - 257698037760*B*a**10*c**20*f**5*exp(10*I*e))*exp(4*I*f*x) + (-206158430208*I*A*a**10*c**20*f**5*exp(12
*I*e) - 137438953472*B*a**10*c**20*f**5*exp(12*I*e))*exp(6*I*f*x) + (-25769803776*I*A*a**10*c**20*f**5*exp(14*
I*e) - 25769803776*B*a**10*c**20*f**5*exp(14*I*e))*exp(8*I*f*x))*exp(-6*I*e)/(13194139533312*a**12*c**24*f**6)
, Ne(a**12*c**24*f**6*exp(6*I*e), 0)), (x*(-(15*A + 5*I*B)/(64*a**2*c**4) + (A*exp(12*I*e) + 6*A*exp(10*I*e) +
15*A*exp(8*I*e) + 20*A*exp(6*I*e) + 15*A*exp(4*I*e) + 6*A*exp(2*I*e) + A - I*B*exp(12*I*e) - 4*I*B*exp(10*I*e
) - 5*I*B*exp(8*I*e) + 5*I*B*exp(4*I*e) + 4*I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(64*a**2*c**4)), True)) + x*(15*
A + 5*I*B)/(64*a**2*c**4)

________________________________________________________________________________________

Giac [A]
time = 1.05, size = 241, normalized size = 1.09 \begin {gather*} \frac {\frac {60 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{4}} + \frac {60 \, {\left (-3 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{4}} - \frac {6 \, {\left (-45 i \, A \tan \left (f x + e\right )^{2} + 15 \, B \tan \left (f x + e\right )^{2} - 110 \, A \tan \left (f x + e\right ) - 42 i \, B \tan \left (f x + e\right ) + 69 i \, A - 31 \, B\right )}}{a^{2} c^{4} {\left (\tan \left (f x + e\right ) - i\right )}^{2}} + \frac {-375 i \, A \tan \left (f x + e\right )^{4} + 125 \, B \tan \left (f x + e\right )^{4} + 1740 \, A \tan \left (f x + e\right )^{3} + 548 i \, B \tan \left (f x + e\right )^{3} + 3114 i \, A \tan \left (f x + e\right )^{2} - 894 \, B \tan \left (f x + e\right )^{2} - 2604 \, A \tan \left (f x + e\right ) - 612 i \, B \tan \left (f x + e\right ) - 903 i \, A + 93 \, B}{a^{2} c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{1536 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(60*(3*I*A - B)*log(tan(f*x + e) + I)/(a^2*c^4) + 60*(-3*I*A + B)*log(tan(f*x + e) - I)/(a^2*c^4) - 6*(
-45*I*A*tan(f*x + e)^2 + 15*B*tan(f*x + e)^2 - 110*A*tan(f*x + e) - 42*I*B*tan(f*x + e) + 69*I*A - 31*B)/(a^2*
c^4*(tan(f*x + e) - I)^2) + (-375*I*A*tan(f*x + e)^4 + 125*B*tan(f*x + e)^4 + 1740*A*tan(f*x + e)^3 + 548*I*B*
tan(f*x + e)^3 + 3114*I*A*tan(f*x + e)^2 - 894*B*tan(f*x + e)^2 - 2604*A*tan(f*x + e) - 612*I*B*tan(f*x + e) -
903*I*A + 93*B)/(a^2*c^4*(tan(f*x + e) + I)^4))/f

________________________________________________________________________________________

Mupad [B]
time = 10.17, size = 247, normalized size = 1.12 \begin {gather*} \frac {\frac {B}{12\,a^2\,c^4}+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-\frac {5\,B}{32\,a^2\,c^4}+\frac {A\,15{}\mathrm {i}}{32\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {5\,A}{32\,a^2\,c^4}+\frac {B\,5{}\mathrm {i}}{96\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {15\,A}{64\,a^2\,c^4}+\frac {B\,5{}\mathrm {i}}{64\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {25\,B}{96\,a^2\,c^4}+\frac {A\,25{}\mathrm {i}}{32\,a^2\,c^4}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {17\,A}{64\,a^2\,c^4}+\frac {B\,17{}\mathrm {i}}{192\,a^2\,c^4}\right )+\frac {A\,1{}\mathrm {i}}{4\,a^2\,c^4}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+{\mathrm {tan}\left (e+f\,x\right )}^5\,2{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}-{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}+\frac {5\,x\,\left (3\,A+B\,1{}\mathrm {i}\right )}{64\,a^2\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e + f*x)^4*((A*15i)/(32*a^2*c^4) - (5*B)/(32*a^2*c^4)) - tan(e + f*x)*((17*A)/(64*a^2*c^4) + (B*17i)/(192
*a^2*c^4)) + tan(e + f*x)^3*((5*A)/(32*a^2*c^4) + (B*5i)/(96*a^2*c^4)) + tan(e + f*x)^5*((15*A)/(64*a^2*c^4) +
(B*5i)/(64*a^2*c^4)) + tan(e + f*x)^2*((A*25i)/(32*a^2*c^4) - (25*B)/(96*a^2*c^4)) + (A*1i)/(4*a^2*c^4) + B/(
12*a^2*c^4))/(f*(tan(e + f*x)*2i - tan(e + f*x)^2 + tan(e + f*x)^3*4i + tan(e + f*x)^4 + tan(e + f*x)^5*2i + t
an(e + f*x)^6 - 1)) + (5*x*(3*A + B*1i))/(64*a^2*c^4)

________________________________________________________________________________________