\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx\) [713]

Optimal result

Integrand size = 41, antiderivative size = 149 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {(2 A+i B) x}{8 a c^3}-\frac {A+i B}{16 a c^3 f (i-\tan (e+f x))}-\frac {A-i B}{12 a c^3 f (i+\tan (e+f x))^3}+\frac {i A}{8 a c^3 f (i+\tan (e+f x))^2}+\frac {3 A+i B}{16 a c^3 f (i+\tan (e+f x))} \]

[Out]

1/8*(2*A+I*B)*x/a/c^3+1/16*(-A-I*B)/a/c^3/f/(I-tan(f*x+e))+1/12*(-A+I*B)/a/c^3/f/(I+tan(f*x+e))^3+1/8*I*A/a/c^
3/f/(I+tan(f*x+e))^2+1/16*(3*A+I*B)/a/c^3/f/(I+tan(f*x+e))

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3669, 78, 209} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=-\frac {A+i B}{16 a c^3 f (-\tan (e+f x)+i)}+\frac {3 A+i B}{16 a c^3 f (\tan (e+f x)+i)}-\frac {A-i B}{12 a c^3 f (\tan (e+f x)+i)^3}+\frac {x (2 A+i B)}{8 a c^3}+\frac {i A}{8 a c^3 f (\tan (e+f x)+i)^2} \]

[In]

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

[Out]

((2*A + I*B)*x)/(8*a*c^3) - (A + I*B)/(16*a*c^3*f*(I - Tan[e + f*x])) - (A - I*B)/(12*a*c^3*f*(I + Tan[e + f*x
])^3) + ((I/8)*A)/(a*c^3*f*(I + Tan[e + f*x])^2) + (3*A + I*B)/(16*a*c^3*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*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^2 (c-i c x)^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {-A-i B}{16 a^2 c^4 (-i+x)^2}+\frac {A-i B}{4 a^2 c^4 (i+x)^4}-\frac {i A}{4 a^2 c^4 (i+x)^3}+\frac {-3 A-i B}{16 a^2 c^4 (i+x)^2}+\frac {2 A+i B}{8 a^2 c^4 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {A+i B}{16 a c^3 f (i-\tan (e+f x))}-\frac {A-i B}{12 a c^3 f (i+\tan (e+f x))^3}+\frac {i A}{8 a c^3 f (i+\tan (e+f x))^2}+\frac {3 A+i B}{16 a c^3 f (i+\tan (e+f x))}+\frac {(2 A+i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a c^3 f} \\ & = \frac {(2 A+i B) x}{8 a c^3}-\frac {A+i B}{16 a c^3 f (i-\tan (e+f x))}-\frac {A-i B}{12 a c^3 f (i+\tan (e+f x))^3}+\frac {i A}{8 a c^3 f (i+\tan (e+f x))^2}+\frac {3 A+i B}{16 a c^3 f (i+\tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.81 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {\sec ^3(e+f x) (9 i A \cos (e+f x)+(-1+2 \cos (2 (e+f x))) ((-i A+2 B) \cos (e+f x)-(2 A+i B) \sin (e+f x))-3 (2 A+i B) \arctan (\tan (e+f x)) \sec (e+f x) (\cos (2 (e+f x))-i \sin (2 (e+f x))))}{24 a c^3 f (-i+\tan (e+f x)) (i+\tan (e+f x))^3} \]

[In]

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

[Out]

(Sec[e + f*x]^3*((9*I)*A*Cos[e + f*x] + (-1 + 2*Cos[2*(e + f*x)])*(((-I)*A + 2*B)*Cos[e + f*x] - (2*A + I*B)*S
in[e + f*x]) - 3*(2*A + I*B)*ArcTan[Tan[e + f*x]]*Sec[e + f*x]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])))/(24*a
*c^3*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.31

method	result	size
risch	\(\frac {i x B}{8 a \,c^{3}}+\frac {x A}{4 a \,c^{3}}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{96 a \,c^{3} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A}{96 a \,c^{3} f}-\frac {{\mathrm e}^{4 i \left (f x +e \right )} B}{32 a \,c^{3} f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} A}{16 a \,c^{3} f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a \,c^{3} f}-\frac {5 i \cos \left (2 f x +2 e \right ) A}{32 a \,c^{3} f}+\frac {i \sin \left (2 f x +2 e \right ) B}{32 a \,c^{3} f}+\frac {7 \sin \left (2 f x +2 e \right ) A}{32 a \,c^{3} f}\)	\(195\)
derivativedivides	\(\frac {i A}{8 a \,c^{3} f \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {i B}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 A}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{3}}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f a \,c^{3}}+\frac {A}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}\)	\(206\)
default	\(\frac {i A}{8 a \,c^{3} f \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {i B}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 A}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{3}}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f a \,c^{3}}+\frac {A}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}\)	\(206\)
norman	\(\frac {\frac {\left (i B +2 A \right ) x}{8 a c}-\frac {4 i A +B}{12 a c f}+\frac {B \tan \left (f x +e \right )^{2}}{4 a c f}+\frac {\left (-i B +6 A \right ) \tan \left (f x +e \right )}{8 a c f}+\frac {\left (i B +2 A \right ) \tan \left (f x +e \right )^{3}}{3 a c f}+\frac {\left (i B +2 A \right ) \tan \left (f x +e \right )^{5}}{8 a c f}+\frac {3 \left (i B +2 A \right ) x \tan \left (f x +e \right )^{2}}{8 a c}+\frac {3 \left (i B +2 A \right ) x \tan \left (f x +e \right )^{4}}{8 a c}+\frac {\left (i B +2 A \right ) x \tan \left (f x +e \right )^{6}}{8 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} c^{2}}\)	\(226\)

[In]

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

[Out]

1/8*I*x/a/c^3*B+1/4*x/a/c^3*A-1/96/a/c^3/f*exp(6*I*(f*x+e))*B-1/96*I/a/c^3/f*exp(6*I*(f*x+e))*A-1/32/a/c^3/f*e
xp(4*I*(f*x+e))*B-1/16*I/a/c^3/f*exp(4*I*(f*x+e))*A-1/32/a/c^3/f*cos(2*f*x+2*e)*B-5/32*I/a/c^3/f*cos(2*f*x+2*e
)*A+1/32*I/a/c^3/f*sin(2*f*x+2*e)*B+7/32/a/c^3/f*sin(2*f*x+2*e)*A

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (12 \, {\left (2 \, A + i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 3 \, {\left (2 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 18 i \, A e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{96 \, a c^{3} f} \]

[In]

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

[Out]

1/96*(12*(2*A + I*B)*f*x*e^(2*I*f*x + 2*I*e) + (-I*A - B)*e^(8*I*f*x + 8*I*e) - 3*(2*I*A + B)*e^(6*I*f*x + 6*I
*e) - 18*I*A*e^(4*I*f*x + 4*I*e) + 3*I*A - 3*B)*e^(-2*I*f*x - 2*I*e)/(a*c^3*f)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (- 294912 i A a^{3} c^{9} f^{3} e^{4 i e} e^{2 i f x} + \left (49152 i A a^{3} c^{9} f^{3} - 49152 B a^{3} c^{9} f^{3}\right ) e^{- 2 i f x} + \left (- 98304 i A a^{3} c^{9} f^{3} e^{6 i e} - 49152 B a^{3} c^{9} f^{3} e^{6 i e}\right ) e^{4 i f x} + \left (- 16384 i A a^{3} c^{9} f^{3} e^{8 i e} - 16384 B a^{3} c^{9} f^{3} e^{8 i e}\right ) e^{6 i f x}\right ) e^{- 2 i e}}{1572864 a^{4} c^{12} f^{4}} & \text {for}\: a^{4} c^{12} f^{4} e^{2 i e} \neq 0 \\x \left (- \frac {2 A + i B}{8 a c^{3}} + \frac {\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 2 i e}}{16 a c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (2 A + i B\right )}{8 a c^{3}} \]

[In]

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

[Out]

Piecewise(((-294912*I*A*a**3*c**9*f**3*exp(4*I*e)*exp(2*I*f*x) + (49152*I*A*a**3*c**9*f**3 - 49152*B*a**3*c**9
*f**3)*exp(-2*I*f*x) + (-98304*I*A*a**3*c**9*f**3*exp(6*I*e) - 49152*B*a**3*c**9*f**3*exp(6*I*e))*exp(4*I*f*x)
+ (-16384*I*A*a**3*c**9*f**3*exp(8*I*e) - 16384*B*a**3*c**9*f**3*exp(8*I*e))*exp(6*I*f*x))*exp(-2*I*e)/(15728
64*a**4*c**12*f**4), Ne(a**4*c**12*f**4*exp(2*I*e), 0)), (x*(-(2*A + I*B)/(8*a*c**3) + (A*exp(8*I*e) + 4*A*exp
(6*I*e) + 6*A*exp(4*I*e) + 4*A*exp(2*I*e) + A - I*B*exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(2*I*e) + I*B)*ex
p(-2*I*e)/(16*a*c**3)), True)) + x*(2*A + I*B)/(8*a*c**3)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.61 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, {\left (-2 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{3}} + \frac {6 \, {\left (2 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{3}} + \frac {6 \, {\left (-2 i \, A \tan \left (f x + e\right ) + B \tan \left (f x + e\right ) - 3 \, A - 2 i \, B\right )}}{a c^{3} {\left (\tan \left (f x + e\right ) - i\right )}} + \frac {22 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 84 \, A \tan \left (f x + e\right )^{2} - 39 i \, B \tan \left (f x + e\right )^{2} - 114 i \, A \tan \left (f x + e\right ) + 45 \, B \tan \left (f x + e\right ) + 60 \, A + 9 i \, B}{a c^{3} {\left (\tan \left (f x + e\right ) + i\right )}^{3}}}{96 \, f} \]

[In]

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

[Out]

-1/96*(6*(-2*I*A + B)*log(tan(f*x + e) + I)/(a*c^3) + 6*(2*I*A - B)*log(tan(f*x + e) - I)/(a*c^3) + 6*(-2*I*A*
tan(f*x + e) + B*tan(f*x + e) - 3*A - 2*I*B)/(a*c^3*(tan(f*x + e) - I)) + (22*I*A*tan(f*x + e)^3 - 11*B*tan(f*
x + e)^3 - 84*A*tan(f*x + e)^2 - 39*I*B*tan(f*x + e)^2 - 114*I*A*tan(f*x + e) + 45*B*tan(f*x + e) + 60*A + 9*I
*B)/(a*c^3*(tan(f*x + e) + I)^3))/f

Mupad [B] (verification not implemented)

Time = 9.24 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {\frac {B}{12\,a\,c^3}+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {B}{4\,a\,c^3}+\frac {A\,1{}\mathrm {i}}{2\,a\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {A}{4\,a\,c^3}+\frac {B\,1{}\mathrm {i}}{8\,a\,c^3}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {A}{12\,a\,c^3}+\frac {B\,1{}\mathrm {i}}{24\,a\,c^3}\right )+\frac {A\,1{}\mathrm {i}}{3\,a\,c^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,2{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}-\frac {x\,\left (-B+A\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a\,c^3} \]

[In]

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

[Out]

(tan(e + f*x)^2*((A*1i)/(2*a*c^3) - B/(4*a*c^3)) - tan(e + f*x)*(A/(12*a*c^3) + (B*1i)/(24*a*c^3)) + tan(e + f
*x)^3*(A/(4*a*c^3) + (B*1i)/(8*a*c^3)) + (A*1i)/(3*a*c^3) + B/(12*a*c^3))/(f*(tan(e + f*x)*2i + tan(e + f*x)^3
*2i + tan(e + f*x)^4 - 1)) - (x*(A*2i - B)*1i)/(8*a*c^3)