3.722 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))} \, dx\)

Optimal. Leaf size=117 \[ \frac{A-i B}{8 a^2 c f (\tan (e+f x)+i)}-\frac{-B+i A}{8 a^2 c f (-\tan (e+f x)+i)^2}+\frac{x (3 A-i B)}{8 a^2 c}-\frac{A}{4 a^2 c f (-\tan (e+f x)+i)} \]

[Out]

((3*A - I*B)*x)/(8*a^2*c) - (I*A - B)/(8*a^2*c*f*(I - Tan[e + f*x])^2) - A/(4*a^2*c*f*(I - Tan[e + f*x])) + (A
- I*B)/(8*a^2*c*f*(I + Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.186812, antiderivative size = 117, normalized size of antiderivative = 1., 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 = {3588, 77, 203} \[ \frac{A-i B}{8 a^2 c f (\tan (e+f x)+i)}-\frac{-B+i A}{8 a^2 c f (-\tan (e+f x)+i)^2}+\frac{x (3 A-i B)}{8 a^2 c}-\frac{A}{4 a^2 c f (-\tan (e+f x)+i)} \]

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])),x]

[Out]

((3*A - I*B)*x)/(8*a^2*c) - (I*A - B)/(8*a^2*c*f*(I - Tan[e + f*x])^2) - A/(4*a^2*c*f*(I - Tan[e + f*x])) + (A
- I*B)/(8*a^2*c*f*(I + Tan[e + f*x]))

Rule 3588

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]

Rule 77

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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))} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^3 (c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{i (A+i B)}{4 a^3 c^2 (-i+x)^3}-\frac{A}{4 a^3 c^2 (-i+x)^2}+\frac{-A+i B}{8 a^3 c^2 (i+x)^2}+\frac{3 A-i B}{8 a^3 c^2 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i A-B}{8 a^2 c f (i-\tan (e+f x))^2}-\frac{A}{4 a^2 c f (i-\tan (e+f x))}+\frac{A-i B}{8 a^2 c f (i+\tan (e+f x))}+\frac{(3 A-i B) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^2 c f}\\ &=\frac{(3 A-i B) x}{8 a^2 c}-\frac{i A-B}{8 a^2 c f (i-\tan (e+f x))^2}-\frac{A}{4 a^2 c f (i-\tan (e+f x))}+\frac{A-i B}{8 a^2 c f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.03392, size = 129, normalized size = 1.1 \[ -\frac{2 (A-3 i B) \cos (2 (e+f x))+(B+3 i A) \sin (3 (e+f x)) \sec (e+f x)-12 A f x \tan (e+f x)+6 i A \tan (e+f x)+12 i A f x-7 A-2 B \tan (e+f x)+4 i B f x \tan (e+f x)+4 B f x+i B}{32 a^2 c f (\tan (e+f x)-i)} \]

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])),x]

[Out]

-(-7*A + I*B + (12*I)*A*f*x + 4*B*f*x + 2*(A - (3*I)*B)*Cos[2*(e + f*x)] + ((3*I)*A + B)*Sec[e + f*x]*Sin[3*(e
+ f*x)] + (6*I)*A*Tan[e + f*x] - 2*B*Tan[e + f*x] - 12*A*f*x*Tan[e + f*x] + (4*I)*B*f*x*Tan[e + f*x])/(32*a^2
*c*f*(-I + Tan[e + f*x]))

________________________________________________________________________________________

Maple [B]  time = 0.089, size = 209, normalized size = 1.8 \begin{align*}{\frac{B}{8\,f{a}^{2}c \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{8}}A}{f{a}^{2}c \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{A}{4\,f{a}^{2}c \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) A}{f{a}^{2}c}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{16\,f{a}^{2}c}}+{\frac{A}{8\,f{a}^{2}c \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{{\frac{i}{8}}B}{f{a}^{2}c \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) A}{f{a}^{2}c}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) B}{16\,f{a}^{2}c}} \end{align*}

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)),x)

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.07083, size = 216, normalized size = 1.85 \begin{align*} \frac{{\left (4 \,{\left (3 \, A - i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A - 2 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (6 i \, A - 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} c f} \end{align*}

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)),x, algorithm="fricas")

[Out]

1/32*(4*(3*A - I*B)*f*x*e^(4*I*f*x + 4*I*e) + (-2*I*A - 2*B)*e^(6*I*f*x + 6*I*e) + (6*I*A - 2*B)*e^(2*I*f*x +
2*I*e) + I*A - B)*e^(-4*I*f*x - 4*I*e)/(a^2*c*f)

________________________________________________________________________________________

Sympy [A]  time = 2.54824, size = 298, normalized size = 2.55 \begin{align*} \begin{cases} \frac{\left (\left (256 i A a^{4} c^{2} f^{2} e^{2 i e} - 256 B a^{4} c^{2} f^{2} e^{2 i e}\right ) e^{- 4 i f x} + \left (1536 i A a^{4} c^{2} f^{2} e^{4 i e} - 512 B a^{4} c^{2} f^{2} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 512 i A a^{4} c^{2} f^{2} e^{8 i e} - 512 B a^{4} c^{2} f^{2} e^{8 i e}\right ) e^{2 i f x}\right ) e^{- 6 i e}}{8192 a^{6} c^{3} f^{3}} & \text{for}\: 8192 a^{6} c^{3} f^{3} e^{6 i e} \neq 0 \\x \left (- \frac{3 A - i B}{8 a^{2} c} + \frac{\left (A e^{6 i e} + 3 A e^{4 i e} + 3 A e^{2 i e} + A - i B e^{6 i e} - i B e^{4 i e} + i B e^{2 i e} + i B\right ) e^{- 4 i e}}{8 a^{2} c}\right ) & \text{otherwise} \end{cases} + \frac{x \left (3 A - i B\right )}{8 a^{2} c} \end{align*}

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)),x)

[Out]

Piecewise((((256*I*A*a**4*c**2*f**2*exp(2*I*e) - 256*B*a**4*c**2*f**2*exp(2*I*e))*exp(-4*I*f*x) + (1536*I*A*a*
*4*c**2*f**2*exp(4*I*e) - 512*B*a**4*c**2*f**2*exp(4*I*e))*exp(-2*I*f*x) + (-512*I*A*a**4*c**2*f**2*exp(8*I*e)
- 512*B*a**4*c**2*f**2*exp(8*I*e))*exp(2*I*f*x))*exp(-6*I*e)/(8192*a**6*c**3*f**3), Ne(8192*a**6*c**3*f**3*ex
p(6*I*e), 0)), (x*(-(3*A - I*B)/(8*a**2*c) + (A*exp(6*I*e) + 3*A*exp(4*I*e) + 3*A*exp(2*I*e) + A - I*B*exp(6*I
*e) - I*B*exp(4*I*e) + I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(8*a**2*c)), True)) + x*(3*A - I*B)/(8*a**2*c)

________________________________________________________________________________________

Giac [A]  time = 1.38387, size = 228, normalized size = 1.95 \begin{align*} \frac{\frac{2 \,{\left (3 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c} + \frac{2 \,{\left (-3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c} - \frac{2 \,{\left (3 \, A \tan \left (f x + e\right ) - i \, B \tan \left (f x + e\right ) + 5 i \, A + 3 \, B\right )}}{a^{2} c{\left (-i \, \tan \left (f x + e\right ) + 1\right )}} + \frac{9 i \, A \tan \left (f x + e\right )^{2} + 3 \, B \tan \left (f x + e\right )^{2} + 26 \, A \tan \left (f x + e\right ) - 6 i \, B \tan \left (f x + e\right ) - 21 i \, A + B}{a^{2} c{\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{32 \, f} \end{align*}

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)),x, algorithm="giac")

[Out]

1/32*(2*(3*I*A + B)*log(tan(f*x + e) + I)/(a^2*c) + 2*(-3*I*A - B)*log(tan(f*x + e) - I)/(a^2*c) - 2*(3*A*tan(
f*x + e) - I*B*tan(f*x + e) + 5*I*A + 3*B)/(a^2*c*(-I*tan(f*x + e) + 1)) + (9*I*A*tan(f*x + e)^2 + 3*B*tan(f*x
+ e)^2 + 26*A*tan(f*x + e) - 6*I*B*tan(f*x + e) - 21*I*A + B)/(a^2*c*(tan(f*x + e) - I)^2))/f