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

Optimal. Leaf size=45 \[ \frac{A x}{2 a c}-\frac{\cos ^2(e+f x) (B-A \tan (e+f x))}{2 a c f} \]

[Out]

(A*x)/(2*a*c) - (Cos[e + f*x]^2*(B - A*Tan[e + f*x]))/(2*a*c*f)

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Rubi [A]  time = 0.127554, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {3588, 73, 639, 205} \[ \frac{A x}{2 a c}-\frac{\cos ^2(e+f x) (B-A \tan (e+f x))}{2 a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/(2*a*c) - (Cos[e + f*x]^2*(B - A*Tan[e + f*x]))/(2*a*c*f)

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 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^2 (c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{\left (a c+a c x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cos ^2(e+f x) (B-A \tan (e+f x))}{2 a c f}+\frac{A \operatorname{Subst}\left (\int \frac{1}{a c+a c x^2} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{A x}{2 a c}-\frac{\cos ^2(e+f x) (B-A \tan (e+f x))}{2 a c f}\\ \end{align*}

Mathematica [A]  time = 0.0813218, size = 43, normalized size = 0.96 \[ \frac{A (2 (e+f x)+\sin (2 (e+f x)))-2 B \cos ^2(e+f x)}{4 a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*Cos[e + f*x]^2 + A*(2*(e + f*x) + Sin[2*(e + f*x)]))/(4*a*c*f)

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Maple [C]  time = 0.062, size = 142, normalized size = 3.2 \begin{align*}{\frac{-{\frac{i}{4}}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{afc}}+{\frac{A}{4\,afc \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{4}}B}{afc \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{4}}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{afc}}+{\frac{A}{4\,afc \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{{\frac{i}{4}}B}{afc \left ( \tan \left ( fx+e \right ) +i \right ) }} \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))/(c-I*c*tan(f*x+e)),x)

[Out]

-1/4*I/f/a/c*A*ln(tan(f*x+e)-I)+1/4/f/a/c/(tan(f*x+e)-I)*A+1/4*I/f/a/c/(tan(f*x+e)-I)*B+1/4*I/f/a/c*A*ln(tan(f
*x+e)+I)+1/4/f/a/c/(tan(f*x+e)+I)*A-1/4*I/f/a/c/(tan(f*x+e)+I)*B

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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))/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [C]  time = 1.69623, size = 144, normalized size = 3.2 \begin{align*} \frac{{\left (4 \, A f x e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, A - B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a 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))/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.933316, size = 167, normalized size = 3.71 \begin{align*} \frac{A x}{2 a c} + \begin{cases} \frac{\left (\left (8 i A a c f - 8 B a c f\right ) e^{- 2 i f x} + \left (- 8 i A a c f e^{4 i e} - 8 B a c f e^{4 i e}\right ) e^{2 i f x}\right ) e^{- 2 i e}}{64 a^{2} c^{2} f^{2}} & \text{for}\: 64 a^{2} c^{2} f^{2} e^{2 i e} \neq 0 \\x \left (- \frac{A}{2 a c} + \frac{\left (A e^{4 i e} + 2 A e^{2 i e} + A - i B e^{4 i e} + i B\right ) e^{- 2 i e}}{4 a c}\right ) & \text{otherwise} \end{cases} \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))/(c-I*c*tan(f*x+e)),x)

[Out]

A*x/(2*a*c) + Piecewise((((8*I*A*a*c*f - 8*B*a*c*f)*exp(-2*I*f*x) + (-8*I*A*a*c*f*exp(4*I*e) - 8*B*a*c*f*exp(4
*I*e))*exp(2*I*f*x))*exp(-2*I*e)/(64*a**2*c**2*f**2), Ne(64*a**2*c**2*f**2*exp(2*I*e), 0)), (x*(-A/(2*a*c) + (
A*exp(4*I*e) + 2*A*exp(2*I*e) + A - I*B*exp(4*I*e) + I*B)*exp(-2*I*e)/(4*a*c)), True))

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Giac [A]  time = 1.45036, size = 72, normalized size = 1.6 \begin{align*} \frac{\frac{{\left (f x + e\right )} A}{a c} + \frac{A \tan \left (f x + e\right ) - B}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a c}}{2 \, 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))/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

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