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

Optimal. Leaf size=48 \[ -\frac{c (A+B \tan (e+f x))^2}{2 a^2 f (-B+i A) (1+i \tan (e+f x))^2} \]

[Out]

-(c*(A + B*Tan[e + f*x])^2)/(2*a^2*(I*A - B)*f*(1 + I*Tan[e + f*x])^2)

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Rubi [A]  time = 0.0813494, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.051, Rules used = {3588, 37} \[ -\frac{c (A+B \tan (e+f x))^2}{2 a^2 f (-B+i A) (1+i \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(A + B*Tan[e + f*x])^2)/(2*a^2*(I*A - B)*f*(1 + I*Tan[e + f*x])^2)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{c (A+B \tan (e+f x))^2}{2 a^2 (i A-B) f (1+i \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.44238, size = 58, normalized size = 1.21 \[ \frac{(c-i c \tan (e+f x)) ((A-3 i B) \tan (e+f x)-3 i A-B)}{8 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 46, normalized size = 1. \begin{align*}{\frac{c}{f{a}^{2}} \left ({\frac{-iB}{\tan \left ( fx+e \right ) -i}}-{\frac{iA-B}{2\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.15748, size = 116, normalized size = 2.42 \begin{align*} \frac{{\left ({\left (2 i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A - B\right )} c\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{8 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.52294, size = 160, normalized size = 3.33 \begin{align*} \begin{cases} \frac{\left (\left (4 i A a^{2} c f e^{2 i e} - 4 B a^{2} c f e^{2 i e}\right ) e^{- 4 i f x} + \left (8 i A a^{2} c f e^{4 i e} + 8 B a^{2} c f e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{32 a^{4} f^{2}} & \text{for}\: 32 a^{4} f^{2} e^{6 i e} \neq 0 \\\frac{x \left (A c e^{2 i e} + A c - i B c e^{2 i e} + i B c\right ) e^{- 4 i e}}{2 a^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((4*I*A*a**2*c*f*exp(2*I*e) - 4*B*a**2*c*f*exp(2*I*e))*exp(-4*I*f*x) + (8*I*A*a**2*c*f*exp(4*I*e) +
8*B*a**2*c*f*exp(4*I*e))*exp(-2*I*f*x))*exp(-6*I*e)/(32*a**4*f**2), Ne(32*a**4*f**2*exp(6*I*e), 0)), (x*(A*c*
exp(2*I*e) + A*c - I*B*c*exp(2*I*e) + I*B*c)*exp(-4*I*e)/(2*a**2), True))

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Giac [A]  time = 1.29276, size = 113, normalized size = 2.35 \begin{align*} -\frac{2 \,{\left (A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - i \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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