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

Optimal. Leaf size=46 \[ \frac{a (A+B \tan (e+f x))^2}{2 c^2 f (B+i A) (1-i \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0752459, antiderivative size = 46, 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} \[ \frac{a (A+B \tan (e+f x))^2}{2 c^2 f (B+i A) (1-i \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.83373, size = 62, normalized size = 1.35 \[ \frac{a (\cos (3 (e+f x))+i \sin (3 (e+f x))) ((B-3 i A) \cos (e+f x)-(A+3 i B) \sin (e+f x))}{8 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.36832, size = 117, normalized size = 2.54 \begin{align*} \frac{{\left (-i \, A - B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A + 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )}}{8 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.49322, size = 155, normalized size = 3.37 \begin{align*} \begin{cases} \frac{\left (- 8 i A a c^{2} f e^{2 i e} + 8 B a c^{2} f e^{2 i e}\right ) e^{2 i f x} + \left (- 4 i A a c^{2} f e^{4 i e} - 4 B a c^{2} f e^{4 i e}\right ) e^{4 i f x}}{32 c^{4} f^{2}} & \text{for}\: 32 c^{4} f^{2} \neq 0 \\\frac{x \left (A a e^{4 i e} + A a e^{2 i e} - i B a e^{4 i e} + i B a e^{2 i e}\right )}{2 c^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.40107, size = 113, normalized size = 2.46 \begin{align*} -\frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{c^{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+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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