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

Optimal. Leaf size=91 \[ -\frac{a^2 (A-3 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{a^2 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^2 B \log (\cos (e+f x))}{c^2 f}-\frac{i a^2 B x}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.151359, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac{a^2 (A-3 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{a^2 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^2 B \log (\cos (e+f x))}{c^2 f}-\frac{i a^2 B x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [B]  time = 3.16926, size = 184, normalized size = 2.02 \[ \frac{a^2 (\cos (2 (e+2 f x))+i \sin (2 (e+2 f x))) \left (-i \cos (2 (e+f x)) \left (A-2 i B \log \left (\cos ^2(e+f x)\right )+8 B f x-i B\right )+A \sin (2 (e+f x))-8 B f x \sin (2 (e+f x))-i B \sin (2 (e+f x))+2 i B \sin (2 (e+f x)) \log \left (\cos ^2(e+f x)\right )+4 B \tan ^{-1}(\tan (3 e+f x)) (\sin (2 (e+f x))+i \cos (2 (e+f x)))+4 B\right )}{4 c^2 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(4*B - I*Cos[2*(e + f*x)]*(A - I*B + 8*B*f*x - (2*I)*B*Log[Cos[e + f*x]^2]) + A*Sin[2*(e + f*x)] - I*B*Si
n[2*(e + f*x)] - 8*B*f*x*Sin[2*(e + f*x)] + (2*I)*B*Log[Cos[e + f*x]^2]*Sin[2*(e + f*x)] + 4*B*ArcTan[Tan[3*e
+ f*x]]*(I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]))*(Cos[2*(e + 2*f*x)] + I*Sin[2*(e + 2*f*x)]))/(4*c^2*f*(Cos[f*
x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.044, size = 116, normalized size = 1.3 \begin{align*}{\frac{3\,i{a}^{2}B}{f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{{a}^{2}A}{f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{iA{a}^{2}}{f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{{a}^{2}B}{f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{{a}^{2}B\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*I/f*a^2/c^2/(tan(f*x+e)+I)*B-1/f*a^2/c^2/(tan(f*x+e)+I)*A+I/f*a^2/c^2/(tan(f*x+e)+I)^2*A+1/f*a^2/c^2/(tan(f*
x+e)+I)^2*B+1/f*a^2/c^2*B*ln(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))^2*(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 = 1.50054, size = 161, normalized size = 1.77 \begin{align*} \frac{{\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, B a^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{4 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.13304, size = 162, normalized size = 1.78 \begin{align*} - \frac{B a^{2} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin{cases} \frac{4 B a^{2} c^{2} f e^{2 i e} e^{2 i f x} + \left (- i A a^{2} c^{2} f e^{4 i e} - B a^{2} c^{2} f e^{4 i e}\right ) e^{4 i f x}}{4 c^{4} f^{2}} & \text{for}\: 4 c^{4} f^{2} \neq 0 \\\frac{x \left (A a^{2} e^{4 i e} - i B a^{2} e^{4 i e} + 2 i B a^{2} e^{2 i e}\right )}{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))**2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**2,x)

[Out]

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

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Giac [B]  time = 1.53521, size = 275, normalized size = 3.02 \begin{align*} \frac{\frac{12 \, B a^{2} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{2}} - \frac{6 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac{6 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{25 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 12 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 112 i \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 198 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 112 i \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 25 \, B a^{2}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{4}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*B*a^2*log(tan(1/2*f*x + 1/2*e) + I)/c^2 - 6*B*a^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^2 - 6*B*a^2*log
(abs(tan(1/2*f*x + 1/2*e) - 1))/c^2 - (25*B*a^2*tan(1/2*f*x + 1/2*e)^4 + 12*A*a^2*tan(1/2*f*x + 1/2*e)^3 + 112
*I*B*a^2*tan(1/2*f*x + 1/2*e)^3 - 198*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 12*A*a^2*tan(1/2*f*x + 1/2*e) - 112*I*B*a
^2*tan(1/2*f*x + 1/2*e) + 25*B*a^2)/(c^2*(tan(1/2*f*x + 1/2*e) + I)^4))/f