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

Optimal. Leaf size=97 \[ \frac{c^2 (A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{c^2 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}-\frac{B c^2 \log (\cos (e+f x))}{a^2 f}+\frac{i B c^2 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.151759, antiderivative size = 97, 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} \[ \frac{c^2 (A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{c^2 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}-\frac{B c^2 \log (\cos (e+f x))}{a^2 f}+\frac{i B c^2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.45383, size = 140, normalized size = 1.44 \[ \frac{c^2 \sec ^2(e+f x) \left (\cos (2 (e+f x)) \left (-i A+2 B \log \left (\cos ^2(e+f x)\right )+B\right )-A \sin (2 (e+f x))-i B \sin (2 (e+f x))+2 i B \sin (2 (e+f x)) \log \left (\cos ^2(e+f x)\right )+4 B \tan ^{-1}(\tan (f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))-4 B\right )}{4 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 116, normalized size = 1.2 \begin{align*}{\frac{-3\,i{c}^{2}B}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{A{c}^{2}}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{B{c}^{2}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{2}}}-{\frac{iA{c}^{2}}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{B{c}^{2}}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \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))^2/(a+I*a*tan(f*x+e))^2,x)

[Out]

-3*I/f*c^2/a^2/(tan(f*x+e)-I)*B-1/f*c^2/a^2/(tan(f*x+e)-I)*A+1/f*c^2/a^2*B*ln(tan(f*x+e)-I)-I/f*c^2/a^2/(tan(f
*x+e)-I)^2*A+1/f*c^2/a^2/(tan(f*x+e)-I)^2*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))*(c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.10993, size = 236, normalized size = 2.43 \begin{align*} \frac{{\left (8 i \, B c^{2} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, B c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 4 \, B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A - B\right )} c^{2}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, 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))^2/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.76239, size = 207, normalized size = 2.13 \begin{align*} \frac{2 i B c^{2} x}{a^{2}} - \frac{B c^{2} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \begin{cases} \frac{\left (4 B a^{2} c^{2} f e^{4 i e} e^{- 2 i f x} + \left (i A a^{2} c^{2} f e^{2 i e} - B a^{2} c^{2} f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{4 a^{4} f^{2}} & \text{for}\: 4 a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac{2 i B c^{2}}{a^{2}} + \frac{\left (A c^{2} + 2 i B c^{2} e^{4 i e} - 2 i B c^{2} e^{2 i e} + i B c^{2}\right ) e^{- 4 i e}}{a^{2}}\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))*(c-I*c*tan(f*x+e))**2/(a+I*a*tan(f*x+e))**2,x)

[Out]

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

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Giac [B]  time = 1.24559, size = 275, normalized size = 2.84 \begin{align*} \frac{\frac{12 \, B c^{2} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a^{2}} - \frac{6 \, B c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, B c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{25 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 12 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 112 i \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 198 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 112 i \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 25 \, B c^{2}}{a^{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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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