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

Optimal. Leaf size=93 \[ \frac{2 a^2 (A-i B)}{c f (\tan (e+f x)+i)}+\frac{a^2 (3 B+i A) \log (\cos (e+f x))}{c f}-\frac{a^2 x (A-3 i B)}{c}-\frac{i a^2 B \tan (e+f x)}{c f} \]

[Out]

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

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Rubi [A]  time = 0.155555, antiderivative size = 93, 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} \[ \frac{2 a^2 (A-i B)}{c f (\tan (e+f x)+i)}+\frac{a^2 (3 B+i A) \log (\cos (e+f x))}{c f}-\frac{a^2 x (A-3 i B)}{c}-\frac{i a^2 B \tan (e+f x)}{c f} \]

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]),x]

[Out]

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

Mathematica [B]  time = 5.07876, size = 418, normalized size = 4.49 \[ \frac{a^2 \sec (e) (\sin (e+f x)-i \cos (e+f x))^2 (A+B \tan (e+f x)) \left (2 f x (A-3 i B) \cos ^3(e) \cos (e+f x)+\cos (e) \cos (e+f x) \left (-i \cos (2 e) \left (6 B f x+(A-3 i B) \log \left (\cos ^2(e+f x)\right )\right )+2 (B+i A) \cos (2 f x)-2 A f x \sin ^2(e)+A f x-2 A \sin (2 f x)+18 i B f x \sin ^2(e)-6 B f x \sin (2 e)+2 i B \sin (2 f x)\right )-2 i \sin (e) \cos ^2(e) \cos (e+f x) \left ((-3 B-i A) \log \left (\cos ^2(e+f x)\right )+f x (5 A-9 i B)\right )-2 (A-3 i B) \cos (e) (\cos (2 e)-i \sin (2 e)) \cos (e+f x) \tan ^{-1}(\tan (3 e+f x))+A f x \cos (3 e) \cos (e+f x)+2 i A f x \sin ^3(e) \cos (e+f x)+2 i A f x \sin (e) \cos (2 e) \cos (e+f x)+2 B \sin (2 e) \sin (f x)+6 B f x \sin ^3(e) \cos (e+f x)+6 B f x \sin (e) \cos (2 e) \cos (e+f x)-6 i B f x \sin (e) \sin (2 e) \cos (e+f x)+2 i B \cos (2 e) \sin (f x)\right )}{2 c f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]

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]),x]

[Out]

(a^2*Sec[e]*(2*(A - (3*I)*B)*f*x*Cos[e]^3*Cos[e + f*x] + A*f*x*Cos[3*e]*Cos[e + f*x] + (2*I)*A*f*x*Cos[2*e]*Co
s[e + f*x]*Sin[e] + 6*B*f*x*Cos[2*e]*Cos[e + f*x]*Sin[e] - (2*I)*Cos[e]^2*Cos[e + f*x]*((5*A - (9*I)*B)*f*x +
((-I)*A - 3*B)*Log[Cos[e + f*x]^2])*Sin[e] + (2*I)*A*f*x*Cos[e + f*x]*Sin[e]^3 + 6*B*f*x*Cos[e + f*x]*Sin[e]^3
- 2*(A - (3*I)*B)*ArcTan[Tan[3*e + f*x]]*Cos[e]*Cos[e + f*x]*(Cos[2*e] - I*Sin[2*e]) - (6*I)*B*f*x*Cos[e + f*
x]*Sin[e]*Sin[2*e] + (2*I)*B*Cos[2*e]*Sin[f*x] + 2*B*Sin[2*e]*Sin[f*x] + Cos[e]*Cos[e + f*x]*(A*f*x + 2*(I*A +
B)*Cos[2*f*x] - I*Cos[2*e]*(6*B*f*x + (A - (3*I)*B)*Log[Cos[e + f*x]^2]) - 2*A*f*x*Sin[e]^2 + (18*I)*B*f*x*Si
n[e]^2 - 6*B*f*x*Sin[2*e] - 2*A*Sin[2*f*x] + (2*I)*B*Sin[2*f*x]))*((-I)*Cos[e + f*x] + Sin[e + f*x])^2*(A + B*
Tan[e + f*x]))/(2*c*f*(Cos[f*x] + I*Sin[f*x])^2*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]  time = 0.04, size = 113, normalized size = 1.2 \begin{align*}{\frac{-i{a}^{2}B\tan \left ( fx+e \right ) }{cf}}-{\frac{2\,iB{a}^{2}}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }}+2\,{\frac{{a}^{2}A}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{iA{a}^{2}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{cf}}-3\,{\frac{{a}^{2}B\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{cf}} \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)),x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.47812, size = 275, normalized size = 2.96 \begin{align*} \frac{{\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, B a^{2} +{\left ({\left (i \, A + 3 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A + 3 \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f e^{\left (2 i \, f x + 2 i \, e\right )} + c 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.77546, size = 144, normalized size = 1.55 \begin{align*} \frac{2 B a^{2} e^{- 2 i e}}{c f \left (e^{2 i f x} + e^{- 2 i e}\right )} + \frac{a^{2} \left (i A + 3 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \frac{\begin{cases} - \frac{i A a^{2} e^{2 i e} e^{2 i f x}}{f} - \frac{B a^{2} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 A a^{2} e^{2 i e} - 2 i B a^{2} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c} \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)),x)

[Out]

2*B*a**2*exp(-2*I*e)/(c*f*(exp(2*I*f*x) + exp(-2*I*e))) + a**2*(I*A + 3*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(c*
f) + Piecewise((-I*A*a**2*exp(2*I*e)*exp(2*I*f*x)/f - B*a**2*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(2*A*a**
2*exp(2*I*e) - 2*I*B*a**2*exp(2*I*e)), True))/c

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Giac [B]  time = 1.44234, size = 385, normalized size = 4.14 \begin{align*} \frac{\frac{2 \,{\left (-i \, A a^{2} - 3 \, B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c} + \frac{{\left (i \, A a^{2} + 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac{{\left (-i \, A a^{2} - 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac{i \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 i \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i \, A a^{2} - 3 \, B a^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c} - \frac{-3 i \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 10 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 22 i \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 i \, A a^{2} + 9 \, B a^{2}}{c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{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)),x, algorithm="giac")

[Out]

(2*(-I*A*a^2 - 3*B*a^2)*log(tan(1/2*f*x + 1/2*e) + I)/c + (I*A*a^2 + 3*B*a^2)*log(abs(tan(1/2*f*x + 1/2*e) + 1
))/c - (-I*A*a^2 - 3*B*a^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c - (I*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*a^2*t
an(1/2*f*x + 1/2*e)^2 - 2*I*B*a^2*tan(1/2*f*x + 1/2*e) - I*A*a^2 - 3*B*a^2)/((tan(1/2*f*x + 1/2*e)^2 - 1)*c) -
(-3*I*A*a^2*tan(1/2*f*x + 1/2*e)^2 - 9*B*a^2*tan(1/2*f*x + 1/2*e)^2 + 10*A*a^2*tan(1/2*f*x + 1/2*e) - 22*I*B*
a^2*tan(1/2*f*x + 1/2*e) + 3*I*A*a^2 + 9*B*a^2)/(c*(tan(1/2*f*x + 1/2*e) + I)^2))/f