3.731 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=99 \[ \frac{c^2 (-3 B+i A)}{2 a^3 f (-\tan (e+f x)+i)^2}+\frac{2 c^2 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac{i B c^2}{a^3 f (-\tan (e+f x)+i)} \]

[Out]

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

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Rubi [A]  time = 0.156943, antiderivative size = 99, 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} \[ \frac{c^2 (-3 B+i A)}{2 a^3 f (-\tan (e+f x)+i)^2}+\frac{2 c^2 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac{i B c^2}{a^3 f (-\tan (e+f x)+i)} \]

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])^3,x]

[Out]

(2*(A + I*B)*c^2)/(3*a^3*f*(I - Tan[e + f*x])^3) + ((I*A - 3*B)*c^2)/(2*a^3*f*(I - Tan[e + f*x])^2) - (I*B*c^2
)/(a^3*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))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{2 (A+i B) c}{a^4 (-i+x)^4}+\frac{(-i A+3 B) c}{a^4 (-i+x)^3}-\frac{i B c}{a^4 (-i+x)^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 (A+i B) c^2}{3 a^3 f (i-\tan (e+f x))^3}+\frac{(i A-3 B) c^2}{2 a^3 f (i-\tan (e+f x))^2}-\frac{i B c^2}{a^3 f (i-\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.7558, size = 79, normalized size = 0.8 \[ -\frac{i c^2 \sec ^2(e+f x) (\cos (2 (e+f x))-i \sin (2 (e+f x))) ((A-5 i B) \tan (e+f x)-5 i A-B)}{24 a^3 f (\tan (e+f x)-i)^3} \]

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])^3,x]

[Out]

((-I/24)*c^2*Sec[e + f*x]^2*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*((-5*I)*A - B + (A - (5*I)*B)*Tan[e + f*x]
))/(a^3*f*(-I + Tan[e + f*x])^3)

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Maple [A]  time = 0.049, size = 69, normalized size = 0.7 \begin{align*}{\frac{{c}^{2}}{f{a}^{3}} \left ({\frac{iB}{\tan \left ( fx+e \right ) -i}}-{\frac{-iA+3\,B}{2\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{2\,iB+2\,A}{3\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}} \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))^2/(a+I*a*tan(f*x+e))^3,x)

[Out]

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

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.051, size = 128, normalized size = 1.29 \begin{align*} \frac{{\left ({\left (3 i \, A + 3 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 2 \, B\right )} c^{2}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/24*((3*I*A + 3*B)*c^2*e^(2*I*f*x + 2*I*e) + (2*I*A - 2*B)*c^2)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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Sympy [A]  time = 2.76566, size = 173, normalized size = 1.75 \begin{align*} \begin{cases} \frac{\left (\left (8 i A a^{3} c^{2} f e^{4 i e} - 8 B a^{3} c^{2} f e^{4 i e}\right ) e^{- 6 i f x} + \left (12 i A a^{3} c^{2} f e^{6 i e} + 12 B a^{3} c^{2} f e^{6 i e}\right ) e^{- 4 i f x}\right ) e^{- 10 i e}}{96 a^{6} f^{2}} & \text{for}\: 96 a^{6} f^{2} e^{10 i e} \neq 0 \\\frac{x \left (A c^{2} e^{2 i e} + A c^{2} - i B c^{2} e^{2 i e} + i B c^{2}\right ) e^{- 6 i e}}{2 a^{3}} & \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))**3,x)

[Out]

Piecewise((((8*I*A*a**3*c**2*f*exp(4*I*e) - 8*B*a**3*c**2*f*exp(4*I*e))*exp(-6*I*f*x) + (12*I*A*a**3*c**2*f*ex
p(6*I*e) + 12*B*a**3*c**2*f*exp(6*I*e))*exp(-4*I*f*x))*exp(-10*I*e)/(96*a**6*f**2), Ne(96*a**6*f**2*exp(10*I*e
), 0)), (x*(A*c**2*exp(2*I*e) + A*c**2 - I*B*c**2*exp(2*I*e) + I*B*c**2)*exp(-6*I*e)/(2*a**3), True))

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Giac [B]  time = 1.48839, size = 223, normalized size = 2.25 \begin{align*} -\frac{2 \,{\left (3 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 i \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 3 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 8 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 i \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 i \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, a^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{6}} \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))^3,x, algorithm="giac")

[Out]

-2/3*(3*A*c^2*tan(1/2*f*x + 1/2*e)^5 - 3*I*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 3*B*c^2*tan(1/2*f*x + 1/2*e)^4 - 8*A
*c^2*tan(1/2*f*x + 1/2*e)^3 - 2*I*B*c^2*tan(1/2*f*x + 1/2*e)^3 + 3*I*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*c^2*ta
n(1/2*f*x + 1/2*e)^2 + 3*A*c^2*tan(1/2*f*x + 1/2*e))/(a^3*f*(tan(1/2*f*x + 1/2*e) - I)^6)