3.686 \(\int \frac{(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=91 \[ \frac{a^2 (A-3 i B)}{3 c^4 f (\tan (e+f x)+i)^3}-\frac{a^2 (B+i A)}{2 c^4 f (\tan (e+f x)+i)^4}+\frac{a^2 B}{2 c^4 f (\tan (e+f x)+i)^2} \]

[Out]

-(a^2*(I*A + B))/(2*c^4*f*(I + Tan[e + f*x])^4) + (a^2*(A - (3*I)*B))/(3*c^4*f*(I + Tan[e + f*x])^3) + (a^2*B)
/(2*c^4*f*(I + Tan[e + f*x])^2)

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Rubi [A]  time = 0.149692, 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} \[ \frac{a^2 (A-3 i B)}{3 c^4 f (\tan (e+f x)+i)^3}-\frac{a^2 (B+i A)}{2 c^4 f (\tan (e+f x)+i)^4}+\frac{a^2 B}{2 c^4 f (\tan (e+f x)+i)^2} \]

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])^4,x]

[Out]

-(a^2*(I*A + B))/(2*c^4*f*(I + Tan[e + f*x])^4) + (a^2*(A - (3*I)*B))/(3*c^4*f*(I + Tan[e + f*x])^3) + (a^2*B)
/(2*c^4*f*(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 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))^4} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x) (A+B x)}{(c-i c x)^5} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{2 a (i A+B)}{c^5 (i+x)^5}-\frac{a (A-3 i B)}{c^5 (i+x)^4}-\frac{a B}{c^5 (i+x)^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a^2 (i A+B)}{2 c^4 f (i+\tan (e+f x))^4}+\frac{a^2 (A-3 i B)}{3 c^4 f (i+\tan (e+f x))^3}+\frac{a^2 B}{2 c^4 f (i+\tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 2.79376, size = 91, normalized size = 1. \[ \frac{a^2 (\cos (6 e+8 f x)+i \sin (6 e+8 f x)) (-3 (A+3 i B) \sin (2 (e+f x))+3 (B-3 i A) \cos (2 (e+f x))-8 i A)}{96 c^4 f (\cos (f x)+i \sin (f x))^2} \]

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])^4,x]

[Out]

(a^2*((-8*I)*A + 3*((-3*I)*A + B)*Cos[2*(e + f*x)] - 3*(A + (3*I)*B)*Sin[2*(e + f*x)])*(Cos[6*e + 8*f*x] + I*S
in[6*e + 8*f*x]))/(96*c^4*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.047, size = 68, normalized size = 0.8 \begin{align*}{\frac{{a}^{2}}{f{c}^{4}} \left ( -{\frac{2\,B+2\,iA}{4\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-{\frac{-A+3\,iB}{3\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{B}{2\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \right ) } \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))^4,x)

[Out]

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

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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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.3503, size = 173, normalized size = 1.9 \begin{align*} \frac{{\left (-3 i \, A - 3 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} - 8 i \, A a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-6 i \, A + 6 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{96 \, c^{4} 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))^4,x, algorithm="fricas")

[Out]

1/96*((-3*I*A - 3*B)*a^2*e^(8*I*f*x + 8*I*e) - 8*I*A*a^2*e^(6*I*f*x + 6*I*e) + (-6*I*A + 6*B)*a^2*e^(4*I*f*x +
4*I*e))/(c^4*f)

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Sympy [A]  time = 1.83493, size = 219, normalized size = 2.41 \begin{align*} \begin{cases} \frac{- 512 i A a^{2} c^{8} f^{2} e^{6 i e} e^{6 i f x} + \left (- 384 i A a^{2} c^{8} f^{2} e^{4 i e} + 384 B a^{2} c^{8} f^{2} e^{4 i e}\right ) e^{4 i f x} + \left (- 192 i A a^{2} c^{8} f^{2} e^{8 i e} - 192 B a^{2} c^{8} f^{2} e^{8 i e}\right ) e^{8 i f x}}{6144 c^{12} f^{3}} & \text{for}\: 6144 c^{12} f^{3} \neq 0 \\\frac{x \left (A a^{2} e^{8 i e} + 2 A a^{2} e^{6 i e} + A a^{2} e^{4 i e} - i B a^{2} e^{8 i e} + i B a^{2} e^{4 i e}\right )}{4 c^{4}} & \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))**4,x)

[Out]

Piecewise(((-512*I*A*a**2*c**8*f**2*exp(6*I*e)*exp(6*I*f*x) + (-384*I*A*a**2*c**8*f**2*exp(4*I*e) + 384*B*a**2
*c**8*f**2*exp(4*I*e))*exp(4*I*f*x) + (-192*I*A*a**2*c**8*f**2*exp(8*I*e) - 192*B*a**2*c**8*f**2*exp(8*I*e))*e
xp(8*I*f*x))/(6144*c**12*f**3), Ne(6144*c**12*f**3, 0)), (x*(A*a**2*exp(8*I*e) + 2*A*a**2*exp(6*I*e) + A*a**2*
exp(4*I*e) - I*B*a**2*exp(8*I*e) + I*B*a**2*exp(4*I*e))/(4*c**4), True))

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Giac [B]  time = 1.48193, size = 271, normalized size = 2.98 \begin{align*} -\frac{2 \,{\left (3 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 6 i \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 17 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 16 i \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 6 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 17 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 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} - 3 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{8}} \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))^4,x, algorithm="giac")

[Out]

-2/3*(3*A*a^2*tan(1/2*f*x + 1/2*e)^7 + 6*I*A*a^2*tan(1/2*f*x + 1/2*e)^6 - 3*B*a^2*tan(1/2*f*x + 1/2*e)^6 - 17*
A*a^2*tan(1/2*f*x + 1/2*e)^5 - 16*I*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 6*B*a^2*tan(1/2*f*x + 1/2*e)^4 + 17*A*a^2*t
an(1/2*f*x + 1/2*e)^3 + 6*I*A*a^2*tan(1/2*f*x + 1/2*e)^2 - 3*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 3*A*a^2*tan(1/2*f*
x + 1/2*e))/(c^4*f*(tan(1/2*f*x + 1/2*e) + I)^8)