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

Optimal. Leaf size=55 \[ -\frac{a (A-i B)}{3 c^3 f (\tan (e+f x)+i)^3}-\frac{a B}{2 c^3 f (\tan (e+f x)+i)^2} \]

[Out]

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

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Rubi [A]  time = 0.0867701, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.051, Rules used = {3588, 43} \[ -\frac{a (A-i B)}{3 c^3 f (\tan (e+f x)+i)^3}-\frac{a B}{2 c^3 f (\tan (e+f x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(A - I*B))/(3*c^3*f*(I + Tan[e + f*x])^3) - (a*B)/(2*c^3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 1.29339, size = 72, normalized size = 1.31 \[ \frac{a (\cos (4 (e+f x))+i \sin (4 (e+f x))) (-2 (A+2 i B) \sin (2 (e+f x))+2 (B-2 i A) \cos (2 (e+f x))-3 i A)}{24 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((-3*I)*A + 2*((-2*I)*A + B)*Cos[2*(e + f*x)] - 2*(A + (2*I)*B)*Sin[2*(e + f*x)])*(Cos[4*(e + f*x)] + I*Sin
[4*(e + f*x)]))/(24*c^3*f)

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Maple [A]  time = 0.043, size = 43, normalized size = 0.8 \begin{align*}{\frac{a}{f{c}^{3}} \left ( -{\frac{B}{2\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{A-iB}{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+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.40369, size = 159, normalized size = 2.89 \begin{align*} \frac{{\left (-i \, A - B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, A a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-3 i \, A + 3 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )}}{24 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((-I*A - B)*a*e^(6*I*f*x + 6*I*e) - 3*I*A*a*e^(4*I*f*x + 4*I*e) + (-3*I*A + 3*B)*a*e^(2*I*f*x + 2*I*e))/(
c^3*f)

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Sympy [A]  time = 1.94791, size = 202, normalized size = 3.67 \begin{align*} \begin{cases} \frac{- 192 i A a c^{6} f^{2} e^{4 i e} e^{4 i f x} + \left (- 192 i A a c^{6} f^{2} e^{2 i e} + 192 B a c^{6} f^{2} e^{2 i e}\right ) e^{2 i f x} + \left (- 64 i A a c^{6} f^{2} e^{6 i e} - 64 B a c^{6} f^{2} e^{6 i e}\right ) e^{6 i f x}}{1536 c^{9} f^{3}} & \text{for}\: 1536 c^{9} f^{3} \neq 0 \\\frac{x \left (A a e^{6 i e} + 2 A a e^{4 i e} + A a e^{2 i e} - i B a e^{6 i e} + i B a e^{2 i e}\right )}{4 c^{3}} & \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))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**3,x)

[Out]

Piecewise(((-192*I*A*a*c**6*f**2*exp(4*I*e)*exp(4*I*f*x) + (-192*I*A*a*c**6*f**2*exp(2*I*e) + 192*B*a*c**6*f**
2*exp(2*I*e))*exp(2*I*f*x) + (-64*I*A*a*c**6*f**2*exp(6*I*e) - 64*B*a*c**6*f**2*exp(6*I*e))*exp(6*I*f*x))/(153
6*c**9*f**3), Ne(1536*c**9*f**3, 0)), (x*(A*a*exp(6*I*e) + 2*A*a*exp(4*I*e) + A*a*exp(2*I*e) - I*B*a*exp(6*I*e
) + I*B*a*exp(2*I*e))/(4*c**3), True))

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Giac [B]  time = 1.37792, size = 201, normalized size = 3.65 \begin{align*} -\frac{2 \,{\left (3 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 6 i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 3 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 10 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 i \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, c^{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+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-2/3*(3*A*a*tan(1/2*f*x + 1/2*e)^5 + 6*I*A*a*tan(1/2*f*x + 1/2*e)^4 - 3*B*a*tan(1/2*f*x + 1/2*e)^4 - 10*A*a*ta
n(1/2*f*x + 1/2*e)^3 - 2*I*B*a*tan(1/2*f*x + 1/2*e)^3 - 6*I*A*a*tan(1/2*f*x + 1/2*e)^2 + 3*B*a*tan(1/2*f*x + 1
/2*e)^2 + 3*A*a*tan(1/2*f*x + 1/2*e))/(c^3*f*(tan(1/2*f*x + 1/2*e) + I)^6)