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

Optimal. Leaf size=149 \[ -\frac{A+i B}{16 a c^3 f (-\tan (e+f x)+i)}+\frac{3 A+i B}{16 a c^3 f (\tan (e+f x)+i)}-\frac{A-i B}{12 a c^3 f (\tan (e+f x)+i)^3}+\frac{x (2 A+i B)}{8 a c^3}+\frac{i A}{8 a c^3 f (\tan (e+f x)+i)^2} \]

[Out]

((2*A + I*B)*x)/(8*a*c^3) - (A + I*B)/(16*a*c^3*f*(I - Tan[e + f*x])) - (A - I*B)/(12*a*c^3*f*(I + Tan[e + f*x
])^3) + ((I/8)*A)/(a*c^3*f*(I + Tan[e + f*x])^2) + (3*A + I*B)/(16*a*c^3*f*(I + Tan[e + f*x]))

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Rubi [A]  time = 0.215936, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3588, 77, 203} \[ -\frac{A+i B}{16 a c^3 f (-\tan (e+f x)+i)}+\frac{3 A+i B}{16 a c^3 f (\tan (e+f x)+i)}-\frac{A-i B}{12 a c^3 f (\tan (e+f x)+i)^3}+\frac{x (2 A+i B)}{8 a c^3}+\frac{i A}{8 a c^3 f (\tan (e+f x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + I*B)*x)/(8*a*c^3) - (A + I*B)/(16*a*c^3*f*(I - Tan[e + f*x])) - (A - I*B)/(12*a*c^3*f*(I + Tan[e + f*x
])^3) + ((I/8)*A)/(a*c^3*f*(I + Tan[e + f*x])^2) + (3*A + I*B)/(16*a*c^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]))))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^2 (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}{16 a^2 c^4 (-i+x)^2}+\frac{A-i B}{4 a^2 c^4 (i+x)^4}-\frac{i A}{4 a^2 c^4 (i+x)^3}+\frac{-3 A-i B}{16 a^2 c^4 (i+x)^2}+\frac{2 A+i B}{8 a^2 c^4 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{A+i B}{16 a c^3 f (i-\tan (e+f x))}-\frac{A-i B}{12 a c^3 f (i+\tan (e+f x))^3}+\frac{i A}{8 a c^3 f (i+\tan (e+f x))^2}+\frac{3 A+i B}{16 a c^3 f (i+\tan (e+f x))}+\frac{(2 A+i B) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a c^3 f}\\ &=\frac{(2 A+i B) x}{8 a c^3}-\frac{A+i B}{16 a c^3 f (i-\tan (e+f x))}-\frac{A-i B}{12 a c^3 f (i+\tan (e+f x))^3}+\frac{i A}{8 a c^3 f (i+\tan (e+f x))^2}+\frac{3 A+i B}{16 a c^3 f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.4001, size = 203, normalized size = 1.36 \[ \frac{(\cos (3 (e+f x))+i \sin (3 (e+f x))) (A+B \tan (e+f x)) (3 (A (-2-8 i f x)+B (4 f x+i)) \cos (2 (e+f x))+2 (A+2 i B) \cos (4 (e+f x))-24 A f x \sin (2 (e+f x))-6 i A \sin (2 (e+f x))-4 i A \sin (4 (e+f x))-18 A-3 B \sin (2 (e+f x))-12 i B f x \sin (2 (e+f x))+2 B \sin (4 (e+f x)))}{96 a c^3 f (\tan (e+f x)-i) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(-18*A + 3*(A*(-2 - (8*I)*f*x) + B*(I + 4*f*x))*Cos[2*(e + f*x)] + 2*
(A + (2*I)*B)*Cos[4*(e + f*x)] - (6*I)*A*Sin[2*(e + f*x)] - 3*B*Sin[2*(e + f*x)] - 24*A*f*x*Sin[2*(e + f*x)] -
(12*I)*B*f*x*Sin[2*(e + f*x)] - (4*I)*A*Sin[4*(e + f*x)] + 2*B*Sin[4*(e + f*x)])*(A + B*Tan[e + f*x]))/(96*a*
c^3*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.068, size = 257, normalized size = 1.7 \begin{align*}{\frac{{\frac{i}{16}}B}{af{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{A}{16\,af{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{16\,af{c}^{3}}}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) A}{af{c}^{3}}}+{\frac{{\frac{i}{8}}A}{af{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{A}{12\,af{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{{\frac{i}{12}}B}{af{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{3\,A}{16\,af{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{{\frac{i}{16}}B}{af{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) B}{16\,af{c}^{3}}}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) A}{af{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*I/f/a/c^3/(tan(f*x+e)-I)*B+1/16/f/a/c^3/(tan(f*x+e)-I)*A+1/16/f/a/c^3*ln(tan(f*x+e)-I)*B-1/8*I/f/a/c^3*ln
(tan(f*x+e)-I)*A+1/8*I*A/a/c^3/f/(tan(f*x+e)+I)^2-1/12/f/a/c^3/(tan(f*x+e)+I)^3*A+1/12*I/f/a/c^3/(tan(f*x+e)+I
)^3*B+3/16/f/a/c^3/(tan(f*x+e)+I)*A+1/16*I/f/a/c^3/(tan(f*x+e)+I)*B-1/16/f/a/c^3*ln(tan(f*x+e)+I)*B+1/8*I/f/a/
c^3*ln(tan(f*x+e)+I)*A

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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))/(a+I*a*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 = 1.1169, size = 258, normalized size = 1.73 \begin{align*} \frac{{\left (12 \,{\left (2 \, A + i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-6 i \, A - 3 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 18 i \, A e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{96 \, a c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(12*(2*A + I*B)*f*x*e^(2*I*f*x + 2*I*e) + (-I*A - B)*e^(8*I*f*x + 8*I*e) + (-6*I*A - 3*B)*e^(6*I*f*x + 6*
I*e) - 18*I*A*e^(4*I*f*x + 4*I*e) + 3*I*A - 3*B)*e^(-2*I*f*x - 2*I*e)/(a*c^3*f)

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Sympy [A]  time = 2.75076, size = 330, normalized size = 2.21 \begin{align*} \begin{cases} \frac{\left (- 294912 i A a^{3} c^{9} f^{3} e^{4 i e} e^{2 i f x} + \left (49152 i A a^{3} c^{9} f^{3} - 49152 B a^{3} c^{9} f^{3}\right ) e^{- 2 i f x} + \left (- 98304 i A a^{3} c^{9} f^{3} e^{6 i e} - 49152 B a^{3} c^{9} f^{3} e^{6 i e}\right ) e^{4 i f x} + \left (- 16384 i A a^{3} c^{9} f^{3} e^{8 i e} - 16384 B a^{3} c^{9} f^{3} e^{8 i e}\right ) e^{6 i f x}\right ) e^{- 2 i e}}{1572864 a^{4} c^{12} f^{4}} & \text{for}\: 1572864 a^{4} c^{12} f^{4} e^{2 i e} \neq 0 \\x \left (- \frac{2 A + i B}{8 a c^{3}} + \frac{\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 2 i e}}{16 a c^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (2 A + i B\right )}{8 a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**3,x)

[Out]

Piecewise(((-294912*I*A*a**3*c**9*f**3*exp(4*I*e)*exp(2*I*f*x) + (49152*I*A*a**3*c**9*f**3 - 49152*B*a**3*c**9
*f**3)*exp(-2*I*f*x) + (-98304*I*A*a**3*c**9*f**3*exp(6*I*e) - 49152*B*a**3*c**9*f**3*exp(6*I*e))*exp(4*I*f*x)
+ (-16384*I*A*a**3*c**9*f**3*exp(8*I*e) - 16384*B*a**3*c**9*f**3*exp(8*I*e))*exp(6*I*f*x))*exp(-2*I*e)/(15728
64*a**4*c**12*f**4), Ne(1572864*a**4*c**12*f**4*exp(2*I*e), 0)), (x*(-(2*A + I*B)/(8*a*c**3) + (A*exp(8*I*e) +
4*A*exp(6*I*e) + 6*A*exp(4*I*e) + 4*A*exp(2*I*e) + A - I*B*exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(2*I*e) +
I*B)*exp(-2*I*e)/(16*a*c**3)), True)) + x*(2*A + I*B)/(8*a*c**3)

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Giac [A]  time = 1.50393, size = 259, normalized size = 1.74 \begin{align*} -\frac{\frac{6 \,{\left (-2 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{3}} + \frac{6 \,{\left (2 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{3}} + \frac{6 \,{\left (-2 i \, A \tan \left (f x + e\right ) + B \tan \left (f x + e\right ) - 3 \, A - 2 i \, B\right )}}{a c^{3}{\left (\tan \left (f x + e\right ) - i\right )}} + \frac{22 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 84 \, A \tan \left (f x + e\right )^{2} - 39 i \, B \tan \left (f x + e\right )^{2} - 114 i \, A \tan \left (f x + e\right ) + 45 \, B \tan \left (f x + e\right ) + 60 \, A + 9 i \, B}{a c^{3}{\left (\tan \left (f x + e\right ) + i\right )}^{3}}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*(-2*I*A + B)*log(tan(f*x + e) + I)/(a*c^3) + 6*(2*I*A - B)*log(tan(f*x + e) - I)/(a*c^3) + 6*(-2*I*A*
tan(f*x + e) + B*tan(f*x + e) - 3*A - 2*I*B)/(a*c^3*(tan(f*x + e) - I)) + (22*I*A*tan(f*x + e)^3 - 11*B*tan(f*
x + e)^3 - 84*A*tan(f*x + e)^2 - 39*I*B*tan(f*x + e)^2 - 114*I*A*tan(f*x + e) + 45*B*tan(f*x + e) + 60*A + 9*I
*B)/(a*c^3*(tan(f*x + e) + I)^3))/f