3.930 \(\int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx\)

Optimal. Leaf size=124 \[ \frac{i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c}{8 a^3 f (c+i c \tan (e+f x))^2}-\frac{i}{16 a^3 f (c-i c \tan (e+f x))}+\frac{3 i}{16 a^3 f (c+i c \tan (e+f x))}+\frac{x}{4 a^3 c} \]

[Out]

x/(4*a^3*c) - (I/16)/(a^3*f*(c - I*c*Tan[e + f*x])) + ((I/12)*c^2)/(a^3*f*(c + I*c*Tan[e + f*x])^3) + ((I/8)*c
)/(a^3*f*(c + I*c*Tan[e + f*x])^2) + ((3*I)/16)/(a^3*f*(c + I*c*Tan[e + f*x]))

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Rubi [A]  time = 0.159861, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3522, 3487, 44, 206} \[ \frac{i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c}{8 a^3 f (c+i c \tan (e+f x))^2}-\frac{i}{16 a^3 f (c-i c \tan (e+f x))}+\frac{3 i}{16 a^3 f (c+i c \tan (e+f x))}+\frac{x}{4 a^3 c} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])),x]

[Out]

x/(4*a^3*c) - (I/16)/(a^3*f*(c - I*c*Tan[e + f*x])) + ((I/12)*c^2)/(a^3*f*(c + I*c*Tan[e + f*x])^3) + ((I/8)*c
)/(a^3*f*(c + I*c*Tan[e + f*x])^2) + ((3*I)/16)/(a^3*f*(c + I*c*Tan[e + f*x]))

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^2 \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 (c+x)^2} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \left (\frac{1}{4 c^2 (c-x)^4}+\frac{1}{4 c^3 (c-x)^3}+\frac{3}{16 c^4 (c-x)^2}+\frac{1}{16 c^4 (c+x)^2}+\frac{1}{4 c^4 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=-\frac{i}{16 a^3 f (c-i c \tan (e+f x))}+\frac{i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c}{8 a^3 f (c+i c \tan (e+f x))^2}+\frac{3 i}{16 a^3 f (c+i c \tan (e+f x))}+\frac{i \operatorname{Subst}\left (\int \frac{1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f}\\ &=\frac{x}{4 a^3 c}-\frac{i}{16 a^3 f (c-i c \tan (e+f x))}+\frac{i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c}{8 a^3 f (c+i c \tan (e+f x))^2}+\frac{3 i}{16 a^3 f (c+i c \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.01488, size = 101, normalized size = 0.81 \[ -\frac{\sec ^2(e+f x) (12 i f x \sin (2 (e+f x))+3 \sin (2 (e+f x))+2 \sin (4 (e+f x))+3 (4 f x+i) \cos (2 (e+f x))-i \cos (4 (e+f x))+9 i)}{48 a^3 c f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])),x]

[Out]

-(Sec[e + f*x]^2*(9*I + 3*(I + 4*f*x)*Cos[2*(e + f*x)] - I*Cos[4*(e + f*x)] + 3*Sin[2*(e + f*x)] + (12*I)*f*x*
Sin[2*(e + f*x)] + 2*Sin[4*(e + f*x)]))/(48*a^3*c*f*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.044, size = 135, normalized size = 1.1 \begin{align*}{\frac{-{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{3}c}}-{\frac{{\frac{i}{8}}}{f{a}^{3}c \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{1}{12\,f{a}^{3}c \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{3}{16\,f{a}^{3}c \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{3}c}}+{\frac{1}{16\,f{a}^{3}c \left ( \tan \left ( fx+e \right ) +i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x)

[Out]

-1/8*I/f/a^3/c*ln(tan(f*x+e)-I)-1/8*I/f/a^3/c/(tan(f*x+e)-I)^2-1/12/f/a^3/c/(tan(f*x+e)-I)^3+3/16/f/a^3/c/(tan
(f*x+e)-I)+1/8*I/f/a^3/c*ln(tan(f*x+e)+I)+1/16/f/a^3/c/(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(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.38087, size = 201, normalized size = 1.62 \begin{align*} \frac{{\left (24 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 18 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 6 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/96*(24*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(8*I*f*x + 8*I*e) + 18*I*e^(4*I*f*x + 4*I*e) + 6*I*e^(2*I*f*x + 2*I*e
) + I)*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)

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Sympy [A]  time = 1.40613, size = 216, normalized size = 1.74 \begin{align*} \begin{cases} \frac{\left (- 24576 i a^{9} c^{3} f^{3} e^{14 i e} e^{2 i f x} + 147456 i a^{9} c^{3} f^{3} e^{10 i e} e^{- 2 i f x} + 49152 i a^{9} c^{3} f^{3} e^{8 i e} e^{- 4 i f x} + 8192 i a^{9} c^{3} f^{3} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{786432 a^{12} c^{4} f^{4}} & \text{for}\: 786432 a^{12} c^{4} f^{4} e^{12 i e} \neq 0 \\x \left (\frac{\left (e^{8 i e} + 4 e^{6 i e} + 6 e^{4 i e} + 4 e^{2 i e} + 1\right ) e^{- 6 i e}}{16 a^{3} c} - \frac{1}{4 a^{3} c}\right ) & \text{otherwise} \end{cases} + \frac{x}{4 a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e)),x)

[Out]

Piecewise(((-24576*I*a**9*c**3*f**3*exp(14*I*e)*exp(2*I*f*x) + 147456*I*a**9*c**3*f**3*exp(10*I*e)*exp(-2*I*f*
x) + 49152*I*a**9*c**3*f**3*exp(8*I*e)*exp(-4*I*f*x) + 8192*I*a**9*c**3*f**3*exp(6*I*e)*exp(-6*I*f*x))*exp(-12
*I*e)/(786432*a**12*c**4*f**4), Ne(786432*a**12*c**4*f**4*exp(12*I*e), 0)), (x*((exp(8*I*e) + 4*exp(6*I*e) + 6
*exp(4*I*e) + 4*exp(2*I*e) + 1)*exp(-6*I*e)/(16*a**3*c) - 1/(4*a**3*c)), True)) + x/(4*a**3*c)

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Giac [A]  time = 1.37959, size = 166, normalized size = 1.34 \begin{align*} -\frac{-\frac{6 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c} + \frac{6 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c} + \frac{3 \,{\left (2 i \, \tan \left (f x + e\right ) - 3\right )}}{a^{3} c{\left (\tan \left (f x + e\right ) + i\right )}} + \frac{-11 i \, \tan \left (f x + e\right )^{3} - 42 \, \tan \left (f x + e\right )^{2} + 57 i \, \tan \left (f x + e\right ) + 30}{a^{3} c{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/48*(-6*I*log(tan(f*x + e) + I)/(a^3*c) + 6*I*log(tan(f*x + e) - I)/(a^3*c) + 3*(2*I*tan(f*x + e) - 3)/(a^3*
c*(tan(f*x + e) + I)) + (-11*I*tan(f*x + e)^3 - 42*tan(f*x + e)^2 + 57*I*tan(f*x + e) + 30)/(a^3*c*(tan(f*x +
e) - I)^3))/f