3.929 \(\int \frac{1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))} \, dx\)

Optimal. Leaf size=87 \[ \frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a^2 c f}+\frac{3 \sin (e+f x) \cos (e+f x)}{8 a^2 c f}+\frac{3 x}{8 a^2 c} \]

[Out]

(3*x)/(8*a^2*c) + ((I/4)*Cos[e + f*x]^4)/(a^2*c*f) + (3*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*c*f) + (Cos[e + f*x]
^3*Sin[e + f*x])/(4*a^2*c*f)

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Rubi [A]  time = 0.110006, antiderivative size = 87, 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, 3486, 2635, 8} \[ \frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a^2 c f}+\frac{3 \sin (e+f x) \cos (e+f x)}{8 a^2 c f}+\frac{3 x}{8 a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])),x]

[Out]

(3*x)/(8*a^2*c) + ((I/4)*Cos[e + f*x]^4)/(a^2*c*f) + (3*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*c*f) + (Cos[e + f*x]
^3*Sin[e + f*x])/(4*a^2*c*f)

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))} \, dx &=\frac{\int \cos ^4(e+f x) (c-i c \tan (e+f x)) \, dx}{a^2 c^2}\\ &=\frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{\int \cos ^4(e+f x) \, dx}{a^2 c}\\ &=\frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a^2 c f}+\frac{3 \int \cos ^2(e+f x) \, dx}{4 a^2 c}\\ &=\frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{3 \cos (e+f x) \sin (e+f x)}{8 a^2 c f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a^2 c f}+\frac{3 \int 1 \, dx}{8 a^2 c}\\ &=\frac{3 x}{8 a^2 c}+\frac{i \cos ^4(e+f x)}{4 a^2 c f}+\frac{3 \cos (e+f x) \sin (e+f x)}{8 a^2 c f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a^2 c f}\\ \end{align*}

Mathematica [A]  time = 1.00201, size = 81, normalized size = 0.93 \[ -\frac{2 \cos (2 (e+f x))-12 f x \tan (e+f x)+6 i \tan (e+f x)+3 i \sin (3 (e+f x)) \sec (e+f x)+12 i f x-7}{32 a^2 c f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])),x]

[Out]

-(-7 + (12*I)*f*x + 2*Cos[2*(e + f*x)] + (3*I)*Sec[e + f*x]*Sin[3*(e + f*x)] + (6*I)*Tan[e + f*x] - 12*f*x*Tan
[e + f*x])/(32*a^2*c*f*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.041, size = 113, normalized size = 1.3 \begin{align*}{\frac{-{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{2}c}}-{\frac{{\frac{i}{8}}}{f{a}^{2}c \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{1}{4\,f{a}^{2}c \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{2}c}}+{\frac{1}{8\,f{a}^{2}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))^2/(c-I*c*tan(f*x+e)),x)

[Out]

-3/16*I/f/a^2/c*ln(tan(f*x+e)-I)-1/8*I/f/a^2/c/(tan(f*x+e)-I)^2+1/4/f/a^2/c/(tan(f*x+e)-I)+3/16*I/f/a^2/c*ln(t
an(f*x+e)+I)+1/8/f/a^2/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))^2/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.32671, size = 165, normalized size = 1.9 \begin{align*} \frac{{\left (12 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/32*(12*f*x*e^(4*I*f*x + 4*I*e) - 2*I*e^(6*I*f*x + 6*I*e) + 6*I*e^(2*I*f*x + 2*I*e) + I)*e^(-4*I*f*x - 4*I*e)
/(a^2*c*f)

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Sympy [A]  time = 0.931928, size = 180, normalized size = 2.07 \begin{align*} \begin{cases} \frac{\left (- 512 i a^{4} c^{2} f^{2} e^{8 i e} e^{2 i f x} + 1536 i a^{4} c^{2} f^{2} e^{4 i e} e^{- 2 i f x} + 256 i a^{4} c^{2} f^{2} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{8192 a^{6} c^{3} f^{3}} & \text{for}\: 8192 a^{6} c^{3} f^{3} e^{6 i e} \neq 0 \\x \left (\frac{\left (e^{6 i e} + 3 e^{4 i e} + 3 e^{2 i e} + 1\right ) e^{- 4 i e}}{8 a^{2} c} - \frac{3}{8 a^{2} c}\right ) & \text{otherwise} \end{cases} + \frac{3 x}{8 a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e)),x)

[Out]

Piecewise(((-512*I*a**4*c**2*f**2*exp(8*I*e)*exp(2*I*f*x) + 1536*I*a**4*c**2*f**2*exp(4*I*e)*exp(-2*I*f*x) + 2
56*I*a**4*c**2*f**2*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(8192*a**6*c**3*f**3), Ne(8192*a**6*c**3*f**3*exp(6*
I*e), 0)), (x*((exp(6*I*e) + 3*exp(4*I*e) + 3*exp(2*I*e) + 1)*exp(-4*I*e)/(8*a**2*c) - 3/(8*a**2*c)), True)) +
3*x/(8*a**2*c)

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Giac [A]  time = 1.45022, size = 159, normalized size = 1.83 \begin{align*} -\frac{\frac{6 i \, \log \left (i \, \tan \left (f x + e\right ) + 1\right )}{a^{2} c} - \frac{6 i \, \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{2} c} + \frac{2 \,{\left (3 \, \tan \left (f x + e\right ) + 5 i\right )}}{a^{2} c{\left (-i \, \tan \left (f x + e\right ) + 1\right )}} + \frac{-9 i \, \tan \left (f x + e\right )^{2} - 26 \, \tan \left (f x + e\right ) + 21 i}{a^{2} c{\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{32 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/32*(6*I*log(I*tan(f*x + e) + 1)/(a^2*c) - 6*I*log(I*tan(f*x + e) - 1)/(a^2*c) + 2*(3*tan(f*x + e) + 5*I)/(a
^2*c*(-I*tan(f*x + e) + 1)) + (-9*I*tan(f*x + e)^2 - 26*tan(f*x + e) + 21*I)/(a^2*c*(tan(f*x + e) - I)^2))/f