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

Optimal. Leaf size=101 \[ -\frac{i}{4 a f \left (c^2-i c^2 \tan (e+f x)\right )}+\frac{i}{8 a f \left (c^2+i c^2 \tan (e+f x)\right )}+\frac{3 x}{8 a c^2}-\frac{i}{8 a f (c-i c \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.144005, antiderivative size = 101, 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} \[ -\frac{i}{4 a f \left (c^2-i c^2 \tan (e+f x)\right )}+\frac{i}{8 a f \left (c^2+i c^2 \tan (e+f x)\right )}+\frac{3 x}{8 a c^2}-\frac{i}{8 a f (c-i c \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/(8*a*c^2) - (I/8)/(a*f*(c - I*c*Tan[e + f*x])^2) - (I/4)/(a*f*(c^2 - I*c^2*Tan[e + f*x])) + (I/8)/(a*f*(
c^2 + I*c^2*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)) (c-i c \tan (e+f x))^2} \, dx &=\frac{\int \frac{\cos ^2(e+f x)}{c-i c \tan (e+f x)} \, dx}{a c}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^3} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{8 c^3 (c-x)^2}+\frac{1}{4 c^2 (c+x)^3}+\frac{1}{4 c^3 (c+x)^2}+\frac{3}{8 c^3 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=-\frac{i}{8 a f (c-i c \tan (e+f x))^2}-\frac{i}{4 a f \left (c^2-i c^2 \tan (e+f x)\right )}+\frac{i}{8 a f \left (c^2+i c^2 \tan (e+f x)\right )}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{8 a c f}\\ &=\frac{3 x}{8 a c^2}-\frac{i}{8 a f (c-i c \tan (e+f x))^2}-\frac{i}{4 a f \left (c^2-i c^2 \tan (e+f x)\right )}+\frac{i}{8 a f \left (c^2+i c^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.730589, size = 102, normalized size = 1.01 \[ -\frac{(\cos (2 (e+f x))+i \sin (2 (e+f x))) (-2 \cos (2 (e+f x))+12 f x \tan (e+f x)+6 i \tan (e+f x)+3 i \sin (3 (e+f x)) \sec (e+f x)+12 i f x+7)}{32 a c^2 f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(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*c^2*f*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.041, size = 113, normalized size = 1.1 \begin{align*}{\frac{-{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{fa{c}^{2}}}+{\frac{1}{8\,fa{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{8}}}{fa{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{fa{c}^{2}}}+{\frac{1}{4\,fa{c}^{2} \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))/(c-I*c*tan(f*x+e))^2,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.42315, size = 165, normalized size = 1.63 \begin{align*} \frac{{\left (12 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{32 \, a c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.835328, size = 173, normalized size = 1.71 \begin{align*} \begin{cases} \frac{\left (- 256 i a^{2} c^{4} f^{2} e^{6 i e} e^{4 i f x} - 1536 i a^{2} c^{4} f^{2} e^{4 i e} e^{2 i f x} + 512 i a^{2} c^{4} f^{2} e^{- 2 i f x}\right ) e^{- 2 i e}}{8192 a^{3} c^{6} f^{3}} & \text{for}\: 8192 a^{3} c^{6} f^{3} e^{2 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^{- 2 i e}}{8 a c^{2}} - \frac{3}{8 a c^{2}}\right ) & \text{otherwise} \end{cases} + \frac{3 x}{8 a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-256*I*a**2*c**4*f**2*exp(6*I*e)*exp(4*I*f*x) - 1536*I*a**2*c**4*f**2*exp(4*I*e)*exp(2*I*f*x) + 51
2*I*a**2*c**4*f**2*exp(-2*I*f*x))*exp(-2*I*e)/(8192*a**3*c**6*f**3), Ne(8192*a**3*c**6*f**3*exp(2*I*e), 0)), (
x*((exp(6*I*e) + 3*exp(4*I*e) + 3*exp(2*I*e) + 1)*exp(-2*I*e)/(8*a*c**2) - 3/(8*a*c**2)), True)) + 3*x/(8*a*c*
*2)

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Giac [A]  time = 1.28757, size = 159, normalized size = 1.57 \begin{align*} -\frac{-\frac{6 i \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a c^{2}} + \frac{6 i \, \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a c^{2}} + \frac{2 \,{\left (3 \, \tan \left (f x + e\right ) - 5 i\right )}}{a c^{2}{\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 c^{2}{\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))/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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