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

Optimal. Leaf size=28 \[ \frac{c^2 \tan (e+f x)}{f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0997602, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 34} \[ \frac{c^2 \tan (e+f x)}{f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]

Rubi steps

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

Mathematica [A]  time = 0.235883, size = 34, normalized size = 1.21 \[ \frac{c^2 (\sin (4 (e+f x))+i \cos (4 (e+f x)))}{4 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 39, normalized size = 1.4 \begin{align*}{\frac{{c}^{2}}{f{a}^{2}} \left ( - \left ( \tan \left ( fx+e \right ) -i \right ) ^{-1}-{\frac{i}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^2,x)

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.44798, size = 54, normalized size = 1.93 \begin{align*} \frac{i \, c^{2} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*c^2*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [A]  time = 0.513613, size = 53, normalized size = 1.89 \begin{align*} \begin{cases} \frac{i c^{2} e^{- 4 i e} e^{- 4 i f x}}{4 a^{2} f} & \text{for}\: 4 a^{2} f e^{4 i e} \neq 0 \\\frac{c^{2} x e^{- 4 i e}}{a^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*c**2*exp(-4*I*e)*exp(-4*I*f*x)/(4*a**2*f), Ne(4*a**2*f*exp(4*I*e), 0)), (c**2*x*exp(-4*I*e)/a**2,
True))

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Giac [B]  time = 1.43299, size = 73, normalized size = 2.61 \begin{align*} -\frac{2 \,{\left (c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(c^2*tan(1/2*f*x + 1/2*e)^3 - c^2*tan(1/2*f*x + 1/2*e))/(a^2*f*(tan(1/2*f*x + 1/2*e) - I)^4)