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

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

[Out]

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

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Rubi [A]  time = 0.101718, 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} \[ \frac{a^2 \tan (e+f x)}{f (c-i c \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.236265, size = 34, normalized size = 1.21 \[ \frac{a^2 (\sin (4 (e+f x))-i \cos (4 (e+f x)))}{4 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 39, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}}{f{c}^{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((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^2,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.48172, size = 54, normalized size = 1.93 \begin{align*} -\frac{i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{4 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.559509, size = 48, normalized size = 1.71 \begin{align*} \begin{cases} - \frac{i a^{2} e^{4 i e} e^{4 i f x}}{4 c^{2} f} & \text{for}\: 4 c^{2} f \neq 0 \\\frac{a^{2} x e^{4 i e}}{c^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.31115, size = 73, normalized size = 2.61 \begin{align*} -\frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{c^{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((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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