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

Optimal. Leaf size=23 \[ \frac{i c}{f (a+i a \tan (e+f x))} \]

[Out]

(I*c)/(f*(a + I*a*Tan[e + f*x]))

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Rubi [A]  time = 0.0771408, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3522, 3487, 32} \[ \frac{i c}{f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*c)/(f*(a + I*a*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{c-i c \tan (e+f x)}{a+i a \tan (e+f x)} \, dx &=(a c) \int \frac{\sec ^2(e+f x)}{(a+i a \tan (e+f x))^2} \, dx\\ &=-\frac{(i c) \operatorname{Subst}\left (\int \frac{1}{(a+x)^2} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{i c}{f (a+i a \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.0989466, size = 32, normalized size = 1.39 \[ \frac{c (\sin (2 (e+f x))+i \cos (2 (e+f x)))}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 20, normalized size = 0.9 \begin{align*}{\frac{c}{fa \left ( \tan \left ( fx+e \right ) -i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.12565, size = 49, normalized size = 2.13 \begin{align*} \frac{i \, c e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*c*e^(-2*I*f*x - 2*I*e)/(a*f)

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Sympy [A]  time = 0.238739, size = 44, normalized size = 1.91 \begin{align*} \begin{cases} \frac{i c e^{- 2 i e} e^{- 2 i f x}}{2 a f} & \text{for}\: 2 a f e^{2 i e} \neq 0 \\\frac{c x e^{- 2 i e}}{a} & \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))/(a+I*a*tan(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.3554, size = 45, normalized size = 1.96 \begin{align*} -\frac{2 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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