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

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

[Out]

((-I)*a)/(f*(c - I*c*Tan[e + f*x]))

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Rubi [A]  time = 0.0773809, 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} \[ -\frac{i a}{f (c-i c \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.103282, size = 32, normalized size = 1.39 \[ \frac{a (\sin (2 (e+f x))-i \cos (2 (e+f x)))}{2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.024, size = 20, normalized size = 0.9 \begin{align*}{\frac{a}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.39965, size = 49, normalized size = 2.13 \begin{align*} -\frac{i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{2 \, c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*a*e^(2*I*f*x + 2*I*e)/(c*f)

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Sympy [A]  time = 0.404668, size = 39, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{i a e^{2 i e} e^{2 i f x}}{2 c f} & \text{for}\: 2 c f \neq 0 \\\frac{a x e^{2 i e}}{c} & \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))/(c-I*c*tan(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.31153, size = 45, normalized size = 1.96 \begin{align*} -\frac{2 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{c 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((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

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