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

Optimal. Leaf size=25 \[ -\frac{i a}{2 f (c-i c \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0735024, antiderivative size = 25, 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} \[ -\frac{i a}{2 f (c-i c \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.627989, size = 51, normalized size = 2.04 \[ \frac{a (3 \cos (e+f x)-i \sin (e+f x)) (\sin (3 (e+f x))-i \cos (3 (e+f x)))}{8 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 22, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{2}}a}{f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \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))^2,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.32476, size = 92, normalized size = 3.68 \begin{align*} \frac{-i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{8 \, c^{2} 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))^2,x, algorithm="fricas")

[Out]

1/8*(-I*a*e^(4*I*f*x + 4*I*e) - 2*I*a*e^(2*I*f*x + 2*I*e))/(c^2*f)

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Sympy [A]  time = 0.744005, size = 90, normalized size = 3.6 \begin{align*} \begin{cases} \frac{- 4 i a c^{2} f e^{4 i e} e^{4 i f x} - 8 i a c^{2} f e^{2 i e} e^{2 i f x}}{32 c^{4} f^{2}} & \text{for}\: 32 c^{4} f^{2} \neq 0 \\\frac{x \left (a e^{4 i e} + a e^{2 i e}\right )}{2 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))/(c-I*c*tan(f*x+e))**2,x)

[Out]

Piecewise(((-4*I*a*c**2*f*exp(4*I*e)*exp(4*I*f*x) - 8*I*a*c**2*f*exp(2*I*e)*exp(2*I*f*x))/(32*c**4*f**2), Ne(3
2*c**4*f**2, 0)), (x*(a*exp(4*I*e) + a*exp(2*I*e))/(2*c**2), True))

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Giac [B]  time = 1.26699, size = 88, normalized size = 3.52 \begin{align*} -\frac{2 \,{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + i \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a \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))/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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