3.943 \(\int \frac{a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=25 \[ -\frac{i a}{3 f (c-i c \tan (e+f x))^3} \]

[Out]

((-I/3)*a)/(f*(c - I*c*Tan[e + f*x])^3)

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Rubi [A]  time = 0.0738695, 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} \[ -\frac{i a}{3 f (c-i c \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])/(c - I*c*Tan[e + f*x])^3,x]

[Out]

((-I/3)*a)/(f*(c - I*c*Tan[e + f*x])^3)

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))^3} \, dx &=(a c) \int \frac{\sec ^2(e+f x)}{(c-i c \tan (e+f x))^4} \, dx\\ &=\frac{(i a) \operatorname{Subst}\left (\int \frac{1}{(c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=-\frac{i a}{3 f (c-i c \tan (e+f x))^3}\\ \end{align*}

Mathematica [B]  time = 0.445941, size = 56, normalized size = 2.24 \[ \frac{a (-2 i \sin (2 (e+f x))+4 \cos (2 (e+f x))+3) (\sin (4 (e+f x))-i \cos (4 (e+f x)))}{24 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])/(c - I*c*Tan[e + f*x])^3,x]

[Out]

(a*(3 + 4*Cos[2*(e + f*x)] - (2*I)*Sin[2*(e + f*x)])*((-I)*Cos[4*(e + f*x)] + Sin[4*(e + f*x)]))/(24*c^3*f)

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Maple [A]  time = 0.026, size = 21, normalized size = 0.8 \begin{align*} -{\frac{a}{3\,f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \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))^3,x)

[Out]

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

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.35444, size = 131, normalized size = 5.24 \begin{align*} \frac{-i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 3 i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{24 \, c^{3} 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))^3,x, algorithm="fricas")

[Out]

1/24*(-I*a*e^(6*I*f*x + 6*I*e) - 3*I*a*e^(4*I*f*x + 4*I*e) - 3*I*a*e^(2*I*f*x + 2*I*e))/(c^3*f)

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Sympy [A]  time = 0.509905, size = 131, normalized size = 5.24 \begin{align*} \begin{cases} \frac{- 64 i a c^{6} f^{2} e^{6 i e} e^{6 i f x} - 192 i a c^{6} f^{2} e^{4 i e} e^{4 i f x} - 192 i a c^{6} f^{2} e^{2 i e} e^{2 i f x}}{1536 c^{9} f^{3}} & \text{for}\: 1536 c^{9} f^{3} \neq 0 \\\frac{x \left (a e^{6 i e} + 2 a e^{4 i e} + a e^{2 i e}\right )}{4 c^{3}} & \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))**3,x)

[Out]

Piecewise(((-64*I*a*c**6*f**2*exp(6*I*e)*exp(6*I*f*x) - 192*I*a*c**6*f**2*exp(4*I*e)*exp(4*I*f*x) - 192*I*a*c*
*6*f**2*exp(2*I*e)*exp(2*I*f*x))/(1536*c**9*f**3), Ne(1536*c**9*f**3, 0)), (x*(a*exp(6*I*e) + 2*a*exp(4*I*e) +
a*exp(2*I*e))/(4*c**3), True))

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Giac [B]  time = 1.35298, size = 130, normalized size = 5.2 \begin{align*} -\frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 6 i \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 10 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 i \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, c^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{6}} \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))^3,x, algorithm="giac")

[Out]

-2/3*(3*a*tan(1/2*f*x + 1/2*e)^5 + 6*I*a*tan(1/2*f*x + 1/2*e)^4 - 10*a*tan(1/2*f*x + 1/2*e)^3 - 6*I*a*tan(1/2*
f*x + 1/2*e)^2 + 3*a*tan(1/2*f*x + 1/2*e))/(c^3*f*(tan(1/2*f*x + 1/2*e) + I)^6)