3.922 \(\int \frac{(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=50 \[ \frac{i c^4 \left (a^2-i a^2 \tan (e+f x)\right )^4}{8 f \left (a^3+i a^3 \tan (e+f x)\right )^4} \]

[Out]

((I/8)*c^4*(a^2 - I*a^2*Tan[e + f*x])^4)/(f*(a^3 + I*a^3*Tan[e + f*x])^4)

________________________________________________________________________________________

Rubi [A]  time = 0.102417, antiderivative size = 50, 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, 37} \[ \frac{i c^4 \left (a^2-i a^2 \tan (e+f x)\right )^4}{8 f \left (a^3+i a^3 \tan (e+f x)\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^4/(a + I*a*Tan[e + f*x])^4,x]

[Out]

((I/8)*c^4*(a^2 - I*a^2*Tan[e + f*x])^4)/(f*(a^3 + I*a^3*Tan[e + f*x])^4)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^4} \, dx &=\left (a^4 c^4\right ) \int \frac{\sec ^8(e+f x)}{(a+i a \tan (e+f x))^8} \, dx\\ &=-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{(a-x)^3}{(a+x)^5} \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i c^4 (1-i \tan (e+f x))^4}{8 f (a+i a \tan (e+f x))^4}\\ \end{align*}

Mathematica [A]  time = 0.2512, size = 34, normalized size = 0.68 \[ \frac{c^4 (\sin (8 (e+f x))+i \cos (8 (e+f x)))}{8 a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^4/(a + I*a*Tan[e + f*x])^4,x]

[Out]

(c^4*(I*Cos[8*(e + f*x)] + Sin[8*(e + f*x)]))/(8*a^4*f)

________________________________________________________________________________________

Maple [A]  time = 0.031, size = 66, normalized size = 1.3 \begin{align*}{\frac{{c}^{4}}{f{a}^{4}} \left ({\frac{-3\,i}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{2\,i}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{4}}}+4\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{-3}- \left ( \tan \left ( fx+e \right ) -i \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^4,x)

[Out]

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

________________________________________________________________________________________

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))^4/(a+I*a*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.36913, size = 54, normalized size = 1.08 \begin{align*} \frac{i \, c^{4} e^{\left (-8 i \, f x - 8 i \, e\right )}}{8 \, a^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.01785, size = 53, normalized size = 1.06 \begin{align*} \begin{cases} \frac{i c^{4} e^{- 8 i e} e^{- 8 i f x}}{8 a^{4} f} & \text{for}\: 8 a^{4} f e^{8 i e} \neq 0 \\\frac{c^{4} x e^{- 8 i e}}{a^{4}} & \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))**4/(a+I*a*tan(f*x+e))**4,x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.59597, size = 119, normalized size = 2.38 \begin{align*} -\frac{2 \,{\left (c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 7 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 7 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-2*(c^4*tan(1/2*f*x + 1/2*e)^7 - 7*c^4*tan(1/2*f*x + 1/2*e)^5 + 7*c^4*tan(1/2*f*x + 1/2*e)^3 - c^4*tan(1/2*f*x
+ 1/2*e))/(a^4*f*(tan(1/2*f*x + 1/2*e) - I)^8)