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

Optimal. Leaf size=25 \[ -\frac{i a}{4 f (c-i c \tan (e+f x))^4} \]

[Out]

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

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Rubi [A]  time = 0.0725323, 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} \[ -\frac{i a}{4 f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*a)/(f*(c - I*c*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 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))^4} \, dx &=(a c) \int \frac{\sec ^2(e+f x)}{(c-i c \tan (e+f x))^5} \, dx\\ &=\frac{(i a) \operatorname{Subst}\left (\int \frac{1}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=-\frac{i a}{4 f (c-i c \tan (e+f x))^4}\\ \end{align*}

Mathematica [B]  time = 0.55047, size = 74, normalized size = 2.96 \[ \frac{a (-i (2 \sin (e+f x)+3 \sin (3 (e+f x)))+10 \cos (e+f x)+5 \cos (3 (e+f x))) (\sin (5 (e+f x))-i \cos (5 (e+f x)))}{64 c^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(10*Cos[e + f*x] + 5*Cos[3*(e + f*x)] - I*(2*Sin[e + f*x] + 3*Sin[3*(e + f*x)]))*((-I)*Cos[5*(e + f*x)] + S
in[5*(e + f*x)]))/(64*c^4*f)

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Maple [A]  time = 0.026, size = 22, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{4}}a}{f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}} \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))^4,x)

[Out]

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

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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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.32627, size = 169, normalized size = 6.76 \begin{align*} \frac{-i \, a e^{\left (8 i \, f x + 8 i \, e\right )} - 4 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{64 \, c^{4} 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))^4,x, algorithm="fricas")

[Out]

1/64*(-I*a*e^(8*I*f*x + 8*I*e) - 4*I*a*e^(6*I*f*x + 6*I*e) - 6*I*a*e^(4*I*f*x + 4*I*e) - 4*I*a*e^(2*I*f*x + 2*
I*e))/(c^4*f)

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Sympy [B]  time = 1.13309, size = 168, normalized size = 6.72 \begin{align*} \begin{cases} \frac{- 8192 i a c^{12} f^{3} e^{8 i e} e^{8 i f x} - 32768 i a c^{12} f^{3} e^{6 i e} e^{6 i f x} - 49152 i a c^{12} f^{3} e^{4 i e} e^{4 i f x} - 32768 i a c^{12} f^{3} e^{2 i e} e^{2 i f x}}{524288 c^{16} f^{4}} & \text{for}\: 524288 c^{16} f^{4} \neq 0 \\\frac{x \left (a e^{8 i e} + 3 a e^{6 i e} + 3 a e^{4 i e} + a e^{2 i e}\right )}{8 c^{4}} & \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))**4,x)

[Out]

Piecewise(((-8192*I*a*c**12*f**3*exp(8*I*e)*exp(8*I*f*x) - 32768*I*a*c**12*f**3*exp(6*I*e)*exp(6*I*f*x) - 4915
2*I*a*c**12*f**3*exp(4*I*e)*exp(4*I*f*x) - 32768*I*a*c**12*f**3*exp(2*I*e)*exp(2*I*f*x))/(524288*c**16*f**4),
Ne(524288*c**16*f**4, 0)), (x*(a*exp(8*I*e) + 3*a*exp(6*I*e) + 3*a*exp(4*I*e) + a*exp(2*I*e))/(8*c**4), True))

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Giac [B]  time = 1.26034, size = 169, normalized size = 6.76 \begin{align*} -\frac{2 \,{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 3 i \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 7 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 8 i \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 7 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 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^{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((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-2*(a*tan(1/2*f*x + 1/2*e)^7 + 3*I*a*tan(1/2*f*x + 1/2*e)^6 - 7*a*tan(1/2*f*x + 1/2*e)^5 - 8*I*a*tan(1/2*f*x +
1/2*e)^4 + 7*a*tan(1/2*f*x + 1/2*e)^3 + 3*I*a*tan(1/2*f*x + 1/2*e)^2 - a*tan(1/2*f*x + 1/2*e))/(c^4*f*(tan(1/
2*f*x + 1/2*e) + I)^8)