3.903 \(\int \frac{(c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=62 \[ \frac{i c^2}{2 f (a+i a \tan (e+f x))^4}-\frac{i a^2 c^2}{3 f \left (a^2+i a^2 \tan (e+f x)\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.108742, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ \frac{i c^2}{2 f (a+i a \tan (e+f x))^4}-\frac{i a^2 c^2}{3 f \left (a^2+i a^2 \tan (e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*c^2)/(f*(a + I*a*Tan[e + f*x])^4) - ((I/3)*a^2*c^2)/(f*(a^2 + I*a^2*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^4} \, dx &=\left (a^2 c^2\right ) \int \frac{\sec ^4(e+f x)}{(a+i a \tan (e+f x))^6} \, dx\\ &=-\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{a-x}{(a+x)^5} \, dx,x,i a \tan (e+f x)\right )}{a f}\\ &=-\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \left (\frac{2 a}{(a+x)^5}-\frac{1}{(a+x)^4}\right ) \, dx,x,i a \tan (e+f x)\right )}{a f}\\ &=\frac{i c^2}{2 f (a+i a \tan (e+f x))^4}-\frac{i c^2}{3 a f (a+i a \tan (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 1.81146, size = 58, normalized size = 0.94 \[ \frac{c^2 (3 i \sin (2 (e+f x))+9 \cos (2 (e+f x))+8) (\sin (6 (e+f x))+i \cos (6 (e+f x)))}{96 a^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(8 + 9*Cos[2*(e + f*x)] + (3*I)*Sin[2*(e + f*x)])*(I*Cos[6*(e + f*x)] + Sin[6*(e + f*x)]))/(96*a^4*f)

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Maple [A]  time = 0.031, size = 39, normalized size = 0.6 \begin{align*}{\frac{{c}^{2}}{f{a}^{4}} \left ({\frac{{\frac{i}{2}}}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{4}}}+{\frac{1}{3\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*c^2/a^4*(1/2*I/(tan(f*x+e)-I)^4+1/3/(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((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.37984, size = 142, normalized size = 2.29 \begin{align*} \frac{{\left (6 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, c^{2}\right )} e^{\left (-8 i \, f x - 8 i \, e\right )}}{96 \, a^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/96*(6*I*c^2*e^(4*I*f*x + 4*I*e) + 8*I*c^2*e^(2*I*f*x + 2*I*e) + 3*I*c^2)*e^(-8*I*f*x - 8*I*e)/(a^4*f)

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Sympy [A]  time = 1.07346, size = 153, normalized size = 2.47 \begin{align*} \begin{cases} \frac{\left (384 i a^{8} c^{2} f^{2} e^{14 i e} e^{- 4 i f x} + 512 i a^{8} c^{2} f^{2} e^{12 i e} e^{- 6 i f x} + 192 i a^{8} c^{2} f^{2} e^{10 i e} e^{- 8 i f x}\right ) e^{- 18 i e}}{6144 a^{12} f^{3}} & \text{for}\: 6144 a^{12} f^{3} e^{18 i e} \neq 0 \\\frac{x \left (c^{2} e^{4 i e} + 2 c^{2} e^{2 i e} + c^{2}\right ) e^{- 8 i e}}{4 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))**2/(a+I*a*tan(f*x+e))**4,x)

[Out]

Piecewise(((384*I*a**8*c**2*f**2*exp(14*I*e)*exp(-4*I*f*x) + 512*I*a**8*c**2*f**2*exp(12*I*e)*exp(-6*I*f*x) +
192*I*a**8*c**2*f**2*exp(10*I*e)*exp(-8*I*f*x))*exp(-18*I*e)/(6144*a**12*f**3), Ne(6144*a**12*f**3*exp(18*I*e)
, 0)), (x*(c**2*exp(4*I*e) + 2*c**2*exp(2*I*e) + c**2)*exp(-8*I*e)/(4*a**4), True))

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Giac [B]  time = 1.46167, size = 189, normalized size = 3.05 \begin{align*} -\frac{2 \,{\left (3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 6 i \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 17 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 16 i \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 17 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 i \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, 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))^2/(a+I*a*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-2/3*(3*c^2*tan(1/2*f*x + 1/2*e)^7 - 6*I*c^2*tan(1/2*f*x + 1/2*e)^6 - 17*c^2*tan(1/2*f*x + 1/2*e)^5 + 16*I*c^2
*tan(1/2*f*x + 1/2*e)^4 + 17*c^2*tan(1/2*f*x + 1/2*e)^3 - 6*I*c^2*tan(1/2*f*x + 1/2*e)^2 - 3*c^2*tan(1/2*f*x +
1/2*e))/(a^4*f*(tan(1/2*f*x + 1/2*e) - I)^8)