3.902 \(\int \frac{(c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=58 \[ \frac{2 i c^2}{3 f (a+i a \tan (e+f x))^3}-\frac{i c^2}{2 a f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.107872, antiderivative size = 58, 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} \[ \frac{2 i c^2}{3 f (a+i a \tan (e+f x))^3}-\frac{i c^2}{2 a f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.5956, size = 53, normalized size = 0.91 \[ \frac{c^2 (5 \cos (e+f x)+i \sin (e+f x)) (\sin (5 (e+f x))+i \cos (5 (e+f x)))}{24 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(5*Cos[e + f*x] + I*Sin[e + f*x])*(I*Cos[5*(e + f*x)] + Sin[5*(e + f*x)]))/(24*a^3*f)

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Maple [A]  time = 0.029, size = 39, normalized size = 0.7 \begin{align*}{\frac{{c}^{2}}{f{a}^{3}} \left ({\frac{{\frac{i}{2}}}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{2}{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))^3,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.33407, size = 101, normalized size = 1.74 \begin{align*} \frac{{\left (3 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{2}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.879648, size = 109, normalized size = 1.88 \begin{align*} \begin{cases} \frac{\left (12 i a^{3} c^{2} f e^{6 i e} e^{- 4 i f x} + 8 i a^{3} c^{2} f e^{4 i e} e^{- 6 i f x}\right ) e^{- 10 i e}}{96 a^{6} f^{2}} & \text{for}\: 96 a^{6} f^{2} e^{10 i e} \neq 0 \\\frac{x \left (c^{2} e^{2 i e} + c^{2}\right ) e^{- 6 i e}}{2 a^{3}} & \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))**3,x)

[Out]

Piecewise(((12*I*a**3*c**2*f*exp(6*I*e)*exp(-4*I*f*x) + 8*I*a**3*c**2*f*exp(4*I*e)*exp(-6*I*f*x))*exp(-10*I*e)
/(96*a**6*f**2), Ne(96*a**6*f**2*exp(10*I*e), 0)), (x*(c**2*exp(2*I*e) + c**2)*exp(-6*I*e)/(2*a**3), True))

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Giac [B]  time = 1.41555, size = 143, normalized size = 2.47 \begin{align*} -\frac{2 \,{\left (3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 i \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 8 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 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^{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((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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