3.899 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=25 \[ \frac{i a (c-i c \tan (e+f x))^2}{2 f} \]

[Out]

((I/2)*a*(c - I*c*Tan[e + f*x])^2)/f

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Rubi [A]  time = 0.059842, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3522, 3486, 3767, 8} \[ \frac{a c^2 \tan (e+f x)}{f}-\frac{i a c^2 \sec ^2(e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.122822, size = 45, normalized size = 1.8 \[ \frac{a c^2 \left (-i \tan ^2(e+f x)-2 \tan ^{-1}(\tan (e+f x))+2 \tan (e+f x)+2 f x\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 27, normalized size = 1.1 \begin{align*}{\frac{a{c}^{2} \left ( \tan \left ( fx+e \right ) -{\frac{i}{2}} \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{f}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.70676, size = 43, normalized size = 1.72 \begin{align*} \frac{-i \, a c^{2} \tan \left (f x + e\right )^{2} + 2 \, a c^{2} \tan \left (f x + e\right )}{2 \, 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.11279, size = 88, normalized size = 3.52 \begin{align*} \frac{2 i \, a c^{2}}{f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 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))^2,x, algorithm="fricas")

[Out]

2*I*a*c^2/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [B]  time = 0.654952, size = 49, normalized size = 1.96 \begin{align*} \frac{2 i a c^{2} e^{- 4 i e}}{f \left (e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}\right )} \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))**2,x)

[Out]

2*I*a*c**2*exp(-4*I*e)/(f*(exp(4*I*f*x) + 2*exp(-2*I*e)*exp(2*I*f*x) + exp(-4*I*e)))

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Giac [A]  time = 1.37031, size = 47, normalized size = 1.88 \begin{align*} \frac{2 i \, a c^{2}}{f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 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))^2,x, algorithm="giac")

[Out]

2*I*a*c^2/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)