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

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

[Out]

((-I/2)*c*(a + I*a*Tan[e + f*x])^2)/f

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Rubi [A]  time = 0.0641839, 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} \[ \frac{a^2 c \tan (e+f x)}{f}+\frac{i a^2 c \sec ^2(e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.121311, size = 45, normalized size = 1.8 \[ \frac{a^2 c \left (i \tan ^2(e+f x)-2 \tan ^{-1}(\tan (e+f x))+2 \tan (e+f x)+2 f x\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*(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 = 0.003, size = 27, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}c \left ({\frac{i}{2}} \left ( \tan \left ( fx+e \right ) \right ) ^{2}+\tan \left ( fx+e \right ) \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.65279, size = 43, normalized size = 1.72 \begin{align*} -\frac{-i \, a^{2} c \tan \left (f x + e\right )^{2} - 2 \, a^{2} c \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))^2*(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.05101, size = 134, normalized size = 5.36 \begin{align*} \frac{4 i \, a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2} c}{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))^2*(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

(4*I*a^2*c*e^(2*I*f*x + 2*I*e) + 2*I*a^2*c)/(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 = 0.981931, size = 75, normalized size = 3. \begin{align*} \frac{\frac{4 i a^{2} c e^{- 2 i e} e^{2 i f x}}{f} + \frac{2 i a^{2} c e^{- 4 i e}}{f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(c-I*c*tan(f*x+e)),x)

[Out]

(4*I*a**2*c*exp(-2*I*e)*exp(2*I*f*x)/f + 2*I*a**2*c*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 [B]  time = 1.42773, size = 72, normalized size = 2.88 \begin{align*} \frac{4 i \, a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2} c}{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))^2*(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

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