3.906 \(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=59 \[ \frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac{2 a^3 c^3 \tan ^3(e+f x)}{3 f}+\frac{a^3 c^3 \tan (e+f x)}{f} \]

[Out]

(a^3*c^3*Tan[e + f*x])/f + (2*a^3*c^3*Tan[e + f*x]^3)/(3*f) + (a^3*c^3*Tan[e + f*x]^5)/(5*f)

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Rubi [A]  time = 0.0681824, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {3522, 3767} \[ \frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac{2 a^3 c^3 \tan ^3(e+f x)}{3 f}+\frac{a^3 c^3 \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^3*Tan[e + f*x])/f + (2*a^3*c^3*Tan[e + f*x]^3)/(3*f) + (a^3*c^3*Tan[e + f*x]^5)/(5*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 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]

Rubi steps

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

Mathematica [A]  time = 0.118951, size = 41, normalized size = 0.69 \[ \frac{a^3 c^3 \left (\frac{1}{5} \tan ^5(e+f x)+\frac{2}{3} \tan ^3(e+f x)+\tan (e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^3*(Tan[e + f*x] + (2*Tan[e + f*x]^3)/3 + Tan[e + f*x]^5/5))/f

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Maple [A]  time = 0.004, size = 38, normalized size = 0.6 \begin{align*}{\frac{{a}^{3}{c}^{3}}{f} \left ({\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{5}}+{\frac{2\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}+\tan \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^3,x)

[Out]

1/f*a^3*c^3*(1/5*tan(f*x+e)^5+2/3*tan(f*x+e)^3+tan(f*x+e))

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Maxima [A]  time = 1.7079, size = 70, normalized size = 1.19 \begin{align*} \frac{3 \, a^{3} c^{3} \tan \left (f x + e\right )^{5} + 10 \, a^{3} c^{3} \tan \left (f x + e\right )^{3} + 15 \, a^{3} c^{3} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^3*c^3*tan(f*x + e)^5 + 10*a^3*c^3*tan(f*x + e)^3 + 15*a^3*c^3*tan(f*x + e))/f

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Fricas [C]  time = 1.44973, size = 308, normalized size = 5.22 \begin{align*} \frac{160 i \, a^{3} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 80 i \, a^{3} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, a^{3} c^{3}}{15 \,{\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(160*I*a^3*c^3*e^(4*I*f*x + 4*I*e) + 80*I*a^3*c^3*e^(2*I*f*x + 2*I*e) + 16*I*a^3*c^3)/(f*e^(10*I*f*x + 10
*I*e) + 5*f*e^(8*I*f*x + 8*I*e) + 10*f*e^(6*I*f*x + 6*I*e) + 10*f*e^(4*I*f*x + 4*I*e) + 5*f*e^(2*I*f*x + 2*I*e
) + f)

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Sympy [C]  time = 6.60708, size = 162, normalized size = 2.75 \begin{align*} \frac{\frac{32 i a^{3} c^{3} e^{- 6 i e} e^{4 i f x}}{3 f} + \frac{16 i a^{3} c^{3} e^{- 8 i e} e^{2 i f x}}{3 f} + \frac{16 i a^{3} c^{3} e^{- 10 i e}}{15 f}}{e^{10 i f x} + 5 e^{- 2 i e} e^{8 i f x} + 10 e^{- 4 i e} e^{6 i f x} + 10 e^{- 6 i e} e^{4 i f x} + 5 e^{- 8 i e} e^{2 i f x} + e^{- 10 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(c-I*c*tan(f*x+e))**3,x)

[Out]

(32*I*a**3*c**3*exp(-6*I*e)*exp(4*I*f*x)/(3*f) + 16*I*a**3*c**3*exp(-8*I*e)*exp(2*I*f*x)/(3*f) + 16*I*a**3*c**
3*exp(-10*I*e)/(15*f))/(exp(10*I*f*x) + 5*exp(-2*I*e)*exp(8*I*f*x) + 10*exp(-4*I*e)*exp(6*I*f*x) + 10*exp(-6*I
*e)*exp(4*I*f*x) + 5*exp(-8*I*e)*exp(2*I*f*x) + exp(-10*I*e))

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Giac [B]  time = 1.85655, size = 501, normalized size = 8.49 \begin{align*} -\frac{15 \, a^{3} c^{3} \tan \left (f x\right )^{5} \tan \left (e\right )^{4} + 15 \, a^{3} c^{3} \tan \left (f x\right )^{4} \tan \left (e\right )^{5} + 10 \, a^{3} c^{3} \tan \left (f x\right )^{5} \tan \left (e\right )^{2} - 30 \, a^{3} c^{3} \tan \left (f x\right )^{4} \tan \left (e\right )^{3} - 30 \, a^{3} c^{3} \tan \left (f x\right )^{3} \tan \left (e\right )^{4} + 10 \, a^{3} c^{3} \tan \left (f x\right )^{2} \tan \left (e\right )^{5} + 3 \, a^{3} c^{3} \tan \left (f x\right )^{5} - 5 \, a^{3} c^{3} \tan \left (f x\right )^{4} \tan \left (e\right ) + 60 \, a^{3} c^{3} \tan \left (f x\right )^{3} \tan \left (e\right )^{2} + 60 \, a^{3} c^{3} \tan \left (f x\right )^{2} \tan \left (e\right )^{3} - 5 \, a^{3} c^{3} \tan \left (f x\right ) \tan \left (e\right )^{4} + 3 \, a^{3} c^{3} \tan \left (e\right )^{5} + 10 \, a^{3} c^{3} \tan \left (f x\right )^{3} - 30 \, a^{3} c^{3} \tan \left (f x\right )^{2} \tan \left (e\right ) - 30 \, a^{3} c^{3} \tan \left (f x\right ) \tan \left (e\right )^{2} + 10 \, a^{3} c^{3} \tan \left (e\right )^{3} + 15 \, a^{3} c^{3} \tan \left (f x\right ) + 15 \, a^{3} c^{3} \tan \left (e\right )}{15 \,{\left (f \tan \left (f x\right )^{5} \tan \left (e\right )^{5} - 5 \, f \tan \left (f x\right )^{4} \tan \left (e\right )^{4} + 10 \, f \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 10 \, f \tan \left (f x\right )^{2} \tan \left (e\right )^{2} + 5 \, f \tan \left (f x\right ) \tan \left (e\right ) - f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(15*a^3*c^3*tan(f*x)^5*tan(e)^4 + 15*a^3*c^3*tan(f*x)^4*tan(e)^5 + 10*a^3*c^3*tan(f*x)^5*tan(e)^2 - 30*a
^3*c^3*tan(f*x)^4*tan(e)^3 - 30*a^3*c^3*tan(f*x)^3*tan(e)^4 + 10*a^3*c^3*tan(f*x)^2*tan(e)^5 + 3*a^3*c^3*tan(f
*x)^5 - 5*a^3*c^3*tan(f*x)^4*tan(e) + 60*a^3*c^3*tan(f*x)^3*tan(e)^2 + 60*a^3*c^3*tan(f*x)^2*tan(e)^3 - 5*a^3*
c^3*tan(f*x)*tan(e)^4 + 3*a^3*c^3*tan(e)^5 + 10*a^3*c^3*tan(f*x)^3 - 30*a^3*c^3*tan(f*x)^2*tan(e) - 30*a^3*c^3
*tan(f*x)*tan(e)^2 + 10*a^3*c^3*tan(e)^3 + 15*a^3*c^3*tan(f*x) + 15*a^3*c^3*tan(e))/(f*tan(f*x)^5*tan(e)^5 - 5
*f*tan(f*x)^4*tan(e)^4 + 10*f*tan(f*x)^3*tan(e)^3 - 10*f*tan(f*x)^2*tan(e)^2 + 5*f*tan(f*x)*tan(e) - f)