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

Optimal. Leaf size=61 \[ -\frac{a^3 c (-B+i A) (1+i \tan (e+f x))^3}{3 f}-\frac{a^3 B c (1+i \tan (e+f x))^4}{4 f} \]

[Out]

-(a^3*(I*A - B)*c*(1 + I*Tan[e + f*x])^3)/(3*f) - (a^3*B*c*(1 + I*Tan[e + f*x])^4)/(4*f)

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Rubi [A]  time = 0.0874857, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.051, Rules used = {3588, 43} \[ -\frac{a^3 c (-B+i A) (1+i \tan (e+f x))^3}{3 f}-\frac{a^3 B c (1+i \tan (e+f x))^4}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(I*A - B)*c*(1 + I*Tan[e + f*x])^3)/(3*f) - (a^3*B*c*(1 + I*Tan[e + f*x])^4)/(4*f)

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

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

Mathematica [B]  time = 3.61148, size = 161, normalized size = 2.64 \[ \frac{a^3 c \sec (e) \sec ^4(e+f x) (3 (B+i A) \cos (e+2 f x)+3 (B+2 i A) \cos (e)+5 A \sin (e+2 f x)-3 A \sin (3 e+2 f x)+2 A \sin (3 e+4 f x)+3 i A \cos (3 e+2 f x)-6 A \sin (e)-i B \sin (e+2 f x)+3 i B \sin (3 e+2 f x)-i B \sin (3 e+4 f x)+3 B \cos (3 e+2 f x)+3 i B \sin (e))}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*Sec[e]*Sec[e + f*x]^4*(3*((2*I)*A + B)*Cos[e] + 3*(I*A + B)*Cos[e + 2*f*x] + (3*I)*A*Cos[3*e + 2*f*x] +
3*B*Cos[3*e + 2*f*x] - 6*A*Sin[e] + (3*I)*B*Sin[e] + 5*A*Sin[e + 2*f*x] - I*B*Sin[e + 2*f*x] - 3*A*Sin[3*e +
2*f*x] + (3*I)*B*Sin[3*e + 2*f*x] + 2*A*Sin[3*e + 4*f*x] - I*B*Sin[3*e + 4*f*x]))/(12*f)

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Maple [A]  time = 0.012, size = 75, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}c}{f} \left ({\frac{2\,i}{3}}B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{4}}+iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}-{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}+{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2}}+A\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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e)),x)

[Out]

1/f*a^3*c*(2/3*I*B*tan(f*x+e)^3-1/4*B*tan(f*x+e)^4+I*A*tan(f*x+e)^2-1/3*A*tan(f*x+e)^3+1/2*B*tan(f*x+e)^2+A*ta
n(f*x+e))

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Maxima [A]  time = 1.65207, size = 99, normalized size = 1.62 \begin{align*} -\frac{3 \, B a^{3} c \tan \left (f x + e\right )^{4} +{\left (4 \, A - 8 i \, B\right )} a^{3} c \tan \left (f x + e\right )^{3} - 6 \,{\left (2 i \, A + B\right )} a^{3} c \tan \left (f x + e\right )^{2} - 12 \, A a^{3} c \tan \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*B*a^3*c*tan(f*x + e)^4 + (4*A - 8*I*B)*a^3*c*tan(f*x + e)^3 - 6*(2*I*A + B)*a^3*c*tan(f*x + e)^2 - 12
*A*a^3*c*tan(f*x + e))/f

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Fricas [B]  time = 1.32911, size = 358, normalized size = 5.87 \begin{align*} \frac{{\left (24 i \, A + 24 \, B\right )} a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (48 i \, A + 24 \, B\right )} a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (32 i \, A + 16 \, B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A + 4 \, B\right )} a^{3} c}{3 \,{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, 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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/3*((24*I*A + 24*B)*a^3*c*e^(6*I*f*x + 6*I*e) + (48*I*A + 24*B)*a^3*c*e^(4*I*f*x + 4*I*e) + (32*I*A + 16*B)*a
^3*c*e^(2*I*f*x + 2*I*e) + (8*I*A + 4*B)*a^3*c)/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*
f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [B]  time = 50.1462, size = 204, normalized size = 3.34 \begin{align*} \frac{\frac{\left (8 i A a^{3} c + 4 B a^{3} c\right ) e^{- 8 i e}}{3 f} + \frac{\left (8 i A a^{3} c + 8 B a^{3} c\right ) e^{- 2 i e} e^{6 i f x}}{f} + \frac{\left (16 i A a^{3} c + 8 B a^{3} c\right ) e^{- 4 i e} e^{4 i f x}}{f} + \frac{\left (32 i A a^{3} c + 16 B a^{3} c\right ) e^{- 6 i e} e^{2 i f x}}{3 f}}{e^{8 i f x} + 4 e^{- 2 i e} e^{6 i f x} + 6 e^{- 4 i e} e^{4 i f x} + 4 e^{- 6 i e} e^{2 i f x} + e^{- 8 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*I*A*a**3*c + 4*B*a**3*c)*exp(-8*I*e)/(3*f) + (8*I*A*a**3*c + 8*B*a**3*c)*exp(-2*I*e)*exp(6*I*f*x)/f + (16*
I*A*a**3*c + 8*B*a**3*c)*exp(-4*I*e)*exp(4*I*f*x)/f + (32*I*A*a**3*c + 16*B*a**3*c)*exp(-6*I*e)*exp(2*I*f*x)/(
3*f))/(exp(8*I*f*x) + 4*exp(-2*I*e)*exp(6*I*f*x) + 6*exp(-4*I*e)*exp(4*I*f*x) + 4*exp(-6*I*e)*exp(2*I*f*x) + e
xp(-8*I*e))

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Giac [B]  time = 1.60286, size = 235, normalized size = 3.85 \begin{align*} \frac{24 i \, A a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + 24 \, B a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, A a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 24 \, B a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 32 i \, A a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 16 \, B a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A a^{3} c + 4 \, B a^{3} c}{3 \,{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, 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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

1/3*(24*I*A*a^3*c*e^(6*I*f*x + 6*I*e) + 24*B*a^3*c*e^(6*I*f*x + 6*I*e) + 48*I*A*a^3*c*e^(4*I*f*x + 4*I*e) + 24
*B*a^3*c*e^(4*I*f*x + 4*I*e) + 32*I*A*a^3*c*e^(2*I*f*x + 2*I*e) + 16*B*a^3*c*e^(2*I*f*x + 2*I*e) + 8*I*A*a^3*c
+ 4*B*a^3*c)/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*
I*e) + f)