3.678 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=99 \[ -\frac{a^2 c^4 (3 B+i A) (1-i \tan (e+f x))^5}{5 f}+\frac{a^2 c^4 (B+i A) (1-i \tan (e+f x))^4}{2 f}+\frac{a^2 B c^4 (1-i \tan (e+f x))^6}{6 f} \]

[Out]

(a^2*(I*A + B)*c^4*(1 - I*Tan[e + f*x])^4)/(2*f) - (a^2*(I*A + 3*B)*c^4*(1 - I*Tan[e + f*x])^5)/(5*f) + (a^2*B
*c^4*(1 - I*Tan[e + f*x])^6)/(6*f)

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Rubi [A]  time = 0.155122, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac{a^2 c^4 (3 B+i A) (1-i \tan (e+f x))^5}{5 f}+\frac{a^2 c^4 (B+i A) (1-i \tan (e+f x))^4}{2 f}+\frac{a^2 B c^4 (1-i \tan (e+f x))^6}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^2*(I*A + B)*c^4*(1 - I*Tan[e + f*x])^4)/(2*f) - (a^2*(I*A + 3*B)*c^4*(1 - I*Tan[e + f*x])^5)/(5*f) + (a^2*B
*c^4*(1 - I*Tan[e + f*x])^6)/(6*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 5.80774, size = 177, normalized size = 1.79 \[ \frac{a^2 c^4 \sec (e) \sec ^6(e+f x) (15 (B-i A) \cos (e+2 f x)+10 (B-3 i A) \cos (e)+30 A \sin (e+2 f x)-15 A \sin (3 e+2 f x)+18 A \sin (3 e+4 f x)+3 A \sin (5 e+6 f x)-15 i A \cos (3 e+2 f x)-30 A \sin (e)-15 i B \sin (3 e+2 f x)+6 i B \sin (3 e+4 f x)+i B \sin (5 e+6 f x)+15 B \cos (3 e+2 f x)-10 i B \sin (e))}{120 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^2*c^4*Sec[e]*Sec[e + f*x]^6*(10*((-3*I)*A + B)*Cos[e] + 15*((-I)*A + B)*Cos[e + 2*f*x] - (15*I)*A*Cos[3*e +
2*f*x] + 15*B*Cos[3*e + 2*f*x] - 30*A*Sin[e] - (10*I)*B*Sin[e] + 30*A*Sin[e + 2*f*x] - 15*A*Sin[3*e + 2*f*x]
- (15*I)*B*Sin[3*e + 2*f*x] + 18*A*Sin[3*e + 4*f*x] + (6*I)*B*Sin[3*e + 4*f*x] + 3*A*Sin[5*e + 6*f*x] + I*B*Si
n[5*e + 6*f*x]))/(120*f)

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Maple [A]  time = 0.011, size = 101, normalized size = 1. \begin{align*}{\frac{{a}^{2}{c}^{4}}{f} \left ( -{\frac{2\,i}{5}}B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{6}}{6}}-{\frac{i}{2}}A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{5}}-{\frac{2\,i}{3}}B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+{\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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x)

[Out]

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

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Maxima [A]  time = 2.43102, size = 158, normalized size = 1.6 \begin{align*} -\frac{10 \, B a^{2} c^{4} \tan \left (f x + e\right )^{6} +{\left (12 \, A + 24 i \, B\right )} a^{2} c^{4} \tan \left (f x + e\right )^{5} + 30 i \, A a^{2} c^{4} \tan \left (f x + e\right )^{4} + 40 i \, B a^{2} c^{4} \tan \left (f x + e\right )^{3} + 30 \,{\left (2 i \, A - B\right )} a^{2} c^{4} \tan \left (f x + e\right )^{2} - 60 \, A a^{2} c^{4} \tan \left (f x + e\right )}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

-1/60*(10*B*a^2*c^4*tan(f*x + e)^6 + (12*A + 24*I*B)*a^2*c^4*tan(f*x + e)^5 + 30*I*A*a^2*c^4*tan(f*x + e)^4 +
40*I*B*a^2*c^4*tan(f*x + e)^3 + 30*(2*I*A - B)*a^2*c^4*tan(f*x + e)^2 - 60*A*a^2*c^4*tan(f*x + e))/f

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Fricas [A]  time = 1.4085, size = 393, normalized size = 3.97 \begin{align*} \frac{{\left (120 i \, A + 120 \, B\right )} a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (144 i \, A - 48 \, B\right )} a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (24 i \, A - 8 \, B\right )} a^{2} c^{4}}{15 \,{\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/15*((120*I*A + 120*B)*a^2*c^4*e^(4*I*f*x + 4*I*e) + (144*I*A - 48*B)*a^2*c^4*e^(2*I*f*x + 2*I*e) + (24*I*A -
8*B)*a^2*c^4)/(f*e^(12*I*f*x + 12*I*e) + 6*f*e^(10*I*f*x + 10*I*e) + 15*f*e^(8*I*f*x + 8*I*e) + 20*f*e^(6*I*f
*x + 6*I*e) + 15*f*e^(4*I*f*x + 4*I*e) + 6*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [B]  time = 40.7263, size = 212, normalized size = 2.14 \begin{align*} \frac{\frac{\left (8 i A a^{2} c^{4} + 8 B a^{2} c^{4}\right ) e^{- 8 i e} e^{4 i f x}}{f} + \frac{\left (24 i A a^{2} c^{4} - 8 B a^{2} c^{4}\right ) e^{- 12 i e}}{15 f} + \frac{\left (48 i A a^{2} c^{4} - 16 B a^{2} c^{4}\right ) e^{- 10 i e} e^{2 i f x}}{5 f}}{e^{12 i f x} + 6 e^{- 2 i e} e^{10 i f x} + 15 e^{- 4 i e} e^{8 i f x} + 20 e^{- 6 i e} e^{6 i f x} + 15 e^{- 8 i e} e^{4 i f x} + 6 e^{- 10 i e} e^{2 i f x} + e^{- 12 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**4,x)

[Out]

((8*I*A*a**2*c**4 + 8*B*a**2*c**4)*exp(-8*I*e)*exp(4*I*f*x)/f + (24*I*A*a**2*c**4 - 8*B*a**2*c**4)*exp(-12*I*e
)/(15*f) + (48*I*A*a**2*c**4 - 16*B*a**2*c**4)*exp(-10*I*e)*exp(2*I*f*x)/(5*f))/(exp(12*I*f*x) + 6*exp(-2*I*e)
*exp(10*I*f*x) + 15*exp(-4*I*e)*exp(8*I*f*x) + 20*exp(-6*I*e)*exp(6*I*f*x) + 15*exp(-8*I*e)*exp(4*I*f*x) + 6*e
xp(-10*I*e)*exp(2*I*f*x) + exp(-12*I*e))

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Giac [B]  time = 1.85636, size = 240, normalized size = 2.42 \begin{align*} \frac{120 i \, A a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 120 \, B a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 144 i \, A a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 48 \, B a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, A a^{2} c^{4} - 8 \, B a^{2} c^{4}}{15 \,{\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

1/15*(120*I*A*a^2*c^4*e^(4*I*f*x + 4*I*e) + 120*B*a^2*c^4*e^(4*I*f*x + 4*I*e) + 144*I*A*a^2*c^4*e^(2*I*f*x + 2
*I*e) - 48*B*a^2*c^4*e^(2*I*f*x + 2*I*e) + 24*I*A*a^2*c^4 - 8*B*a^2*c^4)/(f*e^(12*I*f*x + 12*I*e) + 6*f*e^(10*
I*f*x + 10*I*e) + 15*f*e^(8*I*f*x + 8*I*e) + 20*f*e^(6*I*f*x + 6*I*e) + 15*f*e^(4*I*f*x + 4*I*e) + 6*f*e^(2*I*
f*x + 2*I*e) + f)