3.679 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.150024, 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} \[ -\frac{a^2 c^3 (3 B+i A) (1-i \tan (e+f x))^4}{4 f}+\frac{2 a^2 c^3 (B+i A) (1-i \tan (e+f x))^3}{3 f}+\frac{a^2 B c^3 (1-i \tan (e+f x))^5}{5 f} \]

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])^3,x]

[Out]

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

Mathematica [A]  time = 5.69938, size = 146, normalized size = 1.47 \[ \frac{a^2 c^3 \sec (e) \sec ^5(e+f x) (15 (B-i A) \cos (2 e+f x)+15 (B-i A) \cos (f x)-15 A \sin (2 e+f x)+25 A \sin (2 e+3 f x)+5 A \sin (4 e+5 f x)+35 A \sin (f x)-15 i B \sin (2 e+f x)+5 i B \sin (2 e+3 f x)+i B \sin (4 e+5 f x)-5 i B \sin (f x))}{120 f} \]

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])^3,x]

[Out]

(a^2*c^3*Sec[e]*Sec[e + f*x]^5*(15*((-I)*A + B)*Cos[f*x] + 15*((-I)*A + B)*Cos[2*e + f*x] + 35*A*Sin[f*x] - (5
*I)*B*Sin[f*x] - 15*A*Sin[2*e + f*x] - (15*I)*B*Sin[2*e + f*x] + 25*A*Sin[2*e + 3*f*x] + (5*I)*B*Sin[2*e + 3*f
*x] + 5*A*Sin[4*e + 5*f*x] + I*B*Sin[4*e + 5*f*x]))/(120*f)

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Maple [A]  time = 0.012, size = 101, normalized size = 1. \begin{align*}{\frac{{c}^{3}{a}^{2}}{f} \left ( -{\frac{i}{5}}B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-{\frac{i}{4}}A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-{\frac{i}{3}}B \left ( \tan \left ( fx+e \right ) \right ) ^{3}+{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{4}}-{\frac{i}{2}}A \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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.90972, size = 139, normalized size = 1.4 \begin{align*} -\frac{12 i \, B a^{2} c^{3} \tan \left (f x + e\right )^{5} - 15 \,{\left (-i \, A + B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{4} -{\left (20 \, A - 20 i \, B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{3} - 30 \,{\left (-i \, A + B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{2} - 60 \, A a^{2} c^{3} \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))^3,x, algorithm="maxima")

[Out]

-1/60*(12*I*B*a^2*c^3*tan(f*x + e)^5 - 15*(-I*A + B)*a^2*c^3*tan(f*x + e)^4 - (20*A - 20*I*B)*a^2*c^3*tan(f*x
+ e)^3 - 30*(-I*A + B)*a^2*c^3*tan(f*x + e)^2 - 60*A*a^2*c^3*tan(f*x + e))/f

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Fricas [A]  time = 1.38537, size = 351, normalized size = 3.55 \begin{align*} \frac{{\left (80 i \, A + 80 \, B\right )} a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (100 i \, A - 20 \, B\right )} a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (20 i \, A - 4 \, B\right )} a^{2} 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*((80*I*A + 80*B)*a^2*c^3*e^(4*I*f*x + 4*I*e) + (100*I*A - 20*B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (20*I*A - 4
*B)*a^2*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 [B]  time = 18.0525, size = 197, normalized size = 1.99 \begin{align*} \frac{\frac{\left (16 i A a^{2} c^{3} + 16 B a^{2} c^{3}\right ) e^{- 6 i e} e^{4 i f x}}{3 f} + \frac{\left (20 i A a^{2} c^{3} - 4 B a^{2} c^{3}\right ) e^{- 8 i e} e^{2 i f x}}{3 f} + \frac{\left (20 i A a^{2} c^{3} - 4 B a^{2} c^{3}\right ) 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))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**3,x)

[Out]

((16*I*A*a**2*c**3 + 16*B*a**2*c**3)*exp(-6*I*e)*exp(4*I*f*x)/(3*f) + (20*I*A*a**2*c**3 - 4*B*a**2*c**3)*exp(-
8*I*e)*exp(2*I*f*x)/(3*f) + (20*I*A*a**2*c**3 - 4*B*a**2*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) + e
xp(-10*I*e))

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Giac [A]  time = 1.69948, size = 223, normalized size = 2.25 \begin{align*} \frac{80 i \, A a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 80 \, B a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 100 i \, A a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 20 \, B a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 20 i \, A a^{2} c^{3} - 4 \, B a^{2} 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(80*I*A*a^2*c^3*e^(4*I*f*x + 4*I*e) + 80*B*a^2*c^3*e^(4*I*f*x + 4*I*e) + 100*I*A*a^2*c^3*e^(2*I*f*x + 2*I
*e) - 20*B*a^2*c^3*e^(2*I*f*x + 2*I*e) + 20*I*A*a^2*c^3 - 4*B*a^2*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)