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

Optimal. Leaf size=101 \[ \frac{a^3 c^2 (-3 B+i A) (1+i \tan (e+f x))^4}{4 f}-\frac{2 a^3 c^2 (-B+i A) (1+i \tan (e+f x))^3}{3 f}+\frac{a^3 B c^2 (1+i \tan (e+f x))^5}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.148442, antiderivative size = 101, 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} \[ \frac{a^3 c^2 (-3 B+i A) (1+i \tan (e+f x))^4}{4 f}-\frac{2 a^3 c^2 (-B+i A) (1+i \tan (e+f x))^3}{3 f}+\frac{a^3 B c^2 (1+i \tan (e+f x))^5}{5 f} \]

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])^2,x]

[Out]

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

Mathematica [A]  time = 5.12569, size = 146, normalized size = 1.45 \[ \frac{a^3 c^2 \sec (e) \sec ^5(e+f x) (15 (B+i A) \cos (2 e+f x)+15 (B+i A) \cos (f x)-15 A \sin (2 e+f x)+25 A \sin (2 e+3 f x)+5 A \sin (4 e+5 f x)+35 A \sin (f x)+15 i B \sin (2 e+f x)-5 i B \sin (2 e+3 f x)-i B \sin (4 e+5 f x)+5 i B \sin (f x))}{120 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^2,x]

[Out]

(a^3*c^2*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 = 0.011, size = 101, normalized size = 1. \begin{align*}{\frac{{c}^{2}{a}^{3}}{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))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2,x)

[Out]

1/f*c^2*a^3*(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 = 1.61037, size = 143, normalized size = 1.42 \begin{align*} \frac{12 i \, B a^{3} c^{2} \tan \left (f x + e\right )^{5} - 15 \,{\left (-i \, A - B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{4} +{\left (20 \, A + 20 i \, B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{3} - 30 \,{\left (-i \, A - B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{2} + 60 \, A a^{3} c^{2} \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))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.38261, size = 417, normalized size = 4.13 \begin{align*} \frac{{\left (120 i \, A + 120 \, B\right )} a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (200 i \, A + 40 \, B\right )} a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (100 i \, A + 20 \, B\right )} a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (20 i \, A + 4 \, B\right )} a^{3} c^{2}}{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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/15*((120*I*A + 120*B)*a^3*c^2*e^(6*I*f*x + 6*I*e) + (200*I*A + 40*B)*a^3*c^2*e^(4*I*f*x + 4*I*e) + (100*I*A
+ 20*B)*a^3*c^2*e^(2*I*f*x + 2*I*e) + (20*I*A + 4*B)*a^3*c^2)/(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 = 36.3931, size = 236, normalized size = 2.34 \begin{align*} \frac{\frac{\left (8 i A a^{3} c^{2} + 8 B a^{3} c^{2}\right ) e^{- 4 i e} e^{6 i f x}}{f} + \frac{\left (20 i A a^{3} c^{2} + 4 B a^{3} c^{2}\right ) e^{- 8 i e} e^{2 i f x}}{3 f} + \frac{\left (20 i A a^{3} c^{2} + 4 B a^{3} c^{2}\right ) e^{- 10 i e}}{15 f} + \frac{\left (40 i A a^{3} c^{2} + 8 B a^{3} c^{2}\right ) e^{- 6 i e} e^{4 i f x}}{3 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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**2,x)

[Out]

((8*I*A*a**3*c**2 + 8*B*a**3*c**2)*exp(-4*I*e)*exp(6*I*f*x)/f + (20*I*A*a**3*c**2 + 4*B*a**3*c**2)*exp(-8*I*e)
*exp(2*I*f*x)/(3*f) + (20*I*A*a**3*c**2 + 4*B*a**3*c**2)*exp(-10*I*e)/(15*f) + (40*I*A*a**3*c**2 + 8*B*a**3*c*
*2)*exp(-6*I*e)*exp(4*I*f*x)/(3*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 = 1.71265, size = 274, normalized size = 2.71 \begin{align*} \frac{120 i \, A a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 120 \, B a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 200 i \, A a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 40 \, B a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 100 i \, A a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 20 \, B a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 20 i \, A a^{3} c^{2} + 4 \, B a^{3} c^{2}}{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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/15*(120*I*A*a^3*c^2*e^(6*I*f*x + 6*I*e) + 120*B*a^3*c^2*e^(6*I*f*x + 6*I*e) + 200*I*A*a^3*c^2*e^(4*I*f*x + 4
*I*e) + 40*B*a^3*c^2*e^(4*I*f*x + 4*I*e) + 100*I*A*a^3*c^2*e^(2*I*f*x + 2*I*e) + 20*B*a^3*c^2*e^(2*I*f*x + 2*I
*e) + 20*I*A*a^3*c^2 + 4*B*a^3*c^2)/(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)