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

Optimal. Leaf size=59 \[ \frac{a c^3 (B+i A) (1-i \tan (e+f x))^3}{3 f}-\frac{a B c^3 (1-i \tan (e+f x))^4}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.0932845, antiderivative size = 59, 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} \[ \frac{a c^3 (B+i A) (1-i \tan (e+f x))^3}{3 f}-\frac{a B c^3 (1-i \tan (e+f x))^4}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 3.47084, size = 161, normalized size = 2.73 \[ \frac{a c^3 \sec (e) \sec ^4(e+f x) (3 (B-i A) \cos (e+2 f x)+3 (B-2 i A) \cos (e)+5 A \sin (e+2 f x)-3 A \sin (3 e+2 f x)+2 A \sin (3 e+4 f x)-3 i A \cos (3 e+2 f x)-6 A \sin (e)+i B \sin (e+2 f x)-3 i B \sin (3 e+2 f x)+i B \sin (3 e+4 f x)+3 B \cos (3 e+2 f x)-3 i B \sin (e))}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*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 = 0.011, size = 75, normalized size = 1.3 \begin{align*}{\frac{a{c}^{3}}{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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x)

[Out]

1/f*a*c^3*(-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*t
an(f*x+e))

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Maxima [A]  time = 1.70329, size = 99, normalized size = 1.68 \begin{align*} -\frac{3 \, B a c^{3} \tan \left (f x + e\right )^{4} +{\left (4 \, A + 8 i \, B\right )} a c^{3} \tan \left (f x + e\right )^{3} - 6 \,{\left (-2 i \, A + B\right )} a c^{3} \tan \left (f x + e\right )^{2} - 12 \, A a c^{3} \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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.36608, size = 236, normalized size = 4. \begin{align*} \frac{{\left (8 i \, A + 8 \, B\right )} a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A - 4 \, B\right )} a c^{3}}{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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/3*((8*I*A + 8*B)*a*c^3*e^(2*I*f*x + 2*I*e) + (8*I*A - 4*B)*a*c^3)/(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 = 9.01819, size = 133, normalized size = 2.25 \begin{align*} \frac{\frac{\left (8 i A a c^{3} - 4 B a c^{3}\right ) e^{- 8 i e}}{3 f} + \frac{\left (8 i A a c^{3} + 8 B a c^{3}\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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**3,x)

[Out]

((8*I*A*a*c**3 - 4*B*a*c**3)*exp(-8*I*e)/(3*f) + (8*I*A*a*c**3 + 8*B*a*c**3)*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) + exp(-8*I
*e))

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Giac [B]  time = 1.59732, size = 143, normalized size = 2.42 \begin{align*} \frac{8 i \, A a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, B a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A a c^{3} - 4 \, B a c^{3}}{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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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