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]

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

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Rubi [A]  time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [B]  time = 3.98, 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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fricas [B]  time = 0.66, size = 129, normalized size = 2.11 \[ \frac {{\left (24 i \, A + 24 \, B\right )} a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (48 i \, A + 24 \, B\right )} a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (32 i \, A + 16 \, B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (8 i \, A + 4 \, B\right )} a^{3} c}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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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giac [B]  time = 1.70, size = 174, normalized size = 2.85 \[ \frac {24 i \, A a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + 24 \, B a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, A a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 24 \, B a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 32 i \, A a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 16 \, B a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A a^{3} c + 4 \, B a^{3} c}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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)

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maple [A]  time = 0.02, size = 75, normalized size = 1.23 \[ \frac {a^{3} c \left (\frac {2 i B \left (\tan ^{3}\left (f x +e \right )\right )}{3}-\frac {B \left (\tan ^{4}\left (f x +e \right )\right )}{4}+i A \left (\tan ^{2}\left (f x +e \right )\right )-\frac {A \left (\tan ^{3}\left (f x +e \right )\right )}{3}+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{2}+A \tan \left (f x +e \right )\right )}{f} \]

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 = 0.53, size = 73, normalized size = 1.20 \[ -\frac {3 \, B a^{3} c \tan \left (f x + e\right )^{4} + 4 \, {\left (A - 2 i \, B\right )} a^{3} c \tan \left (f x + e\right )^{3} + {\left (-12 i \, A - 6 \, B\right )} a^{3} c \tan \left (f x + e\right )^{2} - 12 \, A a^{3} c \tan \left (f x + e\right )}{12 \, f} \]

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

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mupad [B]  time = 8.48, size = 72, normalized size = 1.18 \[ \frac {-\frac {B\,c\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4}-\frac {c\,\left (A-B\,2{}\mathrm {i}\right )\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3}+\frac {c\,\left (B+A\,2{}\mathrm {i}\right )\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+A\,c\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i),x)

[Out]

(A*a^3*c*tan(e + f*x) + (a^3*c*tan(e + f*x)^2*(A*2i + B))/2 - (a^3*c*tan(e + f*x)^3*(A - B*2i))/3 - (B*a^3*c*t
an(e + f*x)^4)/4)/f

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sympy [B]  time = 0.64, size = 224, normalized size = 3.67 \[ \frac {- 8 i A a^{3} c - 4 B a^{3} c + \left (- 32 i A a^{3} c e^{2 i e} - 16 B a^{3} c e^{2 i e}\right ) e^{2 i f x} + \left (- 48 i A a^{3} c e^{4 i e} - 24 B a^{3} c e^{4 i e}\right ) e^{4 i f x} + \left (- 24 i A a^{3} c e^{6 i e} - 24 B a^{3} c e^{6 i e}\right ) e^{6 i f x}}{- 3 f e^{8 i e} e^{8 i f x} - 12 f e^{6 i e} e^{6 i f x} - 18 f e^{4 i e} e^{4 i f x} - 12 f e^{2 i e} e^{2 i f x} - 3 f} \]

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 + (-32*I*A*a**3*c*exp(2*I*e) - 16*B*a**3*c*exp(2*I*e))*exp(2*I*f*x) + (-48*I*A*a**
3*c*exp(4*I*e) - 24*B*a**3*c*exp(4*I*e))*exp(4*I*f*x) + (-24*I*A*a**3*c*exp(6*I*e) - 24*B*a**3*c*exp(6*I*e))*e
xp(6*I*f*x))/(-3*f*exp(8*I*e)*exp(8*I*f*x) - 12*f*exp(6*I*e)*exp(6*I*f*x) - 18*f*exp(4*I*e)*exp(4*I*f*x) - 12*
f*exp(2*I*e)*exp(2*I*f*x) - 3*f)

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