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]

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

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Rubi [A]  time = 0.16, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 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]))))

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))^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*}

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Mathematica [A]  time = 6.30, 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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fricas [A]  time = 0.87, size = 138, normalized size = 1.39 \[ \frac {{\left (120 i \, A + 120 \, B\right )} a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (144 i \, A - 48 \, B\right )} a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (24 i \, A - 8 \, B\right )} a^{2} c^{4}}{15 \, {\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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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giac [B]  time = 3.15, size = 178, normalized size = 1.80 \[ \frac {120 i \, A a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 120 \, B a^{2} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 144 i \, A a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 48 \, B a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, A a^{2} c^{4} - 8 \, B a^{2} c^{4}}{15 \, {\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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)

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maple [A]  time = 0.02, size = 101, normalized size = 1.02 \[ \frac {c^{4} a^{2} \left (-\frac {2 i B \left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {B \left (\tan ^{6}\left (f x +e \right )\right )}{6}-\frac {i A \left (\tan ^{4}\left (f x +e \right )\right )}{2}-\frac {A \left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {2 i B \left (\tan ^{3}\left (f x +e \right )\right )}{3}-i A \left (\tan ^{2}\left (f x +e \right )\right )+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{2}+A \tan \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*c^4*a^2*(-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 = 0.89, size = 116, normalized size = 1.17 \[ -\frac {10 \, B a^{2} c^{4} \tan \left (f x + e\right )^{6} + 12 \, {\left (A + 2 i \, B\right )} a^{2} c^{4} \tan \left (f x + e\right )^{5} + 30 i \, A a^{2} c^{4} \tan \left (f x + e\right )^{4} + 40 i \, B a^{2} c^{4} \tan \left (f x + e\right )^{3} - {\left (-60 i \, A + 30 \, B\right )} a^{2} c^{4} \tan \left (f x + e\right )^{2} - 60 \, A a^{2} c^{4} \tan \left (f x + e\right )}{60 \, f} \]

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 + 2*I*B)*a^2*c^4*tan(f*x + e)^5 + 30*I*A*a^2*c^4*tan(f*x + e)^4 + 4
0*I*B*a^2*c^4*tan(f*x + e)^3 - (-60*I*A + 30*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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mupad [B]  time = 8.60, size = 120, normalized size = 1.21 \[ -\frac {\frac {a^2\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-B+A\,2{}\mathrm {i}\right )}{2}-A\,a^2\,c^4\,\mathrm {tan}\left (e+f\,x\right )+\frac {a^2\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (A+B\,2{}\mathrm {i}\right )}{5}+\frac {B\,a^2\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^6}{6}+\frac {A\,a^2\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}}{2}+\frac {B\,a^2\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^3\,2{}\mathrm {i}}{3}}{f} \]

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)^2*(c - c*tan(e + f*x)*1i)^4,x)

[Out]

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

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sympy [B]  time = 1.15, size = 238, normalized size = 2.40 \[ \frac {- 24 A a^{2} c^{4} - 8 i B a^{2} c^{4} + \left (- 144 A a^{2} c^{4} e^{2 i e} - 48 i B a^{2} c^{4} e^{2 i e}\right ) e^{2 i f x} + \left (- 120 A a^{2} c^{4} e^{4 i e} + 120 i B a^{2} c^{4} e^{4 i e}\right ) e^{4 i f x}}{15 i f e^{12 i e} e^{12 i f x} + 90 i f e^{10 i e} e^{10 i f x} + 225 i f e^{8 i e} e^{8 i f x} + 300 i f e^{6 i e} e^{6 i f x} + 225 i f e^{4 i e} e^{4 i f x} + 90 i f e^{2 i e} e^{2 i f x} + 15 i f} \]

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]

(-24*A*a**2*c**4 - 8*I*B*a**2*c**4 + (-144*A*a**2*c**4*exp(2*I*e) - 48*I*B*a**2*c**4*exp(2*I*e))*exp(2*I*f*x)
+ (-120*A*a**2*c**4*exp(4*I*e) + 120*I*B*a**2*c**4*exp(4*I*e))*exp(4*I*f*x))/(15*I*f*exp(12*I*e)*exp(12*I*f*x)
+ 90*I*f*exp(10*I*e)*exp(10*I*f*x) + 225*I*f*exp(8*I*e)*exp(8*I*f*x) + 300*I*f*exp(6*I*e)*exp(6*I*f*x) + 225*
I*f*exp(4*I*e)*exp(4*I*f*x) + 90*I*f*exp(2*I*e)*exp(2*I*f*x) + 15*I*f)

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