\(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx\) [694]

Optimal result

Integrand size = 41, antiderivative size = 101 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=-\frac {2 a^3 (i A-B) c^2 (1+i \tan (e+f x))^3}{3 f}+\frac {a^3 (i A-3 B) c^2 (1+i \tan (e+f x))^4}{4 f}+\frac {a^3 B c^2 (1+i \tan (e+f x))^5}{5 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 78} \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=\frac {a^3 c^2 (-3 B+i A) (1+i \tan (e+f x))^4}{4 f}-\frac {2 a^3 c^2 (-B+i A) (1+i \tan (e+f x))^3}{3 f}+\frac {a^3 B c^2 (1+i \tan (e+f x))^5}{5 f} \]

[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 78

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 3669

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*} \text {integral}& = \frac {(a c) \text {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) \text {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] (verified)

Time = 5.54 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=\frac {a^3 c^2 \sec ^5(e+f x) (30 (i A+B) \cos (e+f x)+2 (25 A+7 i B+6 (5 A-i B) \cos (2 (e+f x))+(5 A-i B) \cos (4 (e+f x))) \sin (e+f x))}{120 f} \]

[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 + f*x]^5*(30*(I*A + B)*Cos[e + f*x] + 2*(25*A + (7*I)*B + 6*(5*A - I*B)*Cos[2*(e + f*x)] + (5*A
- I*B)*Cos[4*(e + f*x)])*Sin[e + f*x]))/(120*f)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {i c^{2} a^{3} \left (\frac {B \tan \left (f x +e \right )^{5}}{5}+\frac {\left (-i B +A \right ) \tan \left (f x +e \right )^{4}}{4}+\frac {\left (i A -2 i \left (i B +A \right )-B \right ) \tan \left (f x +e \right )^{3}}{3}+\frac {\left (-i B +A \right ) \tan \left (f x +e \right )^{2}}{2}-i \tan \left (f x +e \right ) A \right )}{f}\)	\(94\)
default	\(\frac {i c^{2} a^{3} \left (\frac {B \tan \left (f x +e \right )^{5}}{5}+\frac {\left (-i B +A \right ) \tan \left (f x +e \right )^{4}}{4}+\frac {\left (i A -2 i \left (i B +A \right )-B \right ) \tan \left (f x +e \right )^{3}}{3}+\frac {\left (-i B +A \right ) \tan \left (f x +e \right )^{2}}{2}-i \tan \left (f x +e \right ) A \right )}{f}\)	\(94\)
risch	\(\frac {4 c^{2} a^{3} \left (30 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+30 B \,{\mathrm e}^{6 i \left (f x +e \right )}+50 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+10 B \,{\mathrm e}^{4 i \left (f x +e \right )}+25 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+5 B \,{\mathrm e}^{2 i \left (f x +e \right )}+5 i A +B \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\)	\(106\)
norman	\(\frac {A \,a^{3} c^{2} \tan \left (f x +e \right )}{f}+\frac {\left (i A \,a^{3} c^{2}+B \,a^{3} c^{2}\right ) \tan \left (f x +e \right )^{2}}{2 f}+\frac {\left (i A \,a^{3} c^{2}+B \,a^{3} c^{2}\right ) \tan \left (f x +e \right )^{4}}{4 f}+\frac {\left (i B \,a^{3} c^{2}+A \,a^{3} c^{2}\right ) \tan \left (f x +e \right )^{3}}{3 f}+\frac {i B \,a^{3} c^{2} \tan \left (f x +e \right )^{5}}{5 f}\)	\(136\)
parallelrisch	\(\frac {12 i B \,a^{3} c^{2} \tan \left (f x +e \right )^{5}+15 i A \tan \left (f x +e \right )^{4} a^{3} c^{2}+20 i B \tan \left (f x +e \right )^{3} a^{3} c^{2}+15 B \tan \left (f x +e \right )^{4} a^{3} c^{2}+30 i A \tan \left (f x +e \right )^{2} a^{3} c^{2}+20 A \tan \left (f x +e \right )^{3} a^{3} c^{2}+30 B \tan \left (f x +e \right )^{2} a^{3} c^{2}+60 A \tan \left (f x +e \right ) a^{3} c^{2}}{60 f}\)	\(145\)
parts	\(\frac {\left (i A \,a^{3} c^{2}+B \,a^{3} c^{2}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}+\frac {\left (i A \,a^{3} c^{2}+B \,a^{3} c^{2}\right ) \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {\left (i B \,a^{3} c^{2}+2 A \,a^{3} c^{2}\right ) \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {\left (2 i A \,a^{3} c^{2}+2 B \,a^{3} c^{2}\right ) \left (\frac {\tan \left (f x +e \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {\left (2 i B \,a^{3} c^{2}+A \,a^{3} c^{2}\right ) \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+A \,a^{3} c^{2} x +\frac {i B \,a^{3} c^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(289\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.50 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=-\frac {4 \, {\left (30 \, {\left (-i \, A - B\right )} a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (-5 i \, A - B\right )} a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (-5 i \, A - B\right )} a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A - B\right )} a^{3} c^{2}\right )}}{15 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[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]

-4/15*(30*(-I*A - B)*a^3*c^2*e^(6*I*f*x + 6*I*e) + 10*(-5*I*A - B)*a^3*c^2*e^(4*I*f*x + 4*I*e) + 5*(-5*I*A - B
)*a^3*c^2*e^(2*I*f*x + 2*I*e) + (-5*I*A - 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)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (80) = 160\).

Time = 0.41 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.48 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=\frac {20 i A a^{3} c^{2} + 4 B a^{3} c^{2} + \left (100 i A a^{3} c^{2} e^{2 i e} + 20 B a^{3} c^{2} e^{2 i e}\right ) e^{2 i f x} + \left (200 i A a^{3} c^{2} e^{4 i e} + 40 B a^{3} c^{2} e^{4 i e}\right ) e^{4 i f x} + \left (120 i A a^{3} c^{2} e^{6 i e} + 120 B a^{3} c^{2} e^{6 i e}\right ) e^{6 i f x}}{15 f e^{10 i e} e^{10 i f x} + 75 f e^{8 i e} e^{8 i f x} + 150 f e^{6 i e} e^{6 i f x} + 150 f e^{4 i e} e^{4 i f x} + 75 f e^{2 i e} e^{2 i f x} + 15 f} \]

[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]

(20*I*A*a**3*c**2 + 4*B*a**3*c**2 + (100*I*A*a**3*c**2*exp(2*I*e) + 20*B*a**3*c**2*exp(2*I*e))*exp(2*I*f*x) +
(200*I*A*a**3*c**2*exp(4*I*e) + 40*B*a**3*c**2*exp(4*I*e))*exp(4*I*f*x) + (120*I*A*a**3*c**2*exp(6*I*e) + 120*
B*a**3*c**2*exp(6*I*e))*exp(6*I*f*x))/(15*f*exp(10*I*e)*exp(10*I*f*x) + 75*f*exp(8*I*e)*exp(8*I*f*x) + 150*f*e
xp(6*I*e)*exp(6*I*f*x) + 150*f*exp(4*I*e)*exp(4*I*f*x) + 75*f*exp(2*I*e)*exp(2*I*f*x) + 15*f)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=\frac {12 i \, B a^{3} c^{2} \tan \left (f x + e\right )^{5} - 15 \, {\left (-i \, A - B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{4} + 20 \, {\left (A + i \, B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{3} - 30 \, {\left (-i \, A - B\right )} a^{3} c^{2} \tan \left (f x + e\right )^{2} + 60 \, A a^{3} c^{2} \tan \left (f x + e\right )}{60 \, f} \]

[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 + 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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (85) = 170\).

Time = 0.64 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.90 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=-\frac {4 \, {\left (-30 i \, A a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 30 \, B a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 50 i \, A a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 10 \, B a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 25 i \, A a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 \, B a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 i \, A a^{3} c^{2} - B a^{3} c^{2}\right )}}{15 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[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]

-4/15*(-30*I*A*a^3*c^2*e^(6*I*f*x + 6*I*e) - 30*B*a^3*c^2*e^(6*I*f*x + 6*I*e) - 50*I*A*a^3*c^2*e^(4*I*f*x + 4*
I*e) - 10*B*a^3*c^2*e^(4*I*f*x + 4*I*e) - 25*I*A*a^3*c^2*e^(2*I*f*x + 2*I*e) - 5*B*a^3*c^2*e^(2*I*f*x + 2*I*e)
- 5*I*A*a^3*c^2 - 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)

Mupad [B] (verification not implemented)

Time = 8.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^2 \, dx=\frac {A\,a^3\,c^2\,\mathrm {tan}\left (e+f\,x\right )-\frac {a^3\,c^2\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}+\frac {a^3\,c^2\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^3\,c^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\frac {B\,a^3\,c^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}}{5}}{f} \]

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^2,x)

[Out]

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