\(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx\) [679]

Optimal result

Integrand size = 41, antiderivative size = 99 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=\frac {2 a^2 (i A+B) c^3 (1-i \tan (e+f x))^3}{3 f}-\frac {a^2 (i A+3 B) c^3 (1-i \tan (e+f x))^4}{4 f}+\frac {a^2 B c^3 (1-i \tan (e+f x))^5}{5 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 99, 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} \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=-\frac {a^2 c^3 (3 B+i A) (1-i \tan (e+f x))^4}{4 f}+\frac {2 a^2 c^3 (B+i A) (1-i \tan (e+f x))^3}{3 f}+\frac {a^2 B c^3 (1-i \tan (e+f x))^5}{5 f} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.62 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=\frac {a^2 c^3 \left (5 i A+11 B+60 A \tan (e+f x)+30 (-i A+B) \tan ^2(e+f x)+20 (A-i B) \tan ^3(e+f x)+15 (-i A+B) \tan ^4(e+f x)-12 i B \tan ^5(e+f x)\right )}{60 f} \]

[In]

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

[Out]

(a^2*c^3*((5*I)*A + 11*B + 60*A*Tan[e + f*x] + 30*((-I)*A + B)*Tan[e + f*x]^2 + 20*(A - I*B)*Tan[e + f*x]^3 +
15*((-I)*A + B)*Tan[e + f*x]^4 - (12*I)*B*Tan[e + f*x]^5))/(60*f)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {4 c^{3} a^{2} \left (20 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+20 B \,{\mathrm e}^{4 i \left (f x +e \right )}+25 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-5 B \,{\mathrm e}^{2 i \left (f x +e \right )}+5 i A -B \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\)	\(83\)
derivativedivides	\(\frac {i c^{3} a^{2} \left (-\frac {B \tan \left (f x +e \right )^{5}}{5}+\frac {\left (-i B -A \right ) \tan \left (f x +e \right )^{4}}{4}+\frac {\left (i A +2 i \left (i B -A \right )+B \right ) \tan \left (f x +e \right )^{3}}{3}+\frac {\left (-i B -A \right ) \tan \left (f x +e \right )^{2}}{2}-i \tan \left (f x +e \right ) A \right )}{f}\)	\(98\)
default	\(\frac {i c^{3} a^{2} \left (-\frac {B \tan \left (f x +e \right )^{5}}{5}+\frac {\left (-i B -A \right ) \tan \left (f x +e \right )^{4}}{4}+\frac {\left (i A +2 i \left (i B -A \right )+B \right ) \tan \left (f x +e \right )^{3}}{3}+\frac {\left (-i B -A \right ) \tan \left (f x +e \right )^{2}}{2}-i \tan \left (f x +e \right ) A \right )}{f}\)	\(98\)
norman	\(\frac {A \,a^{2} c^{3} \tan \left (f x +e \right )}{f}+\frac {\left (-i A \,a^{2} c^{3}+B \,a^{2} c^{3}\right ) \tan \left (f x +e \right )^{2}}{2 f}+\frac {\left (-i A \,a^{2} c^{3}+B \,a^{2} c^{3}\right ) \tan \left (f x +e \right )^{4}}{4 f}+\frac {\left (-i B \,a^{2} c^{3}+A \,a^{2} c^{3}\right ) \tan \left (f x +e \right )^{3}}{3 f}-\frac {i B \,a^{2} c^{3} \tan \left (f x +e \right )^{5}}{5 f}\)	\(136\)
parallelrisch	\(-\frac {12 i B \,a^{2} c^{3} \tan \left (f x +e \right )^{5}+15 i A \tan \left (f x +e \right )^{4} a^{2} c^{3}+20 i B \tan \left (f x +e \right )^{3} a^{2} c^{3}-15 B \tan \left (f x +e \right )^{4} a^{2} c^{3}+30 i A \tan \left (f x +e \right )^{2} a^{2} c^{3}-20 A \tan \left (f x +e \right )^{3} a^{2} c^{3}-30 B \tan \left (f x +e \right )^{2} a^{2} c^{3}-60 A \tan \left (f x +e \right ) a^{2} c^{3}}{60 f}\)	\(145\)
parts	\(\frac {\left (-2 i A \,a^{2} c^{3}+2 B \,a^{2} c^{3}\right ) \left (\frac {\tan \left (f x +e \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {\left (-2 i B \,a^{2} c^{3}+A \,a^{2} c^{3}\right ) \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {\left (-i A \,a^{2} c^{3}+B \,a^{2} c^{3}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}+\frac {\left (-i A \,a^{2} c^{3}+B \,a^{2} c^{3}\right ) \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {\left (-i B \,a^{2} c^{3}+2 A \,a^{2} c^{3}\right ) \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+A \,a^{2} c^{3} x -\frac {i B \,a^{2} c^{3} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(289\)

[In]

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

4/15*c^3*a^2*(20*I*A*exp(4*I*(f*x+e))+20*B*exp(4*I*(f*x+e))+25*I*A*exp(2*I*(f*x+e))-5*B*exp(2*I*(f*x+e))+5*I*A
-B)/f/(exp(2*I*(f*x+e))+1)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.25 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=-\frac {4 \, {\left (20 \, {\left (-i \, A - B\right )} a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (-5 i \, A + B\right )} a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A + B\right )} a^{2} c^{3}\right )}}{15 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-4/15*(20*(-I*A - B)*a^2*c^3*e^(4*I*f*x + 4*I*e) + 5*(-5*I*A + B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (-5*I*A + B)*a
^2*c^3)/(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. 206 vs. \(2 (80) = 160\).

Time = 0.39 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.08 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=\frac {20 i A a^{2} c^{3} - 4 B a^{2} c^{3} + \left (100 i A a^{2} c^{3} e^{2 i e} - 20 B a^{2} c^{3} e^{2 i e}\right ) e^{2 i f x} + \left (80 i A a^{2} c^{3} e^{4 i e} + 80 B a^{2} c^{3} e^{4 i e}\right ) e^{4 i f x}}{15 f e^{10 i e} e^{10 i f x} + 75 f e^{8 i e} e^{8 i f x} + 150 f e^{6 i e} e^{6 i f x} + 150 f e^{4 i e} e^{4 i f x} + 75 f e^{2 i e} e^{2 i f x} + 15 f} \]

[In]

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**3,x)

[Out]

(20*I*A*a**2*c**3 - 4*B*a**2*c**3 + (100*I*A*a**2*c**3*exp(2*I*e) - 20*B*a**2*c**3*exp(2*I*e))*exp(2*I*f*x) +
(80*I*A*a**2*c**3*exp(4*I*e) + 80*B*a**2*c**3*exp(4*I*e))*exp(4*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*exp(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 = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=-\frac {12 i \, B a^{2} c^{3} \tan \left (f x + e\right )^{5} - 15 \, {\left (-i \, A + B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{4} - 20 \, {\left (A - i \, B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{3} - 30 \, {\left (-i \, A + B\right )} a^{2} c^{3} \tan \left (f x + e\right )^{2} - 60 \, A a^{2} c^{3} \tan \left (f x + e\right )}{60 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.57 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=-\frac {4 \, {\left (-20 i \, A a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 20 \, B a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 25 i \, A a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 5 \, B a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 i \, A a^{2} c^{3} + B a^{2} c^{3}\right )}}{15 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-4/15*(-20*I*A*a^2*c^3*e^(4*I*f*x + 4*I*e) - 20*B*a^2*c^3*e^(4*I*f*x + 4*I*e) - 25*I*A*a^2*c^3*e^(2*I*f*x + 2*
I*e) + 5*B*a^2*c^3*e^(2*I*f*x + 2*I*e) - 5*I*A*a^2*c^3 + B*a^2*c^3)/(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 = 9.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.09 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx=-\frac {-A\,a^2\,c^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {a^2\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^2\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}+\frac {a^2\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\frac {B\,a^2\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}}{5}}{f} \]

[In]

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

[Out]

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