3.666 \(\int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=59 \[ \frac{a c^4 (B+i A) (1-i \tan (e+f x))^4}{4 f}-\frac{a B c^4 (1-i \tan (e+f x))^5}{5 f} \]

[Out]

(a*(I*A + B)*c^4*(1 - I*Tan[e + f*x])^4)/(4*f) - (a*B*c^4*(1 - I*Tan[e + f*x])^5)/(5*f)

________________________________________________________________________________________

Rubi [A]  time = 0.0835138, 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} \[ \frac{a c^4 (B+i A) (1-i \tan (e+f x))^4}{4 f}-\frac{a B c^4 (1-i \tan (e+f x))^5}{5 f} \]

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])^4,x]

[Out]

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

Mathematica [B]  time = 3.47291, size = 226, normalized size = 3.83 \[ \frac{a c^4 \sec (e) \sec ^5(e+f x) (5 (3 B-5 i A) \cos (2 e+f x)+5 (3 B-5 i A) \cos (f x)-25 A \sin (2 e+f x)+15 A \sin (2 e+3 f x)-10 A \sin (4 e+3 f x)+5 A \sin (4 e+5 f x)-10 i A \cos (2 e+3 f x)-10 i A \cos (4 e+3 f x)+25 A \sin (f x)-15 i B \sin (2 e+f x)+5 i B \sin (2 e+3 f x)-10 i B \sin (4 e+3 f x)+3 i B \sin (4 e+5 f x)+10 B \cos (2 e+3 f x)+10 B \cos (4 e+3 f x)+15 i B \sin (f x))}{40 f} \]

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])^4,x]

[Out]

(a*c^4*Sec[e]*Sec[e + f*x]^5*(5*((-5*I)*A + 3*B)*Cos[f*x] + 5*((-5*I)*A + 3*B)*Cos[2*e + f*x] - (10*I)*A*Cos[2
*e + 3*f*x] + 10*B*Cos[2*e + 3*f*x] - (10*I)*A*Cos[4*e + 3*f*x] + 10*B*Cos[4*e + 3*f*x] + 25*A*Sin[f*x] + (15*
I)*B*Sin[f*x] - 25*A*Sin[2*e + f*x] - (15*I)*B*Sin[2*e + f*x] + 15*A*Sin[2*e + 3*f*x] + (5*I)*B*Sin[2*e + 3*f*
x] - 10*A*Sin[4*e + 3*f*x] - (10*I)*B*Sin[4*e + 3*f*x] + 5*A*Sin[4*e + 5*f*x] + (3*I)*B*Sin[4*e + 5*f*x]))/(40
*f)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 99, normalized size = 1.7 \begin{align*}{\frac{a{c}^{4}}{f} \left ({\frac{i}{5}}B \left ( \tan \left ( fx+e \right ) \right ) ^{5}+{\frac{i}{4}}A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-{\frac{3\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{4}}-{\frac{3\,i}{2}}A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-A \left ( \tan \left ( fx+e \right ) \right ) ^{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))^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 2.18342, size = 131, normalized size = 2.22 \begin{align*} \frac{12 i \, B a c^{4} \tan \left (f x + e\right )^{5} - 15 \,{\left (-i \, A + 3 \, B\right )} a c^{4} \tan \left (f x + e\right )^{4} -{\left (60 \, A + 60 i \, B\right )} a c^{4} \tan \left (f x + e\right )^{3} - 30 \,{\left (3 i \, A - B\right )} a c^{4} \tan \left (f x + e\right )^{2} + 60 \, A a c^{4} \tan \left (f x + e\right )}{60 \, 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))^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.29535, size = 282, normalized size = 4.78 \begin{align*} \frac{{\left (20 i \, A + 20 \, B\right )} a c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (20 i \, A - 12 \, B\right )} a c^{4}}{5 \,{\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 )}} \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))^4,x, algorithm="fricas")

[Out]

1/5*((20*I*A + 20*B)*a*c^4*e^(2*I*f*x + 2*I*e) + (20*I*A - 12*B)*a*c^4)/(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]  time = 15.0631, size = 148, normalized size = 2.51 \begin{align*} \frac{\frac{\left (4 i A a c^{4} + 4 B a c^{4}\right ) e^{- 8 i e} e^{2 i f x}}{f} + \frac{\left (20 i A a c^{4} - 12 B a c^{4}\right ) e^{- 10 i e}}{5 f}}{e^{10 i f x} + 5 e^{- 2 i e} e^{8 i f x} + 10 e^{- 4 i e} e^{6 i f x} + 10 e^{- 6 i e} e^{4 i f x} + 5 e^{- 8 i e} e^{2 i f x} + e^{- 10 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))**4,x)

[Out]

((4*I*A*a*c**4 + 4*B*a*c**4)*exp(-8*I*e)*exp(2*I*f*x)/f + (20*I*A*a*c**4 - 12*B*a*c**4)*exp(-10*I*e)/(5*f))/(e
xp(10*I*f*x) + 5*exp(-2*I*e)*exp(8*I*f*x) + 10*exp(-4*I*e)*exp(6*I*f*x) + 10*exp(-6*I*e)*exp(4*I*f*x) + 5*exp(
-8*I*e)*exp(2*I*f*x) + exp(-10*I*e))

________________________________________________________________________________________

Giac [B]  time = 1.72624, size = 161, normalized size = 2.73 \begin{align*} \frac{20 i \, A a c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 20 \, B a c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 20 i \, A a c^{4} - 12 \, B a c^{4}}{5 \,{\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 )}} \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))^4,x, algorithm="giac")

[Out]

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