3.697 \(\int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=119 \[ \frac{a^3 (A-4 i B) \tan (e+f x)}{c f}+\frac{4 a^3 (A-i B)}{c f (\tan (e+f x)+i)}+\frac{4 a^3 (2 B+i A) \log (\cos (e+f x))}{c f}-\frac{4 a^3 x (A-2 i B)}{c}+\frac{a^3 B \tan ^2(e+f x)}{2 c f} \]

[Out]

(-4*a^3*(A - (2*I)*B)*x)/c + (4*a^3*(I*A + 2*B)*Log[Cos[e + f*x]])/(c*f) + (a^3*(A - (4*I)*B)*Tan[e + f*x])/(c
*f) + (a^3*B*Tan[e + f*x]^2)/(2*c*f) + (4*a^3*(A - I*B))/(c*f*(I + Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.17192, antiderivative size = 119, normalized size of antiderivative = 1., 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} \[ \frac{a^3 (A-4 i B) \tan (e+f x)}{c f}+\frac{4 a^3 (A-i B)}{c f (\tan (e+f x)+i)}+\frac{4 a^3 (2 B+i A) \log (\cos (e+f x))}{c f}-\frac{4 a^3 x (A-2 i B)}{c}+\frac{a^3 B \tan ^2(e+f x)}{2 c f} \]

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]

(-4*a^3*(A - (2*I)*B)*x)/c + (4*a^3*(I*A + 2*B)*Log[Cos[e + f*x]])/(c*f) + (a^3*(A - (4*I)*B)*Tan[e + f*x])/(c
*f) + (a^3*B*Tan[e + f*x]^2)/(2*c*f) + (4*a^3*(A - I*B))/(c*f*(I + Tan[e + f*x]))

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 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]))))

Rubi steps

\begin{align*} \int \frac{(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 \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{a^2 (A-4 i B)}{c^2}+\frac{a^2 B x}{c^2}-\frac{4 a^2 (A-i B)}{c^2 (i+x)^2}-\frac{4 i a^2 (A-2 i B)}{c^2 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{4 a^3 (A-2 i B) x}{c}+\frac{4 a^3 (i A+2 B) \log (\cos (e+f x))}{c f}+\frac{a^3 (A-4 i B) \tan (e+f x)}{c f}+\frac{a^3 B \tan ^2(e+f x)}{2 c f}+\frac{4 a^3 (A-i B)}{c f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 9.46192, size = 944, normalized size = 7.93 \[ \frac{x \left (-\frac{2 A \cos ^3(e)}{c}+\frac{4 i B \cos ^3(e)}{c}+\frac{8 i A \sin (e) \cos ^2(e)}{c}+\frac{16 B \sin (e) \cos ^2(e)}{c}+\frac{12 A \sin ^2(e) \cos (e)}{c}-\frac{24 i B \sin ^2(e) \cos (e)}{c}+\frac{2 A \cos (e)}{c}-\frac{4 i B \cos (e)}{c}-\frac{8 i A \sin ^3(e)}{c}-\frac{16 B \sin ^3(e)}{c}-\frac{4 i A \sin (e)}{c}-\frac{8 B \sin (e)}{c}-\frac{2 A \sin ^3(e) \tan (e)}{c}+\frac{4 i B \sin ^3(e) \tan (e)}{c}-\frac{2 A \sin (e) \tan (e)}{c}+\frac{4 i B \sin (e) \tan (e)}{c}-i (A-2 i B) \left (\frac{4 \cos (3 e)}{c}-\frac{4 i \sin (3 e)}{c}\right ) \tan (e)\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{(\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-i B) \left (\frac{2 \cos (e)}{c}-\frac{2 i \sin (e)}{c}\right ) \sin (2 f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-i B) \cos (2 f x) \left (-\frac{2 i \cos (e)}{c}-\frac{2 \sin (e)}{c}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-2 i B) \left (\frac{4 i f x \sin (3 e)}{c}-\frac{4 f x \cos (3 e)}{c}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(i A+2 B) \left (\frac{2 \cos (3 e) \log \left (\cos ^2(e+f x)\right )}{c}-\frac{2 i \log \left (\cos ^2(e+f x)\right ) \sin (3 e)}{c}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{\left (\frac{\cos (3 e)}{c}-\frac{i \sin (3 e)}{c}\right ) (A \sin (f x)-4 i B \sin (f x)) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^3(e+f x)}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{\left (\frac{B \cos (3 e)}{2 c}-\frac{i B \sin (3 e)}{2 c}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^2(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

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 - I*B)*Cos[2*f*x]*Cos[e + f*x]^4*(((-2*I)*Cos[e])/c - (2*Sin[e])/c)*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e
+ f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^2*((B*Cos[3*e])/(2*c)
- ((I/2)*B*Sin[3*e])/c)*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e
+ f*x] + B*Sin[e + f*x])) + ((A - (2*I)*B)*Cos[e + f*x]^4*((-4*f*x*Cos[3*e])/c + ((4*I)*f*x*Sin[3*e])/c)*(a +
I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + ((
I*A + 2*B)*Cos[e + f*x]^4*((2*Cos[3*e]*Log[Cos[e + f*x]^2])/c - ((2*I)*Log[Cos[e + f*x]^2]*Sin[3*e])/c)*(a + I
*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Co
s[e + f*x]^3*(Cos[3*e]/c - (I*Sin[3*e])/c)*(A*Sin[f*x] - (4*I)*B*Sin[f*x])*(a + I*a*Tan[e + f*x])^3*(A + B*Tan
[e + f*x]))/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e
+ f*x])) + ((A - I*B)*Cos[e + f*x]^4*((2*Cos[e])/c - ((2*I)*Sin[e])/c)*Sin[2*f*x]*(a + I*a*Tan[e + f*x])^3*(A
+ B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (x*Cos[e + f*x]^4*((2*A*
Cos[e])/c - ((4*I)*B*Cos[e])/c - (2*A*Cos[e]^3)/c + ((4*I)*B*Cos[e]^3)/c - ((4*I)*A*Sin[e])/c - (8*B*Sin[e])/c
+ ((8*I)*A*Cos[e]^2*Sin[e])/c + (16*B*Cos[e]^2*Sin[e])/c + (12*A*Cos[e]*Sin[e]^2)/c - ((24*I)*B*Cos[e]*Sin[e]
^2)/c - ((8*I)*A*Sin[e]^3)/c - (16*B*Sin[e]^3)/c - (2*A*Sin[e]*Tan[e])/c + ((4*I)*B*Sin[e]*Tan[e])/c - (2*A*Si
n[e]^3*Tan[e])/c + ((4*I)*B*Sin[e]^3*Tan[e])/c - I*(A - (2*I)*B)*((4*Cos[3*e])/c - ((4*I)*Sin[3*e])/c)*Tan[e])
*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/((Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

________________________________________________________________________________________

Maple [A]  time = 0.044, size = 150, normalized size = 1.3 \begin{align*}{\frac{A{a}^{3}\tan \left ( fx+e \right ) }{cf}}-{\frac{4\,iB{a}^{3}\tan \left ( fx+e \right ) }{cf}}+{\frac{B{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,cf}}-{\frac{4\,iB{a}^{3}}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }}+4\,{\frac{A{a}^{3}}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{4\,i{a}^{3}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{cf}}-8\,{\frac{B{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{cf}} \end{align*}

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*A*tan(f*x+e)-4*I/f*a^3/c*B*tan(f*x+e)+1/2*a^3*B*tan(f*x+e)^2/c/f-4*I/f*a^3/c/(tan(f*x+e)+I)*B+4/f*a^
3/c/(tan(f*x+e)+I)*A-4*I/f*a^3/c*A*ln(tan(f*x+e)+I)-8/f*a^3/c*B*ln(tan(f*x+e)+I)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.26026, size = 439, normalized size = 3.69 \begin{align*} \frac{{\left (-2 i \, A - 2 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-4 i \, A - 4 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A + 8 \, B\right )} a^{3} +{\left ({\left (4 i \, A + 8 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (8 i \, A + 16 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 8 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \end{align*}

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]

((-2*I*A - 2*B)*a^3*e^(6*I*f*x + 6*I*e) + (-4*I*A - 4*B)*a^3*e^(4*I*f*x + 4*I*e) + 8*B*a^3*e^(2*I*f*x + 2*I*e)
+ (2*I*A + 8*B)*a^3 + ((4*I*A + 8*B)*a^3*e^(4*I*f*x + 4*I*e) + (8*I*A + 16*B)*a^3*e^(2*I*f*x + 2*I*e) + (4*I*
A + 8*B)*a^3)*log(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(4*I*f*x + 4*I*e) + 2*c*f*e^(2*I*f*x + 2*I*e) + c*f)

________________________________________________________________________________________

Sympy [A]  time = 5.58664, size = 209, normalized size = 1.76 \begin{align*} \frac{4 a^{3} \left (i A + 2 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \frac{\frac{\left (2 i A a^{3} + 8 B a^{3}\right ) e^{- 4 i e}}{c f} + \frac{\left (2 i A a^{3} + 10 B a^{3}\right ) e^{- 2 i e} e^{2 i f x}}{c f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} + \frac{\begin{cases} - \frac{2 i A a^{3} e^{2 i e} e^{2 i f x}}{f} - \frac{2 B a^{3} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (4 A a^{3} e^{2 i e} - 4 i B a^{3} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c} \end{align*}

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]

4*a**3*(I*A + 2*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(c*f) + ((2*I*A*a**3 + 8*B*a**3)*exp(-4*I*e)/(c*f) + (2*I*A
*a**3 + 10*B*a**3)*exp(-2*I*e)*exp(2*I*f*x)/(c*f))/(exp(4*I*f*x) + 2*exp(-2*I*e)*exp(2*I*f*x) + exp(-4*I*e)) +
Piecewise((-2*I*A*a**3*exp(2*I*e)*exp(2*I*f*x)/f - 2*B*a**3*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(4*A*a**
3*exp(2*I*e) - 4*I*B*a**3*exp(2*I*e)), True))/c

________________________________________________________________________________________

Giac [B]  time = 1.68214, size = 437, normalized size = 3.67 \begin{align*} -\frac{2 \,{\left (\frac{4 \,{\left (i \, A a^{3} + 2 \, B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c} - \frac{{\left (2 i \, A a^{3} + 4 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac{2 \,{\left (-i \, A a^{3} - 2 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} + \frac{5 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 8 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 2 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 7 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 10 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 14 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 8 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + i \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{2} c}\right )}}{f} \end{align*}

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]

-2*(4*(I*A*a^3 + 2*B*a^3)*log(tan(1/2*f*x + 1/2*e) + I)/c - (2*I*A*a^3 + 4*B*a^3)*log(abs(tan(1/2*f*x + 1/2*e)
+ 1))/c + 2*(-I*A*a^3 - 2*B*a^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c + (5*A*a^3*tan(1/2*f*x + 1/2*e)^5 - 8*I
*B*a^3*tan(1/2*f*x + 1/2*e)^5 + 2*I*A*a^3*tan(1/2*f*x + 1/2*e)^4 + 7*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 10*A*a^3*t
an(1/2*f*x + 1/2*e)^3 + 14*I*B*a^3*tan(1/2*f*x + 1/2*e)^3 - 2*I*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 7*B*a^3*tan(1/2
*f*x + 1/2*e)^2 + 5*A*a^3*tan(1/2*f*x + 1/2*e) - 8*I*B*a^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^3 + I*
tan(1/2*f*x + 1/2*e)^2 - tan(1/2*f*x + 1/2*e) - I)^2*c))/f