3.698 \(\int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac{4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac{a^3 x (A-5 i B)}{c^2}+\frac{i a^3 B \tan (e+f x)}{c^2 f} \]

[Out]

(a^3*(A - (5*I)*B)*x)/c^2 - (a^3*(I*A + 5*B)*Log[Cos[e + f*x]])/(c^2*f) + (I*a^3*B*Tan[e + f*x])/(c^2*f) + (2*
a^3*(I*A + B))/(c^2*f*(I + Tan[e + f*x])^2) - (4*a^3*(A - (2*I)*B))/(c^2*f*(I + Tan[e + f*x]))

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Rubi [A]  time = 0.178271, antiderivative size = 123, 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} \[ -\frac{4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac{a^3 x (A-5 i B)}{c^2}+\frac{i a^3 B \tan (e+f x)}{c^2 f} \]

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])^2,x]

[Out]

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

Mathematica [B]  time = 9.54447, size = 1063, normalized size = 8.64 \[ \frac{x \left (\frac{A \cos ^3(e)}{2 c^2}-\frac{5 i B \cos ^3(e)}{2 c^2}-\frac{2 i A \sin (e) \cos ^2(e)}{c^2}-\frac{10 B \sin (e) \cos ^2(e)}{c^2}-\frac{3 A \sin ^2(e) \cos (e)}{c^2}+\frac{15 i B \sin ^2(e) \cos (e)}{c^2}-\frac{A \cos (e)}{2 c^2}+\frac{5 i B \cos (e)}{2 c^2}+\frac{2 i A \sin ^3(e)}{c^2}+\frac{10 B \sin ^3(e)}{c^2}+\frac{i A \sin (e)}{c^2}+\frac{5 B \sin (e)}{c^2}+\frac{A \sin ^3(e) \tan (e)}{2 c^2}-\frac{5 i B \sin ^3(e) \tan (e)}{2 c^2}+\frac{A \sin (e) \tan (e)}{2 c^2}-\frac{5 i B \sin (e) \tan (e)}{2 c^2}+i (A-5 i B) \left (\frac{\cos (3 e)}{c^2}-\frac{i \sin (3 e)}{c^2}\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-3 i B) \left (\frac{i \sin (e)}{c^2}-\frac{\cos (e)}{c^2}\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) \left (\frac{\cos (e)}{2 c^2}+\frac{i \sin (e)}{2 c^2}\right ) \sin (4 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{(i A+3 B) \cos (2 f x) \left (\frac{\cos (e)}{c^2}-\frac{i \sin (e)}{c^2}\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-i B) \cos (4 f x) \left (\frac{\sin (e)}{2 c^2}-\frac{i \cos (e)}{2 c^2}\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-5 i B) \left (\frac{f x \cos (3 e)}{c^2}-\frac{i f x \sin (3 e)}{c^2}\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-5 i B) \left (-\frac{i \cos (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}-\frac{\sin (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}\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 B \left (\frac{\cos (3 e)}{c^2}-\frac{i \sin (3 e)}{c^2}\right ) \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))} \]

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])^2,x]

[Out]

((I*A + 3*B)*Cos[2*f*x]*Cos[e + f*x]^4*(Cos[e]/c^2 - (I*Sin[e])/c^2)*(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 - I*B)*Cos[4*f*x]*Cos[e + f*x]^4*(
((-I/2)*Cos[e])/c^2 + Sin[e]/(2*c^2))*(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 - (5*I)*B)*Cos[e + f*x]^4*((f*x*Cos[3*e])/c^2 - (I*f*x*Sin[3*e])/
c^2)*(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 - (5*I)*B)*Cos[e + f*x]^4*(((-I/2)*Cos[3*e]*Log[Cos[e + f*x]^2])/c^2 - (Log[Cos[e + f*x]^2]*Sin[3*
e])/(2*c^2))*(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*S
in[e + f*x])) + (I*B*Cos[e + f*x]^3*(Cos[3*e]/c^2 - (I*Sin[3*e])/c^2)*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*S
in[e + f*x])) + ((A - (3*I)*B)*Cos[e + f*x]^4*(-(Cos[e]/c^2) + (I*Sin[e])/c^2)*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])) + ((A - I*B)*Cos[e
+ f*x]^4*(Cos[e]/(2*c^2) + ((I/2)*Sin[e])/c^2)*Sin[4*f*x]*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(C
os[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (x*Cos[e + f*x]^4*(-(A*Cos[e])/(2*c^2) + (((5*I)/
2)*B*Cos[e])/c^2 + (A*Cos[e]^3)/(2*c^2) - (((5*I)/2)*B*Cos[e]^3)/c^2 + (I*A*Sin[e])/c^2 + (5*B*Sin[e])/c^2 - (
(2*I)*A*Cos[e]^2*Sin[e])/c^2 - (10*B*Cos[e]^2*Sin[e])/c^2 - (3*A*Cos[e]*Sin[e]^2)/c^2 + ((15*I)*B*Cos[e]*Sin[e
]^2)/c^2 + ((2*I)*A*Sin[e]^3)/c^2 + (10*B*Sin[e]^3)/c^2 + (A*Sin[e]*Tan[e])/(2*c^2) - (((5*I)/2)*B*Sin[e]*Tan[
e])/c^2 + (A*Sin[e]^3*Tan[e])/(2*c^2) - (((5*I)/2)*B*Sin[e]^3*Tan[e])/c^2 + I*(A - (5*I)*B)*(Cos[3*e]/c^2 - (I
*Sin[3*e])/c^2)*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]))

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Maple [A]  time = 0.043, size = 160, normalized size = 1.3 \begin{align*}{\frac{iB{a}^{3}\tan \left ( fx+e \right ) }{{c}^{2}f}}+{\frac{8\,i{a}^{3}B}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) }}-4\,{\frac{A{a}^{3}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{iA{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{2}f}}+5\,{\frac{B{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{2}f}}+{\frac{2\,i{a}^{3}A}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+2\,{\frac{B{a}^{3}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \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))^2,x)

[Out]

I*a^3*B*tan(f*x+e)/c^2/f+8*I/f*a^3/c^2/(tan(f*x+e)+I)*B-4/f*a^3/c^2/(tan(f*x+e)+I)*A+I/f*a^3/c^2*A*ln(tan(f*x+
e)+I)+5/f*a^3/c^2*B*ln(tan(f*x+e)+I)+2*I/f*a^3/c^2/(tan(f*x+e)+I)^2*A+2/f*a^3/c^2/(tan(f*x+e)+I)^2*B

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.42335, size = 352, normalized size = 2.86 \begin{align*} \frac{{\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A + 5 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (2 i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, B a^{3} +{\left ({\left (-2 i \, A - 10 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-2 i \, A - 10 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \,{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \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))^2,x, algorithm="fricas")

[Out]

1/2*((-I*A - B)*a^3*e^(6*I*f*x + 6*I*e) + (I*A + 5*B)*a^3*e^(4*I*f*x + 4*I*e) + (2*I*A + 6*B)*a^3*e^(2*I*f*x +
 2*I*e) - 4*B*a^3 + ((-2*I*A - 10*B)*a^3*e^(2*I*f*x + 2*I*e) + (-2*I*A - 10*B)*a^3)*log(e^(2*I*f*x + 2*I*e) +
1))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)

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Sympy [A]  time = 4.13402, size = 228, normalized size = 1.85 \begin{align*} - \frac{2 B a^{3} e^{- 2 i e}}{c^{2} f \left (e^{2 i f x} + e^{- 2 i e}\right )} - \frac{a^{3} \left (i A + 5 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \frac{\begin{cases} - \frac{i A a^{3} e^{4 i e} e^{4 i f x}}{2 f} + \frac{i A a^{3} e^{2 i e} e^{2 i f x}}{f} - \frac{B a^{3} e^{4 i e} e^{4 i f x}}{2 f} + \frac{3 B a^{3} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 A a^{3} e^{4 i e} - 2 A a^{3} e^{2 i e} - 2 i B a^{3} e^{4 i e} + 6 i B a^{3} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{2}} \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))**2,x)

[Out]

-2*B*a**3*exp(-2*I*e)/(c**2*f*(exp(2*I*f*x) + exp(-2*I*e))) - a**3*(I*A + 5*B)*log(exp(2*I*f*x) + exp(-2*I*e))
/(c**2*f) + Piecewise((-I*A*a**3*exp(4*I*e)*exp(4*I*f*x)/(2*f) + I*A*a**3*exp(2*I*e)*exp(2*I*f*x)/f - B*a**3*e
xp(4*I*e)*exp(4*I*f*x)/(2*f) + 3*B*a**3*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(2*A*a**3*exp(4*I*e) - 2*A*a*
*3*exp(2*I*e) - 2*I*B*a**3*exp(4*I*e) + 6*I*B*a**3*exp(2*I*e)), True))/c**2

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Giac [B]  time = 1.5781, size = 485, normalized size = 3.94 \begin{align*} \frac{\frac{12 \,{\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{2}} + \frac{6 \,{\left (-i \, A a^{3} - 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac{6 \,{\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{6 \,{\left (-i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i \, A a^{3} + 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac{25 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 125 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 100 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 548 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 198 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 894 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 100 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 548 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 25 i \, A a^{3} + 125 \, B a^{3}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{4}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(12*(I*A*a^3 + 5*B*a^3)*log(tan(1/2*f*x + 1/2*e) + I)/c^2 + 6*(-I*A*a^3 - 5*B*a^3)*log(abs(tan(1/2*f*x + 1
/2*e) + 1))/c^2 - 6*(I*A*a^3 + 5*B*a^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^2 - 6*(-I*A*a^3*tan(1/2*f*x + 1/2
*e)^2 - 5*B*a^3*tan(1/2*f*x + 1/2*e)^2 + 2*I*B*a^3*tan(1/2*f*x + 1/2*e) + I*A*a^3 + 5*B*a^3)/((tan(1/2*f*x + 1
/2*e)^2 - 1)*c^2) - (25*I*A*a^3*tan(1/2*f*x + 1/2*e)^4 + 125*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 100*A*a^3*tan(1/2*
f*x + 1/2*e)^3 + 548*I*B*a^3*tan(1/2*f*x + 1/2*e)^3 - 198*I*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 894*B*a^3*tan(1/2*f
*x + 1/2*e)^2 + 100*A*a^3*tan(1/2*f*x + 1/2*e) - 548*I*B*a^3*tan(1/2*f*x + 1/2*e) + 25*I*A*a^3 + 125*B*a^3)/(c
^2*(tan(1/2*f*x + 1/2*e) + I)^4))/f