3.674 \(\int \frac{(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac{a (B+i A)}{4 c^4 f (\tan (e+f x)+i)^4}-\frac{i a B}{3 c^4 f (\tan (e+f x)+i)^3} \]

[Out]

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

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Rubi [A]  time = 0.0878122, antiderivative size = 57, 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} \[ -\frac{a (B+i A)}{4 c^4 f (\tan (e+f x)+i)^4}-\frac{i a B}{3 c^4 f (\tan (e+f x)+i)^3} \]

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

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

Mathematica [A]  time = 1.53203, size = 97, normalized size = 1.7 \[ \frac{a (\cos (5 (e+f x))+i \sin (5 (e+f x))) (-(3 A+5 i B) (2 \sin (e+f x)+3 \sin (3 (e+f x)))+2 (B-15 i A) \cos (e+f x)+3 (3 B-5 i A) \cos (3 (e+f x)))}{192 c^4 f} \]

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*(2*((-15*I)*A + B)*Cos[e + f*x] + 3*((-5*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (5*I)*B)*(2*Sin[e + f*x] + 3
*Sin[3*(e + f*x)]))*(Cos[5*(e + f*x)] + I*Sin[5*(e + f*x)]))/(192*c^4*f)

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Maple [A]  time = 0.045, size = 44, normalized size = 0.8 \begin{align*}{\frac{a}{f{c}^{4}} \left ( -{\frac{iA+B}{4\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-{\frac{{\frac{i}{3}}B}{ \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \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*(-1/4*(I*A+B)/(tan(f*x+e)+I)^4-1/3*I*B/(tan(f*x+e)+I)^3)

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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))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.4196, size = 236, normalized size = 4.14 \begin{align*} \frac{{\left (-3 i \, A - 3 \, B\right )} a e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-12 i \, A - 4 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-18 i \, A + 6 \, B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-12 i \, A + 12 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )}}{192 \, c^{4} 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="fricas")

[Out]

1/192*((-3*I*A - 3*B)*a*e^(8*I*f*x + 8*I*e) + (-12*I*A - 4*B)*a*e^(6*I*f*x + 6*I*e) + (-18*I*A + 6*B)*a*e^(4*I
*f*x + 4*I*e) + (-12*I*A + 12*B)*a*e^(2*I*f*x + 2*I*e))/(c^4*f)

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Sympy [B]  time = 2.75311, size = 306, normalized size = 5.37 \begin{align*} \begin{cases} \frac{\left (- 98304 i A a c^{12} f^{3} e^{2 i e} + 98304 B a c^{12} f^{3} e^{2 i e}\right ) e^{2 i f x} + \left (- 147456 i A a c^{12} f^{3} e^{4 i e} + 49152 B a c^{12} f^{3} e^{4 i e}\right ) e^{4 i f x} + \left (- 98304 i A a c^{12} f^{3} e^{6 i e} - 32768 B a c^{12} f^{3} e^{6 i e}\right ) e^{6 i f x} + \left (- 24576 i A a c^{12} f^{3} e^{8 i e} - 24576 B a c^{12} f^{3} e^{8 i e}\right ) e^{8 i f x}}{1572864 c^{16} f^{4}} & \text{for}\: 1572864 c^{16} f^{4} \neq 0 \\\frac{x \left (A a e^{8 i e} + 3 A a e^{6 i e} + 3 A a e^{4 i e} + A a e^{2 i e} - i B a e^{8 i e} - i B a e^{6 i e} + i B a e^{4 i e} + i B a e^{2 i e}\right )}{8 c^{4}} & \text{otherwise} \end{cases} \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]

Piecewise((((-98304*I*A*a*c**12*f**3*exp(2*I*e) + 98304*B*a*c**12*f**3*exp(2*I*e))*exp(2*I*f*x) + (-147456*I*A
*a*c**12*f**3*exp(4*I*e) + 49152*B*a*c**12*f**3*exp(4*I*e))*exp(4*I*f*x) + (-98304*I*A*a*c**12*f**3*exp(6*I*e)
- 32768*B*a*c**12*f**3*exp(6*I*e))*exp(6*I*f*x) + (-24576*I*A*a*c**12*f**3*exp(8*I*e) - 24576*B*a*c**12*f**3*
exp(8*I*e))*exp(8*I*f*x))/(1572864*c**16*f**4), Ne(1572864*c**16*f**4, 0)), (x*(A*a*exp(8*I*e) + 3*A*a*exp(6*I
*e) + 3*A*a*exp(4*I*e) + A*a*exp(2*I*e) - I*B*a*exp(8*I*e) - I*B*a*exp(6*I*e) + I*B*a*exp(4*I*e) + I*B*a*exp(2
*I*e))/(8*c**4), True))

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Giac [B]  time = 1.39136, size = 288, normalized size = 5.05 \begin{align*} -\frac{2 \,{\left (3 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 9 i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 3 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 21 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 4 i \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 24 i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 8 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 21 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 4 i \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 i \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{8}} \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]

-2/3*(3*A*a*tan(1/2*f*x + 1/2*e)^7 + 9*I*A*a*tan(1/2*f*x + 1/2*e)^6 - 3*B*a*tan(1/2*f*x + 1/2*e)^6 - 21*A*a*ta
n(1/2*f*x + 1/2*e)^5 - 4*I*B*a*tan(1/2*f*x + 1/2*e)^5 - 24*I*A*a*tan(1/2*f*x + 1/2*e)^4 + 8*B*a*tan(1/2*f*x +
1/2*e)^4 + 21*A*a*tan(1/2*f*x + 1/2*e)^3 + 4*I*B*a*tan(1/2*f*x + 1/2*e)^3 + 9*I*A*a*tan(1/2*f*x + 1/2*e)^2 - 3
*B*a*tan(1/2*f*x + 1/2*e)^2 - 3*A*a*tan(1/2*f*x + 1/2*e))/(c^4*f*(tan(1/2*f*x + 1/2*e) + I)^8)