3.701 \(\int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx\)

Optimal. Leaf size=122 \[ -\frac{a^3 (A-5 i B)}{3 c^5 f (\tan (e+f x)+i)^3}+\frac{a^3 (2 B+i A)}{c^5 f (\tan (e+f x)+i)^4}+\frac{4 a^3 (A-i B)}{5 c^5 f (\tan (e+f x)+i)^5}-\frac{a^3 B}{2 c^5 f (\tan (e+f x)+i)^2} \]

[Out]

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

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Rubi [A]  time = 0.175211, antiderivative size = 122, 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} \[ -\frac{a^3 (A-5 i B)}{3 c^5 f (\tan (e+f x)+i)^3}+\frac{a^3 (2 B+i A)}{c^5 f (\tan (e+f x)+i)^4}+\frac{4 a^3 (A-i B)}{5 c^5 f (\tan (e+f x)+i)^5}-\frac{a^3 B}{2 c^5 f (\tan (e+f x)+i)^2} \]

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])^5,x]

[Out]

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

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))^5} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (-\frac{4 a^2 (A-i B)}{c^6 (i+x)^6}-\frac{4 i a^2 (A-2 i B)}{c^6 (i+x)^5}+\frac{a^2 (A-5 i B)}{c^6 (i+x)^4}+\frac{a^2 B}{c^6 (i+x)^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{4 a^3 (A-i B)}{5 c^5 f (i+\tan (e+f x))^5}+\frac{a^3 (i A+2 B)}{c^5 f (i+\tan (e+f x))^4}-\frac{a^3 (A-5 i B)}{3 c^5 f (i+\tan (e+f x))^3}-\frac{a^3 B}{2 c^5 f (i+\tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 3.91606, size = 91, normalized size = 0.75 \[ \frac{a^3 (\cos (8 e+11 f x)+i \sin (8 e+11 f x)) (-4 (A+4 i B) \sin (2 (e+f x))+4 (B-4 i A) \cos (2 (e+f x))-15 i A)}{240 c^5 f (\cos (f x)+i \sin (f x))^3} \]

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])^5,x]

[Out]

(a^3*((-15*I)*A + 4*((-4*I)*A + B)*Cos[2*(e + f*x)] - 4*(A + (4*I)*B)*Sin[2*(e + f*x)])*(Cos[8*e + 11*f*x] + I
*Sin[8*e + 11*f*x]))/(240*c^5*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.052, size = 87, normalized size = 0.7 \begin{align*}{\frac{{a}^{3}}{f{c}^{5}} \left ( -{\frac{-4\,iA-8\,B}{4\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-{\frac{A-5\,iB}{3\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}-{\frac{B}{2\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{4\,iB-4\,A}{5\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{5}}} \right ) } \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))^5,x)

[Out]

1/f*a^3/c^5*(-1/4*(-4*I*A-8*B)/(tan(f*x+e)+I)^4-1/3*(A-5*I*B)/(tan(f*x+e)+I)^3-1/2*B/(tan(f*x+e)+I)^2-1/5*(4*I
*B-4*A)/(tan(f*x+e)+I)^5)

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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))^5,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.27815, size = 181, normalized size = 1.48 \begin{align*} \frac{{\left (-6 i \, A - 6 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} - 15 i \, A a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-10 i \, A + 10 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{240 \, c^{5} 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))^5,x, algorithm="fricas")

[Out]

1/240*((-6*I*A - 6*B)*a^3*e^(10*I*f*x + 10*I*e) - 15*I*A*a^3*e^(8*I*f*x + 8*I*e) + (-10*I*A + 10*B)*a^3*e^(6*I
*f*x + 6*I*e))/(c^5*f)

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Sympy [A]  time = 2.51532, size = 219, normalized size = 1.8 \begin{align*} \begin{cases} \frac{- 960 i A a^{3} c^{10} f^{2} e^{8 i e} e^{8 i f x} + \left (- 640 i A a^{3} c^{10} f^{2} e^{6 i e} + 640 B a^{3} c^{10} f^{2} e^{6 i e}\right ) e^{6 i f x} + \left (- 384 i A a^{3} c^{10} f^{2} e^{10 i e} - 384 B a^{3} c^{10} f^{2} e^{10 i e}\right ) e^{10 i f x}}{15360 c^{15} f^{3}} & \text{for}\: 15360 c^{15} f^{3} \neq 0 \\\frac{x \left (A a^{3} e^{10 i e} + 2 A a^{3} e^{8 i e} + A a^{3} e^{6 i e} - i B a^{3} e^{10 i e} + i B a^{3} e^{6 i e}\right )}{4 c^{5}} & \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))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**5,x)

[Out]

Piecewise(((-960*I*A*a**3*c**10*f**2*exp(8*I*e)*exp(8*I*f*x) + (-640*I*A*a**3*c**10*f**2*exp(6*I*e) + 640*B*a*
*3*c**10*f**2*exp(6*I*e))*exp(6*I*f*x) + (-384*I*A*a**3*c**10*f**2*exp(10*I*e) - 384*B*a**3*c**10*f**2*exp(10*
I*e))*exp(10*I*f*x))/(15360*c**15*f**3), Ne(15360*c**15*f**3, 0)), (x*(A*a**3*exp(10*I*e) + 2*A*a**3*exp(8*I*e
) + A*a**3*exp(6*I*e) - I*B*a**3*exp(10*I*e) + I*B*a**3*exp(6*I*e))/(4*c**5), True))

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Giac [B]  time = 1.67002, size = 417, normalized size = 3.42 \begin{align*} -\frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} + 30 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 15 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 140 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 10 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 170 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 65 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 282 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 12 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 170 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 65 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 140 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 10 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 30 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{15 \, c^{5} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{10}} \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))^5,x, algorithm="giac")

[Out]

-2/15*(15*A*a^3*tan(1/2*f*x + 1/2*e)^9 + 30*I*A*a^3*tan(1/2*f*x + 1/2*e)^8 - 15*B*a^3*tan(1/2*f*x + 1/2*e)^8 -
140*A*a^3*tan(1/2*f*x + 1/2*e)^7 + 10*I*B*a^3*tan(1/2*f*x + 1/2*e)^7 - 170*I*A*a^3*tan(1/2*f*x + 1/2*e)^6 + 6
5*B*a^3*tan(1/2*f*x + 1/2*e)^6 + 282*A*a^3*tan(1/2*f*x + 1/2*e)^5 - 12*I*B*a^3*tan(1/2*f*x + 1/2*e)^5 + 170*I*
A*a^3*tan(1/2*f*x + 1/2*e)^4 - 65*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 140*A*a^3*tan(1/2*f*x + 1/2*e)^3 + 10*I*B*a^3
*tan(1/2*f*x + 1/2*e)^3 - 30*I*A*a^3*tan(1/2*f*x + 1/2*e)^2 + 15*B*a^3*tan(1/2*f*x + 1/2*e)^2 + 15*A*a^3*tan(1
/2*f*x + 1/2*e))/(c^5*f*(tan(1/2*f*x + 1/2*e) + I)^10)