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

Optimal. Leaf size=185 \[ \frac{2 A-i B}{16 a^3 c^2 f (\tan (e+f x)+i)}-\frac{-B+3 i A}{32 a^3 c^2 f (-\tan (e+f x)+i)^2}+\frac{B+i A}{32 a^3 c^2 f (\tan (e+f x)+i)^2}+\frac{A+i B}{24 a^3 c^2 f (-\tan (e+f x)+i)^3}+\frac{x (5 A-i B)}{16 a^3 c^2}-\frac{3 A}{16 a^3 c^2 f (-\tan (e+f x)+i)} \]

[Out]

((5*A - I*B)*x)/(16*a^3*c^2) + (A + I*B)/(24*a^3*c^2*f*(I - Tan[e + f*x])^3) - ((3*I)*A - B)/(32*a^3*c^2*f*(I
- Tan[e + f*x])^2) - (3*A)/(16*a^3*c^2*f*(I - Tan[e + f*x])) + (I*A + B)/(32*a^3*c^2*f*(I + Tan[e + f*x])^2) +
(2*A - I*B)/(16*a^3*c^2*f*(I + Tan[e + f*x]))

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Rubi [A]  time = 0.239247, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3588, 77, 203} \[ \frac{2 A-i B}{16 a^3 c^2 f (\tan (e+f x)+i)}-\frac{-B+3 i A}{32 a^3 c^2 f (-\tan (e+f x)+i)^2}+\frac{B+i A}{32 a^3 c^2 f (\tan (e+f x)+i)^2}+\frac{A+i B}{24 a^3 c^2 f (-\tan (e+f x)+i)^3}+\frac{x (5 A-i B)}{16 a^3 c^2}-\frac{3 A}{16 a^3 c^2 f (-\tan (e+f x)+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A - I*B)*x)/(16*a^3*c^2) + (A + I*B)/(24*a^3*c^2*f*(I - Tan[e + f*x])^3) - ((3*I)*A - B)/(32*a^3*c^2*f*(I
- Tan[e + f*x])^2) - (3*A)/(16*a^3*c^2*f*(I - Tan[e + f*x])) + (I*A + B)/(32*a^3*c^2*f*(I + Tan[e + f*x])^2) +
(2*A - I*B)/(16*a^3*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]))))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^4 (c-i c x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{A+i B}{8 a^4 c^3 (-i+x)^4}+\frac{i (3 A+i B)}{16 a^4 c^3 (-i+x)^3}-\frac{3 A}{16 a^4 c^3 (-i+x)^2}-\frac{i (A-i B)}{16 a^4 c^3 (i+x)^3}+\frac{-2 A+i B}{16 a^4 c^3 (i+x)^2}+\frac{5 A-i B}{16 a^4 c^3 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{A+i B}{24 a^3 c^2 f (i-\tan (e+f x))^3}-\frac{3 i A-B}{32 a^3 c^2 f (i-\tan (e+f x))^2}-\frac{3 A}{16 a^3 c^2 f (i-\tan (e+f x))}+\frac{i A+B}{32 a^3 c^2 f (i+\tan (e+f x))^2}+\frac{2 A-i B}{16 a^3 c^2 f (i+\tan (e+f x))}+\frac{(5 A-i B) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^3 c^2 f}\\ &=\frac{(5 A-i B) x}{16 a^3 c^2}+\frac{A+i B}{24 a^3 c^2 f (i-\tan (e+f x))^3}-\frac{3 i A-B}{32 a^3 c^2 f (i-\tan (e+f x))^2}-\frac{3 A}{16 a^3 c^2 f (i-\tan (e+f x))}+\frac{i A+B}{32 a^3 c^2 f (i+\tan (e+f x))^2}+\frac{2 A-i B}{16 a^3 c^2 f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.38725, size = 217, normalized size = 1.17 \[ \frac{\sec ^3(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (12 (A (-5+10 i f x)+B (2 f x-i)) \cos (e+f x)+3 (5 A-9 i B) \cos (3 (e+f x))+60 i A \sin (e+f x)-120 A f x \sin (e+f x)+45 i A \sin (3 (e+f x))+5 i A \sin (5 (e+f x))+A \cos (5 (e+f x))-12 B \sin (e+f x)+24 i B f x \sin (e+f x)+9 B \sin (3 (e+f x))+B \sin (5 (e+f x))-5 i B \cos (5 (e+f x)))}{384 a^3 c^2 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(12*(A*(-5 + (10*I)*f*x) + B*(-I + 2*f*x))*Cos[e + f*x
] + 3*(5*A - (9*I)*B)*Cos[3*(e + f*x)] + A*Cos[5*(e + f*x)] - (5*I)*B*Cos[5*(e + f*x)] + (60*I)*A*Sin[e + f*x]
- 12*B*Sin[e + f*x] - 120*A*f*x*Sin[e + f*x] + (24*I)*B*f*x*Sin[e + f*x] + (45*I)*A*Sin[3*(e + f*x)] + 9*B*Si
n[3*(e + f*x)] + (5*I)*A*Sin[5*(e + f*x)] + B*Sin[5*(e + f*x)]))/(384*a^3*c^2*f*(-I + Tan[e + f*x])^3)

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Maple [A]  time = 0.073, size = 303, normalized size = 1.6 \begin{align*}{\frac{3\,A}{16\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{3\,i}{32}}A}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{B}{32\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{5\,i}{32}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) A}{f{a}^{3}{c}^{2}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{32\,f{a}^{3}{c}^{2}}}-{\frac{A}{24\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{24}}B}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{{\frac{i}{32}}A}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{B}{32\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{{\frac{i}{16}}B}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{A}{8\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{{\frac{5\,i}{32}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) A}{f{a}^{3}{c}^{2}}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) B}{32\,f{a}^{3}{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x)

[Out]

3/16/f/a^3/c^2*A/(tan(f*x+e)-I)-3/32*I/f/a^3/c^2/(tan(f*x+e)-I)^2*A+1/32/f/a^3/c^2/(tan(f*x+e)-I)^2*B-5/32*I/f
/a^3/c^2*ln(tan(f*x+e)-I)*A-1/32/f/a^3/c^2*ln(tan(f*x+e)-I)*B-1/24/f/a^3/c^2/(tan(f*x+e)-I)^3*A-1/24*I/f/a^3/c
^2/(tan(f*x+e)-I)^3*B+1/32*I/f/a^3/c^2/(tan(f*x+e)+I)^2*A+1/32/f/a^3/c^2/(tan(f*x+e)+I)^2*B-1/16*I/f/a^3/c^2/(
tan(f*x+e)+I)*B+1/8/f/a^3/c^2/(tan(f*x+e)+I)*A+5/32*I/f/a^3/c^2*ln(tan(f*x+e)+I)*A+1/32/f/a^3/c^2*ln(tan(f*x+e
)+I)*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+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.06795, size = 335, normalized size = 1.81 \begin{align*} \frac{{\left (24 \,{\left (5 \, A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-3 i \, A - 3 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-30 i \, A - 18 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (60 i \, A - 12 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (15 i \, A - 9 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/384*(24*(5*A - I*B)*f*x*e^(6*I*f*x + 6*I*e) + (-3*I*A - 3*B)*e^(10*I*f*x + 10*I*e) + (-30*I*A - 18*B)*e^(8*I
*f*x + 8*I*e) + (60*I*A - 12*B)*e^(4*I*f*x + 4*I*e) + (15*I*A - 9*B)*e^(2*I*f*x + 2*I*e) + 2*I*A - 2*B)*e^(-6*
I*f*x - 6*I*e)/(a^3*c^2*f)

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Sympy [A]  time = 6.11453, size = 454, normalized size = 2.45 \begin{align*} \begin{cases} \frac{\left (\left (33554432 i A a^{12} c^{8} f^{4} e^{6 i e} - 33554432 B a^{12} c^{8} f^{4} e^{6 i e}\right ) e^{- 6 i f x} + \left (251658240 i A a^{12} c^{8} f^{4} e^{8 i e} - 150994944 B a^{12} c^{8} f^{4} e^{8 i e}\right ) e^{- 4 i f x} + \left (1006632960 i A a^{12} c^{8} f^{4} e^{10 i e} - 201326592 B a^{12} c^{8} f^{4} e^{10 i e}\right ) e^{- 2 i f x} + \left (- 503316480 i A a^{12} c^{8} f^{4} e^{14 i e} - 301989888 B a^{12} c^{8} f^{4} e^{14 i e}\right ) e^{2 i f x} + \left (- 50331648 i A a^{12} c^{8} f^{4} e^{16 i e} - 50331648 B a^{12} c^{8} f^{4} e^{16 i e}\right ) e^{4 i f x}\right ) e^{- 12 i e}}{6442450944 a^{15} c^{10} f^{5}} & \text{for}\: 6442450944 a^{15} c^{10} f^{5} e^{12 i e} \neq 0 \\x \left (- \frac{5 A - i B}{16 a^{3} c^{2}} + \frac{\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{32 a^{3} c^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (5 A - i B\right )}{16 a^{3} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)

[Out]

Piecewise((((33554432*I*A*a**12*c**8*f**4*exp(6*I*e) - 33554432*B*a**12*c**8*f**4*exp(6*I*e))*exp(-6*I*f*x) +
(251658240*I*A*a**12*c**8*f**4*exp(8*I*e) - 150994944*B*a**12*c**8*f**4*exp(8*I*e))*exp(-4*I*f*x) + (100663296
0*I*A*a**12*c**8*f**4*exp(10*I*e) - 201326592*B*a**12*c**8*f**4*exp(10*I*e))*exp(-2*I*f*x) + (-503316480*I*A*a
**12*c**8*f**4*exp(14*I*e) - 301989888*B*a**12*c**8*f**4*exp(14*I*e))*exp(2*I*f*x) + (-50331648*I*A*a**12*c**8
*f**4*exp(16*I*e) - 50331648*B*a**12*c**8*f**4*exp(16*I*e))*exp(4*I*f*x))*exp(-12*I*e)/(6442450944*a**15*c**10
*f**5), Ne(6442450944*a**15*c**10*f**5*exp(12*I*e), 0)), (x*(-(5*A - I*B)/(16*a**3*c**2) + (A*exp(10*I*e) + 5*
A*exp(8*I*e) + 10*A*exp(6*I*e) + 10*A*exp(4*I*e) + 5*A*exp(2*I*e) + A - I*B*exp(10*I*e) - 3*I*B*exp(8*I*e) - 2
*I*B*exp(6*I*e) + 2*I*B*exp(4*I*e) + 3*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(32*a**3*c**2)), True)) + x*(5*A - I*
B)/(16*a**3*c**2)

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Giac [A]  time = 1.27442, size = 296, normalized size = 1.6 \begin{align*} -\frac{\frac{6 \,{\left (-5 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{2}} + \frac{6 \,{\left (5 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{2}} + \frac{3 \,{\left (-15 i \, A \tan \left (f x + e\right )^{2} - 3 \, B \tan \left (f x + e\right )^{2} + 38 \, A \tan \left (f x + e\right ) - 10 i \, B \tan \left (f x + e\right ) + 25 i \, A + 9 \, B\right )}}{a^{3} c^{2}{\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{2}} + \frac{-55 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 201 \, A \tan \left (f x + e\right )^{2} + 33 i \, B \tan \left (f x + e\right )^{2} + 255 i \, A \tan \left (f x + e\right ) + 27 \, B \tan \left (f x + e\right ) + 117 \, A + 3 i \, B}{a^{3} c^{2}{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{192 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/192*(6*(-5*I*A - B)*log(tan(f*x + e) + I)/(a^3*c^2) + 6*(5*I*A + B)*log(tan(f*x + e) - I)/(a^3*c^2) + 3*(-1
5*I*A*tan(f*x + e)^2 - 3*B*tan(f*x + e)^2 + 38*A*tan(f*x + e) - 10*I*B*tan(f*x + e) + 25*I*A + 9*B)/(a^3*c^2*(
-I*tan(f*x + e) + 1)^2) + (-55*I*A*tan(f*x + e)^3 - 11*B*tan(f*x + e)^3 - 201*A*tan(f*x + e)^2 + 33*I*B*tan(f*
x + e)^2 + 255*I*A*tan(f*x + e) + 27*B*tan(f*x + e) + 117*A + 3*I*B)/(a^3*c^2*(tan(f*x + e) - I)^3))/f