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

Optimal. Leaf size=183 \[ -\frac{2 A+i B}{16 a^2 c^3 f (-\tan (e+f x)+i)}-\frac{-B+i A}{32 a^2 c^3 f (-\tan (e+f x)+i)^2}+\frac{B+3 i A}{32 a^2 c^3 f (\tan (e+f x)+i)^2}-\frac{A-i B}{24 a^2 c^3 f (\tan (e+f x)+i)^3}+\frac{x (5 A+i B)}{16 a^2 c^3}+\frac{3 A}{16 a^2 c^3 f (\tan (e+f x)+i)} \]

[Out]

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

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Rubi [A]  time = 0.238882, antiderivative size = 183, 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} \[ -\frac{2 A+i B}{16 a^2 c^3 f (-\tan (e+f x)+i)}-\frac{-B+i A}{32 a^2 c^3 f (-\tan (e+f x)+i)^2}+\frac{B+3 i A}{32 a^2 c^3 f (\tan (e+f x)+i)^2}-\frac{A-i B}{24 a^2 c^3 f (\tan (e+f x)+i)^3}+\frac{x (5 A+i B)}{16 a^2 c^3}+\frac{3 A}{16 a^2 c^3 f (\tan (e+f x)+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.16704, size = 217, normalized size = 1.19 \[ \frac{\sec ^2(e+f x) (\cos (3 (e+f x))+i \sin (3 (e+f x))) (12 (A (-10 f x+5 i)-2 i B f x+B) \cos (e+f x)+3 (9 B-5 i A) \cos (3 (e+f x))-60 A \sin (e+f x)+120 i A f x \sin (e+f x)-45 A \sin (3 (e+f x))-5 A \sin (5 (e+f x))-i A \cos (5 (e+f x))+12 i B \sin (e+f x)-24 B f x \sin (e+f x)-9 i B \sin (3 (e+f x))-i B \sin (5 (e+f x))+5 B \cos (5 (e+f x)))}{384 a^2 c^3 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.075, size = 303, normalized size = 1.7 \begin{align*}{\frac{{\frac{i}{16}}B}{f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{A}{8\,f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{B}{32\,f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{32}}A}{f{a}^{2}{c}^{3} \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}^{2}{c}^{3}}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{32\,f{a}^{2}{c}^{3}}}+{\frac{3\,A}{16\,f{a}^{2}{c}^{3} \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}^{2}{c}^{3}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) B}{32\,f{a}^{2}{c}^{3}}}-{\frac{A}{24\,f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{{\frac{i}{24}}B}{f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{{\frac{3\,i}{32}}A}{f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{B}{32\,f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{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))^2/(c-I*c*tan(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.11237, size = 336, normalized size = 1.84 \begin{align*} \frac{{\left (24 \,{\left (5 \, A + i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A - 2 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-15 i \, A - 9 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-60 i \, A - 12 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (30 i \, A - 18 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{3} 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))^2/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.87608, size = 456, normalized size = 2.49 \begin{align*} \begin{cases} \frac{\left (\left (50331648 i A a^{8} c^{12} f^{4} e^{2 i e} - 50331648 B a^{8} c^{12} f^{4} e^{2 i e}\right ) e^{- 4 i f x} + \left (503316480 i A a^{8} c^{12} f^{4} e^{4 i e} - 301989888 B a^{8} c^{12} f^{4} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 1006632960 i A a^{8} c^{12} f^{4} e^{8 i e} - 201326592 B a^{8} c^{12} f^{4} e^{8 i e}\right ) e^{2 i f x} + \left (- 251658240 i A a^{8} c^{12} f^{4} e^{10 i e} - 150994944 B a^{8} c^{12} f^{4} e^{10 i e}\right ) e^{4 i f x} + \left (- 33554432 i A a^{8} c^{12} f^{4} e^{12 i e} - 33554432 B a^{8} c^{12} f^{4} e^{12 i e}\right ) e^{6 i f x}\right ) e^{- 6 i e}}{6442450944 a^{10} c^{15} f^{5}} & \text{for}\: 6442450944 a^{10} c^{15} f^{5} e^{6 i e} \neq 0 \\x \left (- \frac{5 A + i B}{16 a^{2} c^{3}} + \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^{- 4 i e}}{32 a^{2} c^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (5 A + i B\right )}{16 a^{2} c^{3}} \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))**2/(c-I*c*tan(f*x+e))**3,x)

[Out]

Piecewise((((50331648*I*A*a**8*c**12*f**4*exp(2*I*e) - 50331648*B*a**8*c**12*f**4*exp(2*I*e))*exp(-4*I*f*x) +
(503316480*I*A*a**8*c**12*f**4*exp(4*I*e) - 301989888*B*a**8*c**12*f**4*exp(4*I*e))*exp(-2*I*f*x) + (-10066329
60*I*A*a**8*c**12*f**4*exp(8*I*e) - 201326592*B*a**8*c**12*f**4*exp(8*I*e))*exp(2*I*f*x) + (-251658240*I*A*a**
8*c**12*f**4*exp(10*I*e) - 150994944*B*a**8*c**12*f**4*exp(10*I*e))*exp(4*I*f*x) + (-33554432*I*A*a**8*c**12*f
**4*exp(12*I*e) - 33554432*B*a**8*c**12*f**4*exp(12*I*e))*exp(6*I*f*x))*exp(-6*I*e)/(6442450944*a**10*c**15*f*
*5), Ne(6442450944*a**10*c**15*f**5*exp(6*I*e), 0)), (x*(-(5*A + I*B)/(16*a**2*c**3) + (A*exp(10*I*e) + 5*A*ex
p(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(-4*I*e)/(32*a**2*c**3)), True)) + x*(5*A + I*B)/(
16*a**2*c**3)

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Giac [A]  time = 1.39957, size = 296, normalized size = 1.62 \begin{align*} -\frac{\frac{6 \,{\left (-5 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{3}} + \frac{6 \,{\left (5 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{3}} + \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^{2} c^{3}{\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^{2} c^{3}{\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))^2/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/192*(6*(-5*I*A + B)*log(tan(f*x + e) + I)/(a^2*c^3) + 6*(5*I*A - B)*log(tan(f*x + e) - I)/(a^2*c^3) + 3*(15
*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^2*c^3*(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^2*c^3*(tan(f*x + e) + I)^3))/f