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

Optimal. Leaf size=71 \[ -\frac{\cos ^4(e+f x) (B-A \tan (e+f x))}{4 a^2 c^2 f}+\frac{3 A \sin (e+f x) \cos (e+f x)}{8 a^2 c^2 f}+\frac{3 A x}{8 a^2 c^2} \]

[Out]

(3*A*x)/(8*a^2*c^2) + (3*A*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*c^2*f) - (Cos[e + f*x]^4*(B - A*Tan[e + f*x]))/(4
*a^2*c^2*f)

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Rubi [A]  time = 0.138077, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3588, 73, 639, 199, 205} \[ -\frac{\cos ^4(e+f x) (B-A \tan (e+f x))}{4 a^2 c^2 f}+\frac{3 A \sin (e+f x) \cos (e+f x)}{8 a^2 c^2 f}+\frac{3 A x}{8 a^2 c^2} \]

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])^2),x]

[Out]

(3*A*x)/(8*a^2*c^2) + (3*A*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*c^2*f) - (Cos[e + f*x]^4*(B - A*Tan[e + f*x]))/(4
*a^2*c^2*f)

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 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

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

Mathematica [A]  time = 0.124274, size = 53, normalized size = 0.75 \[ \frac{A (12 (e+f x)+8 \sin (2 (e+f x))+\sin (4 (e+f x)))-8 B \cos ^4(e+f x)}{32 a^2 c^2 f} \]

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])^2),x]

[Out]

(-8*B*Cos[e + f*x]^4 + A*(12*(e + f*x) + 8*Sin[2*(e + f*x)] + Sin[4*(e + f*x)]))/(32*a^2*c^2*f)

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Maple [C]  time = 0.06, size = 236, normalized size = 3.3 \begin{align*}{\frac{3\,A}{16\,f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{16}}B}{f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{16}}A}{f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{B}{16\,f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{3\,i}{16}}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{2}{c}^{2}}}+{\frac{3\,A}{16\,f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{{\frac{i}{16}}B}{f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{{\frac{3\,i}{16}}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{2}{c}^{2}}}+{\frac{{\frac{i}{16}}A}{f{a}^{2}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{B}{16\,f{a}^{2}{c}^{2} \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))^2,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [C]  time = 1.1672, size = 251, normalized size = 3.54 \begin{align*} \frac{{\left (24 \, A f x e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-8 i \, A - 4 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (8 i \, A - 4 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{64 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/64*(24*A*f*x*e^(4*I*f*x + 4*I*e) + (-I*A - B)*e^(8*I*f*x + 8*I*e) + (-8*I*A - 4*B)*e^(6*I*f*x + 6*I*e) + (8*
I*A - 4*B)*e^(2*I*f*x + 2*I*e) + I*A - B)*e^(-4*I*f*x - 4*I*e)/(a^2*c^2*f)

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Sympy [A]  time = 3.35578, size = 362, normalized size = 5.1 \begin{align*} \frac{3 A x}{8 a^{2} c^{2}} + \begin{cases} \frac{\left (\left (16384 i A a^{6} c^{6} f^{3} e^{2 i e} - 16384 B a^{6} c^{6} f^{3} e^{2 i e}\right ) e^{- 4 i f x} + \left (131072 i A a^{6} c^{6} f^{3} e^{4 i e} - 65536 B a^{6} c^{6} f^{3} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 131072 i A a^{6} c^{6} f^{3} e^{8 i e} - 65536 B a^{6} c^{6} f^{3} e^{8 i e}\right ) e^{2 i f x} + \left (- 16384 i A a^{6} c^{6} f^{3} e^{10 i e} - 16384 B a^{6} c^{6} f^{3} e^{10 i e}\right ) e^{4 i f x}\right ) e^{- 6 i e}}{1048576 a^{8} c^{8} f^{4}} & \text{for}\: 1048576 a^{8} c^{8} f^{4} e^{6 i e} \neq 0 \\x \left (- \frac{3 A}{8 a^{2} c^{2}} + \frac{\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{16 a^{2} c^{2}}\right ) & \text{otherwise} \end{cases} \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))**2,x)

[Out]

3*A*x/(8*a**2*c**2) + Piecewise((((16384*I*A*a**6*c**6*f**3*exp(2*I*e) - 16384*B*a**6*c**6*f**3*exp(2*I*e))*ex
p(-4*I*f*x) + (131072*I*A*a**6*c**6*f**3*exp(4*I*e) - 65536*B*a**6*c**6*f**3*exp(4*I*e))*exp(-2*I*f*x) + (-131
072*I*A*a**6*c**6*f**3*exp(8*I*e) - 65536*B*a**6*c**6*f**3*exp(8*I*e))*exp(2*I*f*x) + (-16384*I*A*a**6*c**6*f*
*3*exp(10*I*e) - 16384*B*a**6*c**6*f**3*exp(10*I*e))*exp(4*I*f*x))*exp(-6*I*e)/(1048576*a**8*c**8*f**4), Ne(10
48576*a**8*c**8*f**4*exp(6*I*e), 0)), (x*(-3*A/(8*a**2*c**2) + (A*exp(8*I*e) + 4*A*exp(6*I*e) + 6*A*exp(4*I*e)
+ 4*A*exp(2*I*e) + A - I*B*exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(16*a**2*c**2)
), True))

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Giac [A]  time = 1.33025, size = 90, normalized size = 1.27 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} A}{a^{2} c^{2}} + \frac{3 \, A \tan \left (f x + e\right )^{3} + 5 \, A \tan \left (f x + e\right ) - 2 \, B}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2} c^{2}}}{8 \, 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))^2,x, algorithm="giac")

[Out]

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