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

Optimal. Leaf size=80 \[ \frac{B+i A}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{x (A-i B)}{4 a^2}+\frac{-B+i A}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0653431, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac{B+i A}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{x (A-i B)}{4 a^2}+\frac{-B+i A}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx &=\frac{i A-B}{4 f (a+i a \tan (e+f x))^2}+\frac{(A-i B) \int \frac{1}{a+i a \tan (e+f x)} \, dx}{2 a}\\ &=\frac{i A-B}{4 f (a+i a \tan (e+f x))^2}+\frac{i A+B}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{(A-i B) \int 1 \, dx}{4 a^2}\\ &=\frac{(A-i B) x}{4 a^2}+\frac{i A-B}{4 f (a+i a \tan (e+f x))^2}+\frac{i A+B}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.503802, size = 94, normalized size = 1.18 \[ -\frac{\sec ^2(e+f x) ((4 i A f x+A+4 B f x+i B) \sin (2 (e+f x))+(A (4 f x+i)+B (-1-4 i f x)) \cos (2 (e+f x))+4 i A)}{16 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[e + f*x]^2*((4*I)*A + (B*(-1 - (4*I)*f*x) + A*(I + 4*f*x))*Cos[2*(e + f*x)] + (A + I*B + (4*I)*A*f*x + 4
*B*f*x)*Sin[2*(e + f*x)]))/(16*a^2*f*(-I + Tan[e + f*x])^2)

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Maple [B]  time = 0.042, size = 162, normalized size = 2. \begin{align*}{\frac{A}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{4}}B}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) A}{f{a}^{2}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{8\,f{a}^{2}}}-{\frac{{\frac{i}{4}}A}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{B}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{B\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{8\,f{a}^{2}}}+{\frac{{\frac{i}{8}}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{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,x)

[Out]

1/4/f/a^2/(tan(f*x+e)-I)*A-1/4*I/f/a^2/(tan(f*x+e)-I)*B-1/8*I/f/a^2*ln(tan(f*x+e)-I)*A-1/8/f/a^2*ln(tan(f*x+e)
-I)*B-1/4*I/f/a^2/(tan(f*x+e)-I)^2*A+1/4/f/a^2/(tan(f*x+e)-I)^2*B+1/8/f/a^2*B*ln(tan(f*x+e)+I)+1/8*I/f/a^2*A*l
n(tan(f*x+e)+I)

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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,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.0329, size = 150, normalized size = 1.88 \begin{align*} \frac{{\left (4 \,{\left (A - i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, A e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{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,x, algorithm="fricas")

[Out]

1/16*(4*(A - I*B)*f*x*e^(4*I*f*x + 4*I*e) + 4*I*A*e^(2*I*f*x + 2*I*e) + I*A - B)*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [A]  time = 1.18336, size = 163, normalized size = 2.04 \begin{align*} \begin{cases} \frac{\left (16 i A a^{2} f e^{4 i e} e^{- 2 i f x} + \left (4 i A a^{2} f e^{2 i e} - 4 B a^{2} f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{64 a^{4} f^{2}} & \text{for}\: 64 a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac{A - i B}{4 a^{2}} + \frac{\left (A e^{4 i e} + 2 A e^{2 i e} + A - i B e^{4 i e} + i B\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (A - i B\right )}{4 a^{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))**2,x)

[Out]

Piecewise(((16*I*A*a**2*f*exp(4*I*e)*exp(-2*I*f*x) + (4*I*A*a**2*f*exp(2*I*e) - 4*B*a**2*f*exp(2*I*e))*exp(-4*
I*f*x))*exp(-6*I*e)/(64*a**4*f**2), Ne(64*a**4*f**2*exp(6*I*e), 0)), (x*(-(A - I*B)/(4*a**2) + (A*exp(4*I*e) +
2*A*exp(2*I*e) + A - I*B*exp(4*I*e) + I*B)*exp(-4*I*e)/(4*a**2)), True)) + x*(A - I*B)/(4*a**2)

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Giac [A]  time = 1.34665, size = 158, normalized size = 1.98 \begin{align*} -\frac{\frac{2 \,{\left (-i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2}} - \frac{2 \,{\left (-i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2}} - \frac{3 i \, A \tan \left (f x + e\right )^{2} + 3 \, B \tan \left (f x + e\right )^{2} + 10 \, A \tan \left (f x + e\right ) - 10 i \, B \tan \left (f x + e\right ) - 11 i \, A - 3 \, B}{a^{2}{\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{16 \, 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,x, algorithm="giac")

[Out]

-1/16*(2*(-I*A - B)*log(tan(f*x + e) + I)/a^2 - 2*(-I*A - B)*log(tan(f*x + e) - I)/a^2 - (3*I*A*tan(f*x + e)^2
+ 3*B*tan(f*x + e)^2 + 10*A*tan(f*x + e) - 10*I*B*tan(f*x + e) - 11*I*A - 3*B)/(a^2*(tan(f*x + e) - I)^2))/f