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

Optimal. Leaf size=104 \[ \frac{A (d+i c)+B (c+3 i d)}{4 a^2 f (1+i \tan (e+f x))}+\frac{x (A-i B) (c-i d)}{4 a^2}+\frac{(-B+i A) (c+i d)}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

((A - I*B)*(c - I*d)*x)/(4*a^2) + (B*(c + (3*I)*d) + A*(I*c + d))/(4*a^2*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*
(c + I*d))/(4*f*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.238247, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3590, 3526, 8} \[ \frac{A (d+i c)+B (c+3 i d)}{4 a^2 f (1+i \tan (e+f x))}+\frac{x (A-i B) (c-i d)}{4 a^2}+\frac{(-B+i A) (c+i d)}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - I*B)*(c - I*d)*x)/(4*a^2) + (B*(c + (3*I)*d) + A*(I*c + d))/(4*a^2*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*
(c + I*d))/(4*f*(a + I*a*Tan[e + f*x])^2)

Rule 3590

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

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 8

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

Rubi steps

\begin{align*} \int \frac{(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}-\frac{i \int \frac{a (B (c+i d)+A (i c+d))+2 a B d \tan (e+f x)}{a+i a \tan (e+f x)} \, dx}{2 a^2}\\ &=\frac{B (c+3 i d)+A (i c+d)}{4 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}+\frac{((A-i B) (c-i d)) \int 1 \, dx}{4 a^2}\\ &=\frac{(A-i B) (c-i d) x}{4 a^2}+\frac{B (c+3 i d)+A (i c+d)}{4 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c+i d)}{4 f (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.68907, size = 201, normalized size = 1.93 \[ -\frac{(A+B \tan (e+f x)) (c+d \tan (e+f x)) (\sin (2 (e+f x)) (A (4 i c f x+c+4 d f x+i d)+B (4 c f x+i c-4 i d f x-d))+\cos (2 (e+f x)) (A (c (4 f x+i)+d (-1-4 i f x))-B (4 i c f x+c+d (4 f x+i)))+4 i (A c+B d))}{16 a^2 f (\tan (e+f x)-i)^2 (A \cos (e+f x)+B \sin (e+f x)) (c \cos (e+f x)+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.045, size = 338, normalized size = 3.3 \begin{align*}{\frac{-{\frac{i}{4}}Bc}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{8}}B\ln \left ( \tan \left ( fx+e \right ) +i \right ) d}{f{a}^{2}}}+{\frac{Ac}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{3\,Bd}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{Ad}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{Bc}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{4}}Ad}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{8}}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) c}{f{a}^{2}}}+{\frac{{\frac{i}{4}}Bd}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{4}}Ac}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) Ad}{8\,f{a}^{2}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) Bc}{8\,f{a}^{2}}}+{\frac{A\ln \left ( \tan \left ( fx+e \right ) +i \right ) d}{8\,f{a}^{2}}}+{\frac{B\ln \left ( \tan \left ( fx+e \right ) +i \right ) c}{8\,f{a}^{2}}}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) Bd}{f{a}^{2}}}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) Ac}{f{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))*(c+d*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 = 1.34732, size = 224, normalized size = 2.15 \begin{align*} \frac{{\left ({\left (4 \,{\left (A - i \, B\right )} c +{\left (-4 i \, A - 4 \, B\right )} d\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (i \, A - B\right )} c -{\left (A + i \, B\right )} d +{\left (4 i \, A c + 4 i \, B d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\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))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.8764, size = 298, normalized size = 2.87 \begin{align*} \begin{cases} \frac{\left (\left (16 i A a^{2} c f e^{4 i e} + 16 i B a^{2} d f e^{4 i e}\right ) e^{- 2 i f x} + \left (4 i A a^{2} c f e^{2 i e} - 4 A a^{2} d f e^{2 i e} - 4 B a^{2} c f e^{2 i e} - 4 i B a^{2} d 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 c - i A d - i B c - B d}{4 a^{2}} + \frac{\left (A c e^{4 i e} + 2 A c e^{2 i e} + A c - i A d e^{4 i e} + i A d - i B c e^{4 i e} + i B c - B d e^{4 i e} + 2 B d e^{2 i e} - B d\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (A c - i A d - i B c - B d\right )}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((16*I*A*a**2*c*f*exp(4*I*e) + 16*I*B*a**2*d*f*exp(4*I*e))*exp(-2*I*f*x) + (4*I*A*a**2*c*f*exp(2*I*
e) - 4*A*a**2*d*f*exp(2*I*e) - 4*B*a**2*c*f*exp(2*I*e) - 4*I*B*a**2*d*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*c - I*A*d - I*B*c - B*d)/(4*a**2) + (A*c*exp(4*I*e)
+ 2*A*c*exp(2*I*e) + A*c - I*A*d*exp(4*I*e) + I*A*d - I*B*c*exp(4*I*e) + I*B*c - B*d*exp(4*I*e) + 2*B*d*exp(2*
I*e) - B*d)*exp(-4*I*e)/(4*a**2)), True)) + x*(A*c - I*A*d - I*B*c - B*d)/(4*a**2)

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Giac [B]  time = 1.23019, size = 270, normalized size = 2.6 \begin{align*} -\frac{\frac{2 \,{\left (-i \, A c - B c - A d + i \, B d\right )} \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2}} + \frac{2 \,{\left (i \, A c + B c + A d - i \, B d\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2}} + \frac{-3 i \, A c \tan \left (f x + e\right )^{2} - 3 \, B c \tan \left (f x + e\right )^{2} - 3 \, A d \tan \left (f x + e\right )^{2} + 3 i \, B d \tan \left (f x + e\right )^{2} - 10 \, A c \tan \left (f x + e\right ) + 10 i \, B c \tan \left (f x + e\right ) + 10 i \, A d \tan \left (f x + e\right ) - 6 \, B d \tan \left (f x + e\right ) + 11 i \, A c + 3 \, B c + 3 \, A d + 5 i \, B d}{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))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/16*(2*(-I*A*c - B*c - A*d + I*B*d)*log(-I*tan(f*x + e) + 1)/a^2 + 2*(I*A*c + B*c + A*d - I*B*d)*log(-I*tan(
f*x + e) - 1)/a^2 + (-3*I*A*c*tan(f*x + e)^2 - 3*B*c*tan(f*x + e)^2 - 3*A*d*tan(f*x + e)^2 + 3*I*B*d*tan(f*x +
e)^2 - 10*A*c*tan(f*x + e) + 10*I*B*c*tan(f*x + e) + 10*I*A*d*tan(f*x + e) - 6*B*d*tan(f*x + e) + 11*I*A*c +
3*B*c + 3*A*d + 5*I*B*d)/(a^2*(tan(f*x + e) - I)^2))/f