3.1069 \(\int \frac{c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0625688, 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} \[ \frac{d+i c}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{x (c-i d)}{4 a^2}+\frac{-d+i c}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)*x)/(4*a^2) + (I*c - d)/(4*f*(a + I*a*Tan[e + f*x])^2) + (I*c + d)/(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{c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx &=\frac{i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac{(c-i d) \int \frac{1}{a+i a \tan (e+f x)} \, dx}{2 a}\\ &=\frac{i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac{i c+d}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{(c-i d) \int 1 \, dx}{4 a^2}\\ &=\frac{(c-i d) x}{4 a^2}+\frac{i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac{i c+d}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.628871, size = 94, normalized size = 1.18 \[ -\frac{\sec ^2(e+f x) ((4 i c f x+c+4 d f x+i d) \sin (2 (e+f x))+(c (4 f x+i)+d (-1-4 i f x)) \cos (2 (e+f x))+4 i c)}{16 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 162, normalized size = 2. \begin{align*}{\frac{-{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) c}{f{a}^{2}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) d}{8\,f{a}^{2}}}+{\frac{c}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{4}}d}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{4}}c}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{d}{4\,f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) d}{8\,f{a}^{2}}}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) c}{f{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.54935, size = 150, normalized size = 1.88 \begin{align*} \frac{{\left (4 \,{\left (c - i \, d\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d\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((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.786315, size = 163, normalized size = 2.04 \begin{align*} \begin{cases} \frac{\left (16 i a^{2} c f e^{4 i e} e^{- 2 i f x} + \left (4 i a^{2} c f e^{2 i e} - 4 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{c - i d}{4 a^{2}} + \frac{\left (c e^{4 i e} + 2 c e^{2 i e} + c - i d e^{4 i e} + i d\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (c - i d\right )}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.38344, size = 158, normalized size = 1.98 \begin{align*} -\frac{\frac{2 \,{\left (-i \, c - d\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2}} - \frac{2 \,{\left (-i \, c - d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2}} - \frac{3 i \, c \tan \left (f x + e\right )^{2} + 3 \, d \tan \left (f x + e\right )^{2} + 10 \, c \tan \left (f x + e\right ) - 10 i \, d \tan \left (f x + e\right ) - 11 i \, c - 3 \, 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((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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