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

Optimal. Leaf size=59 \[ \frac{3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac{x}{4 a^2}-\frac{i}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0818277, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3540, 3526, 8} \[ \frac{3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac{x}{4 a^2}-\frac{i}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[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{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{i}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{a-2 i a \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{2 a^2}\\ &=\frac{3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac{i}{4 d (a+i a \tan (c+d x))^2}-\frac{\int 1 \, dx}{4 a^2}\\ &=-\frac{x}{4 a^2}+\frac{3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac{i}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.188476, size = 68, normalized size = 1.15 \[ \frac{\sec ^2(c+d x) ((1+4 i d x) \sin (2 (c+d x))+(4 d x+i) \cos (2 (c+d x))-4 i)}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 79, normalized size = 1.3 \begin{align*}{\frac{{\frac{i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{2}d}}+{\frac{3}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/d/a^2/(tan(d*x+c)-I)^2+1/8*I/d/a^2*ln(tan(d*x+c)-I)+3/4/a^2/d/(tan(d*x+c)-I)-1/8*I/d/a^2*ln(tan(d*x+c)+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(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.24418, size = 127, normalized size = 2.15 \begin{align*} -\frac{{\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.442861, size = 119, normalized size = 2.02 \begin{align*} \begin{cases} \frac{\left (16 i a^{2} d e^{4 i c} e^{- 2 i d x} - 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text{for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac{\left (e^{4 i c} - 2 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} + \frac{1}{4 a^{2}}\right ) & \text{otherwise} \end{cases} - \frac{x}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((16*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(64*a**4*d**2*exp(6*I*c), 0)), (x*(-(exp(4*I*c) - 2*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) + 1/(4*a**2))
, True)) - x/(4*a**2)

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Giac [A]  time = 1.4551, size = 97, normalized size = 1.64 \begin{align*} -\frac{\frac{2 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{2 i \, \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a^{2}} + \frac{3 i \, \tan \left (d x + c\right )^{2} - 6 \, \tan \left (d x + c\right ) + 5 i}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*I*log(-I*tan(d*x + c) + 1)/a^2 - 2*I*log(-I*tan(d*x + c) - 1)/a^2 + (3*I*tan(d*x + c)^2 - 6*tan(d*x +
c) + 5*I)/(a^2*(tan(d*x + c) - I)^2))/d