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

Optimal. Leaf size=71 \[ \frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{\log (\sin (c+d x))}{a^2 d}-\frac{3 i x}{4 a^2}+\frac{1}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.140743, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3559, 3596, 3531, 3475} \[ \frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{\log (\sin (c+d x))}{a^2 d}-\frac{3 i x}{4 a^2}+\frac{1}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3559

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

Rule 3596

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

Rule 3531

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{1}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot (c+d x) (4 a-2 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{1}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot (c+d x) \left (8 a^2-6 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{3 i x}{4 a^2}+\frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{1}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot (c+d x) \, dx}{a^2}\\ &=-\frac{3 i x}{4 a^2}+\frac{\log (\sin (c+d x))}{a^2 d}+\frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{1}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.323348, size = 135, normalized size = 1.9 \[ \frac{\sec ^2(c+d x) \left (4 d x \sin (2 (c+d x))+i \sin (2 (c+d x))-8 i \sin (2 (c+d x)) \log \left (\sin ^2(c+d x)\right )+\cos (2 (c+d x)) \left (-8 \log \left (\sin ^2(c+d x)\right )-4 i d x-1\right )+16 i \tan ^{-1}(\tan (d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-8\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(-8 + Cos[2*(c + d*x)]*(-1 - (4*I)*d*x - 8*Log[Sin[c + d*x]^2]) + (16*I)*ArcTan[Tan[d*x]]*(Cos
[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + I*Sin[2*(c + d*x)] + 4*d*x*Sin[2*(c + d*x)] - (8*I)*Log[Sin[c + d*x]^2]*
Sin[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.079, size = 91, normalized size = 1.3 \begin{align*}{\frac{-{\frac{3\,i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{1}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{8\,{a}^{2}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*I/d/a^2/(tan(d*x+c)-I)-1/4/d/a^2/(tan(d*x+c)-I)^2-7/8/d/a^2*ln(tan(d*x+c)-I)-1/8/d/a^2*ln(tan(d*x+c)+I)+1
/d/a^2*ln(tan(d*x+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(cot(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 2.20427, size = 201, normalized size = 2.83 \begin{align*} \frac{{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\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(cot(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 8*e^(2*I*d*x + 2*I
*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

________________________________________________________________________________________

Sympy [A]  time = 1.18548, size = 150, normalized size = 2.11 \begin{align*} \begin{cases} \frac{\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text{for}\: 32 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac{\left (7 i e^{4 i c} + 4 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} + \frac{7 i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} - \frac{7 i x}{4 a^{2}} + \frac{\log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2)
, Ne(32*a**4*d**2*exp(6*I*c), 0)), (x*(-(7*I*exp(4*I*c) + 4*I*exp(2*I*c) + I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a*
*2)), True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)

________________________________________________________________________________________

Giac [A]  time = 1.24016, size = 111, normalized size = 1.56 \begin{align*} -\frac{\frac{2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{16 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac{21 \, \tan \left (d x + c\right )^{2} - 54 i \, \tan \left (d x + c\right ) - 37}{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(cot(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*log(tan(d*x + c) + I)/a^2 + 14*log(tan(d*x + c) - I)/a^2 - 16*log(abs(tan(d*x + c)))/a^2 - (21*tan(d*
x + c)^2 - 54*I*tan(d*x + c) - 37)/(a^2*(tan(d*x + c) - I)^2))/d