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

Optimal. Leaf size=97 \[ -\frac{9 \cot (c+d x)}{4 a^2 d}-\frac{2 i \log (\sin (c+d x))}{a^2 d}+\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{9 x}{4 a^2}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(-9*x)/(4*a^2) - (9*Cot[c + d*x])/(4*a^2*d) - ((2*I)*Log[Sin[c + d*x]])/(a^2*d) + Cot[c + d*x]/(a^2*d*(1 + I*T
an[c + d*x])) + Cot[c + d*x]/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.201486, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3559, 3596, 3529, 3531, 3475} \[ -\frac{9 \cot (c+d x)}{4 a^2 d}-\frac{2 i \log (\sin (c+d x))}{a^2 d}+\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{9 x}{4 a^2}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*x)/(4*a^2) - (9*Cot[c + d*x])/(4*a^2*d) - ((2*I)*Log[Sin[c + d*x]])/(a^2*d) + Cot[c + d*x]/(a^2*d*(1 + I*T
an[c + d*x])) + Cot[c + d*x]/(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 3529

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 + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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 ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot ^2(c+d x) (5 a-3 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot ^2(c+d x) \left (18 a^2-16 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{9 \cot (c+d x)}{4 a^2 d}+\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot (c+d x) \left (-16 i a^2-18 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{9 x}{4 a^2}-\frac{9 \cot (c+d x)}{4 a^2 d}+\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 i) \int \cot (c+d x) \, dx}{a^2}\\ &=-\frac{9 x}{4 a^2}-\frac{9 \cot (c+d x)}{4 a^2 d}-\frac{2 i \log (\sin (c+d x))}{a^2 d}+\frac{\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac{\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 1.69457, size = 276, normalized size = 2.85 \[ -\frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-36 i d x \sin (2 c)+i \sin (2 c) \sin (4 d x)+64 d x \cos ^2(c)+32 i d x \cot (c)+16 \sin (2 c) \log \left (\sin ^2(c+d x)\right )-\sin (2 c) \cos (4 d x)+8 i \csc (c) \cos (2 c-d x) \csc (c+d x)-8 i \csc (c) \cos (2 c+d x) \csc (c+d x)-8 \csc (c) \sin (2 c-d x) \csc (c+d x)+8 \csc (c) \sin (2 c+d x) \csc (c+d x)-32 (\cos (2 c)+i \sin (2 c)) \tan ^{-1}(\tan (d x))-i \cos (2 c) \left (32 d x \cot (c)-i \left (16 i \log \left (\sin ^2(c+d x)\right )+36 d x+\sin (4 d x)\right )+\cos (4 d x)\right )-32 d x-12 \sin (2 d x)-12 i \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*(-32*d*x + 64*d*x*Cos[c]^2 - (12*I)*Cos[2*d*x] + (32*I)*d*x*Cot[c]
+ (8*I)*Cos[2*c - d*x]*Csc[c]*Csc[c + d*x] - (8*I)*Cos[2*c + d*x]*Csc[c]*Csc[c + d*x] - 32*ArcTan[Tan[d*x]]*(C
os[2*c] + I*Sin[2*c]) - (36*I)*d*x*Sin[2*c] - Cos[4*d*x]*Sin[2*c] + 16*Log[Sin[c + d*x]^2]*Sin[2*c] - 12*Sin[2
*d*x] + I*Sin[2*c]*Sin[4*d*x] - I*Cos[2*c]*(Cos[4*d*x] + 32*d*x*Cot[c] - I*(36*d*x + (16*I)*Log[Sin[c + d*x]^2
] + Sin[4*d*x])) - 8*Csc[c]*Csc[c + d*x]*Sin[2*c - d*x] + 8*Csc[c]*Csc[c + d*x]*Sin[2*c + d*x]))/(16*a^2*d*(-I
+ Tan[c + d*x])^2)

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Maple [A]  time = 0.071, size = 111, normalized size = 1.1 \begin{align*}{\frac{{\frac{i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{17\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{2}d}}-{\frac{5}{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}}-{\frac{1}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{2\,i\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)^2/(a+I*a*tan(d*x+c))^2,x)

[Out]

1/4*I/d/a^2/(tan(d*x+c)-I)^2+17/8*I/d/a^2*ln(tan(d*x+c)-I)-5/4/a^2/d/(tan(d*x+c)-I)-1/8*I/d/a^2*ln(tan(d*x+c)+
I)-1/d/a^2/tan(d*x+c)-2*I/d/a^2*ln(tan(d*x+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(cot(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 = 2.27193, size = 333, normalized size = 3.43 \begin{align*} -\frac{68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (68 \, d x - 44 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (-32 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 32 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i}{16 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(68*d*x*e^(6*I*d*x + 6*I*c) - (68*d*x - 44*I)*e^(4*I*d*x + 4*I*c) - (-32*I*e^(6*I*d*x + 6*I*c) + 32*I*e^
(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 11*I*e^(2*I*d*x + 2*I*c) - I)/(a^2*d*e^(6*I*d*x + 6*I*c) - a
^2*d*e^(4*I*d*x + 4*I*c))

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Sympy [A]  time = 3.85442, size = 136, normalized size = 1.4 \begin{align*} - \frac{\left (\begin{cases} 17 x e^{4 i c} + \frac{3 i e^{2 i c} e^{- 2 i d x}}{d} + \frac{i e^{- 4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x \left (17 e^{4 i c} + 6 e^{2 i c} + 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} - \frac{2 i \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} - \frac{2 i e^{- 2 i c}}{a^{2} d \left (e^{2 i d x} - e^{- 2 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Piecewise((17*x*exp(4*I*c) + 3*I*exp(2*I*c)*exp(-2*I*d*x)/d + I*exp(-4*I*d*x)/(4*d), Ne(d, 0)), (x*(17*exp(4*
I*c) + 6*exp(2*I*c) + 1), True))*exp(-4*I*c)/(4*a**2) - 2*I*log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d) - 2*I*exp
(-2*I*c)/(a**2*d*(exp(2*I*d*x) - exp(-2*I*c)))

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Giac [A]  time = 1.31388, size = 147, normalized size = 1.52 \begin{align*} -\frac{\frac{32 i \, \log \left (i \, \tan \left (d x + c\right )\right )}{a^{2}} - \frac{34 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{16 \,{\left (-2 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{2} \tan \left (d x + c\right )} + \frac{51 i \, \tan \left (d x + c\right )^{2} + 122 \, \tan \left (d x + c\right ) - 75 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(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(32*I*log(I*tan(d*x + c))/a^2 - 34*I*log(I*tan(d*x + c) + 1)/a^2 + 2*I*log(-I*tan(d*x + c) + 1)/a^2 + 16
*(-2*I*tan(d*x + c) + 1)/(a^2*tan(d*x + c)) + (51*I*tan(d*x + c)^2 + 122*tan(d*x + c) - 75*I)/(a^2*(tan(d*x +
c) - I)^2))/d