3.65 \(\int \frac{\cot ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=122 \[ -\frac{2 \cot ^2(c+d x)}{a^2 d}+\frac{15 i \cot (c+d x)}{4 a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{15 i x}{4 a^2}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(((15*I)/4)*x)/a^2 + (((15*I)/4)*Cot[c + d*x])/(a^2*d) - (2*Cot[c + d*x]^2)/(a^2*d) - (4*Log[Sin[c + d*x]])/(a
^2*d) + (5*Cot[c + d*x]^2)/(4*a^2*d*(1 + I*Tan[c + d*x])) + Cot[c + d*x]^2/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.233879, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 \cot ^2(c+d x)}{a^2 d}+\frac{15 i \cot (c+d x)}{4 a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{15 i x}{4 a^2}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((15*I)/4)*x)/a^2 + (((15*I)/4)*Cot[c + d*x])/(a^2*d) - (2*Cot[c + d*x]^2)/(a^2*d) - (4*Log[Sin[c + d*x]])/(a
^2*d) + (5*Cot[c + d*x]^2)/(4*a^2*d*(1 + I*Tan[c + d*x])) + Cot[c + d*x]^2/(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 ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot ^3(c+d x) (6 a-4 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot ^3(c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{2 \cot ^2(c+d x)}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{15 i \cot (c+d x)}{4 a^2 d}-\frac{2 \cot ^2(c+d x)}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot (c+d x) \left (-32 a^2+30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{15 i x}{4 a^2}+\frac{15 i \cot (c+d x)}{4 a^2 d}-\frac{2 \cot ^2(c+d x)}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{4 \int \cot (c+d x) \, dx}{a^2}\\ &=\frac{15 i x}{4 a^2}+\frac{15 i \cot (c+d x)}{4 a^2 d}-\frac{2 \cot ^2(c+d x)}{a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 1.26648, size = 319, normalized size = 2.61 \[ \frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (60 d x \sin (2 c)-\sin (2 c) \sin (4 d x)+128 i d x \cos ^2(c)-60 i d x \cos (2 c)+\cos (2 c) \cos (4 d x)-64 d x \cot (c)+32 i \sin (2 c) \log \left (\sin ^2(c+d x)\right )-i \sin (2 c) \cos (4 d x)-i \cos (2 c) \sin (4 d x)+64 d x \cos (2 c) \cot (c)+8 \cos (2 c) \csc ^2(c+d x)-16 \csc (c) \cos (2 c-d x) \csc (c+d x)+16 \csc (c) \cos (2 c+d x) \csc (c+d x)+8 i \sin (2 c) \csc ^2(c+d x)-16 i \csc (c) \sin (2 c-d x) \csc (c+d x)+16 i \csc (c) \sin (2 c+d x) \csc (c+d x)+32 \cos (2 c) \log \left (\sin ^2(c+d x)\right )+64 (\sin (2 c)-i \cos (2 c)) \tan ^{-1}(\tan (d x))-64 i d x-16 i \sin (2 d x)+16 \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*((-64*I)*d*x + (128*I)*d*x*Cos[c]^2 - (60*I)*d*x*Cos[2*c] + 16*Cos[2
*d*x] + Cos[2*c]*Cos[4*d*x] - 64*d*x*Cot[c] + 64*d*x*Cos[2*c]*Cot[c] - 16*Cos[2*c - d*x]*Csc[c]*Csc[c + d*x] +
16*Cos[2*c + d*x]*Csc[c]*Csc[c + d*x] + 8*Cos[2*c]*Csc[c + d*x]^2 + 32*Cos[2*c]*Log[Sin[c + d*x]^2] + 60*d*x*
Sin[2*c] - I*Cos[4*d*x]*Sin[2*c] + (8*I)*Csc[c + d*x]^2*Sin[2*c] + (32*I)*Log[Sin[c + d*x]^2]*Sin[2*c] + 64*Ar
cTan[Tan[d*x]]*((-I)*Cos[2*c] + Sin[2*c]) - (16*I)*Sin[2*d*x] - I*Cos[2*c]*Sin[4*d*x] - Sin[2*c]*Sin[4*d*x] -
(16*I)*Csc[c]*Csc[c + d*x]*Sin[2*c - d*x] + (16*I)*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 = 0.084, size = 125, normalized size = 1. \begin{align*}{\frac{{\frac{7\,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{31\,\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{1}{2\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{2\,i}{{a}^{2}d\tan \left ( dx+c \right ) }}-4\,{\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)^3/(a+I*a*tan(d*x+c))^2,x)

[Out]

7/4*I/d/a^2/(tan(d*x+c)-I)+1/4/d/a^2/(tan(d*x+c)-I)^2+31/8/d/a^2*ln(tan(d*x+c)-I)+1/8/d/a^2*ln(tan(d*x+c)+I)-1
/2/d/a^2/tan(d*x+c)^2+2*I/d/a^2/tan(d*x+c)-4/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)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.25422, size = 450, normalized size = 3.69 \begin{align*} \frac{124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \,{\left (31 i \, d x + 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 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 ) - 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, 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)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(124*I*d*x*e^(8*I*d*x + 8*I*c) - 8*(31*I*d*x + 6)*e^(6*I*d*x + 6*I*c) + (124*I*d*x + 95)*e^(4*I*d*x + 4*I
*c) - 64*(e^(8*I*d*x + 8*I*c) - 2*e^(6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 14
*e^(2*I*d*x + 2*I*c) - 1)/(a^2*d*e^(8*I*d*x + 8*I*c) - 2*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 = 2.45356, size = 221, normalized size = 1.81 \begin{align*} \frac{- \frac{2 e^{- 2 i c} e^{2 i d x}}{a^{2} d} + \frac{4 e^{- 4 i c}}{a^{2} d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \begin{cases} \frac{\left (- 16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text{for}\: 16 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac{\left (31 i e^{4 i c} + 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac{31 i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{31 i x}{4 a^{2}} - \frac{4 \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)**3/(a+I*a*tan(d*x+c))**2,x)

[Out]

(-2*exp(-2*I*c)*exp(2*I*d*x)/(a**2*d) + 4*exp(-4*I*c)/(a**2*d))/(exp(4*I*d*x) - 2*exp(-2*I*c)*exp(2*I*d*x) + e
xp(-4*I*c)) + Piecewise(((-16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(
16*a**4*d**2), Ne(16*a**4*d**2*exp(6*I*c), 0)), (x*((31*I*exp(4*I*c) + 8*I*exp(2*I*c) + I)*exp(-4*I*c)/(4*a**2
) - 31*I/(4*a**2)), True)) + 31*I*x/(4*a**2) - 4*log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 1.32373, size = 149, normalized size = 1.22 \begin{align*} \frac{\frac{4 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{124 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{128 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{3 \, \tan \left (d x + c\right )^{4} + 114 i \, \tan \left (d x + c\right )^{3} + 173 \, \tan \left (d x + c\right )^{2} - 32 i \, \tan \left (d x + c\right ) + 16}{{\left (\tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right )\right )}^{2} a^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/32*(4*log(tan(d*x + c) + I)/a^2 + 124*log(tan(d*x + c) - I)/a^2 - 128*log(abs(tan(d*x + c)))/a^2 + (3*tan(d*
x + c)^4 + 114*I*tan(d*x + c)^3 + 173*tan(d*x + c)^2 - 32*I*tan(d*x + c) + 16)/((tan(d*x + c)^2 - I*tan(d*x +
c))^2*a^2))/d