3.1.65 \(\int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [65]

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

[Out]

15/4*I*x/a^2+15/4*I*cot(d*x+c)/a^2/d-2*cot(d*x+c)^2/a^2/d-4*ln(sin(d*x+c))/a^2/d+5/4*cot(d*x+c)^2/a^2/d/(1+I*t
an(d*x+c))+1/4*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.16, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 = {3640, 3677, 3610, 3612, 3556} \begin {gather*} -\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} \end {gather*}

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 3556

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

Rule 3610

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 3612

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 3640

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 3677

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]

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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(122)=244\).
time = 1.62, size = 319, normalized size = 2.61 \begin {gather*} \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-64 i d x+128 i d x \cos ^2(c)-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 ^2(c+d x)+32 \cos (2 c) \log \left (\sin ^2(c+d x)\right )+60 d x \sin (2 c)-i \cos (4 d x) \sin (2 c)+8 i \csc ^2(c+d x) \sin (2 c)+32 i \log \left (\sin ^2(c+d x)\right ) \sin (2 c)+64 \text {ArcTan}(\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)\right )}{16 a^2 d (-i+\tan (c+d x))^2} \end {gather*}

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 = 0.30, size = 90, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {\frac {7 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {1}{2 \tan \left (d x +c \right )^{2}}+\frac {2 i}{\tan \left (d x +c \right )}-4 \ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(90\)
default	\(\frac {\frac {7 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {1}{2 \tan \left (d x +c \right )^{2}}+\frac {2 i}{\tan \left (d x +c \right )}-4 \ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(90\)
risch	\(\frac {31 i x}{4 a^{2}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {8 i c}{a^{2} d}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(105\)
norman	\(\frac {-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{d a}-\frac {1}{2 d a}-\frac {2 \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {15 i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {15 i x \left (\tan ^{4}\left (d x +c \right )\right )}{2 a}+\frac {15 i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {2 i \tan \left (d x +c \right )}{d a}+\frac {25 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}+\frac {15 i \left (\tan ^{5}\left (d x +c \right )\right )}{4 d a}}{\tan \left (d x +c \right )^{2} a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {4 \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(195\)

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,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(7/4*I/(tan(d*x+c)-I)+1/4/(tan(d*x+c)-I)^2+31/8*ln(tan(d*x+c)-I)-1/2/tan(d*x+c)^2+2*I/tan(d*x+c)-4*ln(
tan(d*x+c))+1/8*ln(tan(d*x+c)+I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.38, size = 151, normalized size = 1.24 \begin {gather*} \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 {gather*}

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 = 0.46, size = 216, normalized size = 1.77 \begin {gather*} \frac {- 2 e^{2 i c} e^{2 i d x} + 4}{a^{2} d e^{4 i c} e^{4 i d x} - 2 a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \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}\: 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 {gather*}

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) + 4)/(a**2*d*exp(4*I*c)*exp(4*I*d*x) - 2*a**2*d*exp(2*I*c)*exp(2*I*d*x) + a**2*d)
+ 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(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 = 1.61, size = 109, normalized size = 0.89 \begin {gather*} \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 (\tan \left (d x + c\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 {gather*}

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(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

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Mupad [B]
time = 4.01, size = 135, normalized size = 1.11 \begin {gather*} \frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {\mathrm {tan}\left (c+d\,x\right )}{a^2}-\frac {15\,{\mathrm {tan}\left (c+d\,x\right )}^3}{4\,a^2}+\frac {1{}\mathrm {i}}{2\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,11{}\mathrm {i}}{2\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(31*log(tan(c + d*x) - 1i))/(8*a^2*d) + log(tan(c + d*x) + 1i)/(8*a^2*d) + (tan(c + d*x)/a^2 + 1i/(2*a^2) + (t
an(c + d*x)^2*11i)/(2*a^2) - (15*tan(c + d*x)^3)/(4*a^2))/(d*(2*tan(c + d*x)^3 - tan(c + d*x)^2*1i + tan(c + d
*x)^4*1i)) - (4*log(tan(c + d*x)))/(a^2*d)

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