\(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [74]

Optimal result

Integrand size = 24, antiderivative size = 133 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {25 x}{8 a^3}-\frac {25 \cot (c+d x)}{8 a^3 d}-\frac {3 i \log (\sin (c+d x))}{a^3 d}+\frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-25/8*x/a^3-25/8*cot(d*x+c)/a^3/d-3*I*ln(sin(d*x+c))/a^3/d+1/6*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^3+11/24*cot(d*x
+c)/a/d/(a+I*a*tan(d*x+c))^2+3/2*cot(d*x+c)/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3640, 3677, 3610, 3612, 3556} \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {25 \cot (c+d x)}{8 a^3 d}-\frac {3 i \log (\sin (c+d x))}{a^3 d}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {25 x}{8 a^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

(-25*x)/(8*a^3) - (25*Cot[c + d*x])/(8*a^3*d) - ((3*I)*Log[Sin[c + d*x]])/(a^3*d) + Cot[c + d*x]/(6*d*(a + I*a
*Tan[c + d*x])^3) + (11*Cot[c + d*x])/(24*a*d*(a + I*a*Tan[c + d*x])^2) + (3*Cot[c + d*x])/(2*d*(a^3 + I*a^3*T
an[c + d*x]))

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*} \text {integral}& = \frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^2(c+d x) (7 a-4 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^2(c+d x) \left (39 a^2-33 i a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (150 a^3-144 i a^3 \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {25 \cot (c+d x)}{8 a^3 d}+\frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-144 i a^3-150 a^3 \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {25 x}{8 a^3}-\frac {25 \cot (c+d x)}{8 a^3 d}+\frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(3 i) \int \cot (c+d x) \, dx}{a^3} \\ & = -\frac {25 x}{8 a^3}-\frac {25 \cot (c+d x)}{8 a^3 d}-\frac {3 i \log (\sin (c+d x))}{a^3 d}+\frac {\cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.90 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {4 a^4 \cot (c+d x)+(a+i a \tan (c+d x)) \left (11 a^3 \cot (c+d x)+3 a (a+i a \tan (c+d x)) \left (-25 a (i+\cot (c+d x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+12 a (\cot (c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x))) (-i+\tan (c+d x)))\right )\right )}{24 a^4 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

(4*a^4*Cot[c + d*x] + (a + I*a*Tan[c + d*x])*(11*a^3*Cot[c + d*x] + 3*a*(a + I*a*Tan[c + d*x])*(-25*a*(I + Cot
[c + d*x])*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 12*a*(Cot[c + d*x] + 2*(Log[Cos[c + d*x]] + Log[
Tan[c + d*x]])*(-I + Tan[c + d*x])))))/(24*a^4*d*(a + I*a*Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {49 x}{8 a^{3}}-\frac {23 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {7 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}-\frac {6 c}{a^{3} d}-\frac {2 i}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {3 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}\)	\(114\)
derivativedivides	\(-\frac {\cot \left (d x +c \right )}{a^{3} d}+\frac {3 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 d \,a^{3}}+\frac {\frac {25 \pi }{16}-\frac {25 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )}{8}}{a^{3} d}+\frac {9 i}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {1}{6 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{3}}-\frac {31}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )}\)	\(115\)
default	\(-\frac {\cot \left (d x +c \right )}{a^{3} d}+\frac {3 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 d \,a^{3}}+\frac {\frac {25 \pi }{16}-\frac {25 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )}{8}}{a^{3} d}+\frac {9 i}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {1}{6 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{3}}-\frac {31}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )}\)	\(115\)
norman	\(\frac {-\frac {1}{a d}-\frac {25 \left (\tan ^{4}\left (d x +c \right )\right )}{3 a d}-\frac {25 \left (\tan ^{6}\left (d x +c \right )\right )}{8 a d}-\frac {25 x \tan \left (d x +c \right )}{8 a}-\frac {75 x \left (\tan ^{3}\left (d x +c \right )\right )}{8 a}-\frac {75 x \left (\tan ^{5}\left (d x +c \right )\right )}{8 a}-\frac {25 x \left (\tan ^{7}\left (d x +c \right )\right )}{8 a}-\frac {55 \left (\tan ^{2}\left (d x +c \right )\right )}{8 a d}-\frac {3 i \left (\tan ^{5}\left (d x +c \right )\right )}{2 a d}-\frac {35 i \tan \left (d x +c \right )}{12 d a}-\frac {15 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}}{\tan \left (d x +c \right ) a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}-\frac {3 i \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}+\frac {3 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}\)	\(222\)

[In]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-49/8*x/a^3-23/16*I/a^3/d*exp(-2*I*(d*x+c))-7/32*I/a^3/d*exp(-4*I*(d*x+c))-1/48*I/a^3/d*exp(-6*I*(d*x+c))-6/a^
3/d*c-2*I/d/a^3/(exp(2*I*(d*x+c))-1)-3*I/a^3/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {588 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 6 \, {\left (98 \, d x - 55 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 288 \, {\left (i \, e^{\left (8 i \, d x + 8 i \, c\right )} - i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 117 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 19 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/96*(588*d*x*e^(8*I*d*x + 8*I*c) - 6*(98*d*x - 55*I)*e^(6*I*d*x + 6*I*c) + 288*(I*e^(8*I*d*x + 8*I*c) - I*e^
(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 117*I*e^(4*I*d*x + 4*I*c) - 19*I*e^(2*I*d*x + 2*I*c) - 2*I)/
(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.64 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 35328 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 5376 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 49 e^{6 i c} - 23 e^{4 i c} - 7 e^{2 i c} - 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {49}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{3} d e^{2 i c} e^{2 i d x} - a^{3} d} - \frac {49 x}{8 a^{3}} - \frac {3 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \]

[In]

integrate(cot(d*x+c)**2/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((-35328*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) - 5376*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 512*I*a
**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((-49*exp
(6*I*c) - 23*exp(4*I*c) - 7*exp(2*I*c) - 1)*exp(-6*I*c)/(8*a**3) + 49/(8*a**3)), True)) - 2*I/(a**3*d*exp(2*I*
c)*exp(2*I*d*x) - a**3*d) - 49*x/(8*a**3) - 3*I*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.85 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {294 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {288 i \, \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac {96 \, {\left (-3 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{3} \tan \left (d x + c\right )} + \frac {539 \, \tan \left (d x + c\right )^{3} - 1821 i \, \tan \left (d x + c\right )^{2} - 2085 \, \tan \left (d x + c\right ) + 819 i}{a^{3} {\left (i \, \tan \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

[In]

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

[Out]

-1/96*(6*I*log(tan(d*x + c) + I)/a^3 - 294*I*log(tan(d*x + c) - I)/a^3 + 288*I*log(tan(d*x + c))/a^3 + 96*(-3*
I*tan(d*x + c) + 1)/(a^3*tan(d*x + c)) + (539*tan(d*x + c)^3 - 1821*I*tan(d*x + c)^2 - 2085*tan(d*x + c) + 819
*I)/(a^3*(I*tan(d*x + c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 4.34 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,49{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,3{}\mathrm {i}}{a^3\,d}+\frac {\frac {71\,\mathrm {tan}\left (c+d\,x\right )}{12\,a^3}-\frac {25\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,a^3}-\frac {1{}\mathrm {i}}{a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,63{}\mathrm {i}}{8\,a^3}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*49i)/(16*a^3*d) - (log(tan(c + d*x) + 1i)*1i)/(16*a^3*d) - (log(tan(c + d*x))*3i)/(a^3
*d) + ((71*tan(c + d*x))/(12*a^3) - 1i/a^3 + (tan(c + d*x)^2*63i)/(8*a^3) - (25*tan(c + d*x)^3)/(8*a^3))/(d*(t
an(c + d*x)*1i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*3i + tan(c + d*x)^4))