\(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [64]

Optimal result

Integrand size = 24, antiderivative size = 97 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {9 x}{4 a^2}-\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-9/4*x/a^2-9/4*cot(d*x+c)/a^2/d-2*I*ln(sin(d*x+c))/a^2/d+cot(d*x+c)/a^2/d/(1+I*tan(d*x+c))+1/4*cot(d*x+c)/d/(a
+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3640, 3677, 3610, 3612, 3556} \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 x}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[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 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)}{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 [C] (verified)

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

Time = 1.38 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {\frac {4 \cot ^2(c+d x)}{i+\cot (c+d x)}-9 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-8 i (\log (\cos (c+d x))+\log (\tan (c+d x)))}{a^2}+\frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2}}{4 d} \]

[In]

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

[Out]

(((4*Cot[c + d*x]^2)/(I + Cot[c + d*x]) - 9*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - (8
*I)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))/a^2 + Cot[c + d*x]/(a + I*a*Tan[c + d*x])^2)/(4*d)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {17 x}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {4 c}{a^{2} d}-\frac {2 i}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(96\)
derivativedivides	\(\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}-\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {5}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{a^{2} d \tan \left (d x +c \right )}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(108\)
default	\(\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}-\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {5}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{a^{2} d \tan \left (d x +c \right )}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(108\)
norman	\(\frac {-\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{d a}-\frac {1}{a d}-\frac {9 \left (\tan ^{4}\left (d x +c \right )\right )}{4 a d}-\frac {9 x \tan \left (d x +c \right )}{4 a}-\frac {9 x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}-\frac {9 x \left (\tan ^{5}\left (d x +c \right )\right )}{4 a}-\frac {15 \left (\tan ^{2}\left (d x +c \right )\right )}{4 a d}-\frac {3 i \tan \left (d x +c \right )}{2 d a}}{\tan \left (d x +c \right ) a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(175\)

[In]

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

[Out]

-17/4*x/a^2-3/4*I/a^2/d*exp(-2*I*(d*x+c))-1/16*I/a^2/d*exp(-4*I*(d*x+c))-4/a^2/d*c-2*I/d/a^2/(exp(2*I*(d*x+c))
-1)-2*I/a^2/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, {\left (17 \, d x - 11 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 32 \, {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

[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) - 4*(17*d*x - 11*I)*e^(4*I*d*x + 4*I*c) + 32*(I*e^(6*I*d*x + 6*I*c) - 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))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.86 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 48 i a^{2} d e^{4 i c} e^{- 2 i d x} - 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 17 e^{4 i c} - 6 e^{2 i c} - 1\right ) e^{- 4 i c}}{4 a^{2}} + \frac {17}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{2} d e^{2 i c} e^{2 i d x} - a^{2} d} - \frac {17 x}{4 a^{2}} - \frac {2 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

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

[Out]

Piecewise(((-48*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*
d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((-17*exp(4*I*c) - 6*exp(2*I*c) - 1)*exp(-4*I*c)/(4*a**2) + 17/(4*a**2
)), True)) - 2*I/(a**2*d*exp(2*I*c)*exp(2*I*d*x) - a**2*d) - 17*x/(4*a**2) - 2*I*log(exp(2*I*d*x) - exp(-2*I*c
))/(a**2*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.75 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac {34 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {32 i \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}} + \frac {16 \, {\left (-2 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{2} \tan \left (d x + c\right )} + \frac {51 i \, \tan \left (d x + c\right )^{2} + 122 \, \tan \left (d x + c\right ) - 75 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.68 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.29 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {7\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}-\frac {1{}\mathrm {i}}{a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,9{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^2-\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,17{}\mathrm {i}}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,2{}\mathrm {i}}{a^2\,d} \]

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*17i)/(8*a^2*d) - ((7*tan(c + d*x))/(2*a^2) - 1i/a^2 + (tan(c + d*x)^2*9i)/(4*a^2))/(d*
(2*tan(c + d*x)^2 - tan(c + d*x)*1i + tan(c + d*x)^3*1i)) - (log(tan(c + d*x) + 1i)*1i)/(8*a^2*d) - (log(tan(c
+ d*x))*2i)/(a^2*d)