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

3.1.64.1 Optimal result

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

output

-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
 

3.1.64.2 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 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {\frac {4 \cot ^2(c+d x)}{i+\cot (c+d x)}-9 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-8 i (\log (\cos (c+d x))+\log (\tan (c+d x)))}{a^2}+\frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2}}{4 d} \]

input

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

output

(((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)
 

3.1.64.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {3042, 4042, 3042, 4079, 27, 3042, 4012, 25, 3042, 4014, 3042, 25, 3956}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (5 a-3 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 a-3 i a \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)}dx}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int 2 \cot ^2(c+d x) \left (9 a^2-8 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \cot ^2(c+d x) \left (9 a^2-8 i a^2 \tan (c+d x)\right )dx}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {9 a^2-8 i a^2 \tan (c+d x)}{\tan (c+d x)^2}dx}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {-\frac {9 a^2 \cot (c+d x)}{d}+\int -\cot (c+d x) \left (9 \tan (c+d x) a^2+8 i a^2\right )dx}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {-\frac {9 a^2 \cot (c+d x)}{d}-\int \cot (c+d x) \left (9 \tan (c+d x) a^2+8 i a^2\right )dx}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {9 a^2 \cot (c+d x)}{d}-\int \frac {9 \tan (c+d x) a^2+8 i a^2}{\tan (c+d x)}dx}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {\frac {-8 i a^2 \int \cot (c+d x)dx-\frac {9 a^2 \cot (c+d x)}{d}-9 a^2 x}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-8 i a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {9 a^2 \cot (c+d x)}{d}-9 a^2 x}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {8 i a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {9 a^2 \cot (c+d x)}{d}-9 a^2 x}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {-\frac {9 a^2 \cot (c+d x)}{d}-\frac {8 i a^2 \log (-\sin (c+d x))}{d}-9 a^2 x}{a^2}+\frac {4 \cot (c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

input

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

output

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

3.1.64.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4012

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] + Simp[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 4014

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] + Simp[(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] && N 
eQ[a*c + b*d, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(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 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(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] + Simp[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[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] /; Free 
Q[{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]
 

3.1.64.4 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	\(\frac {-\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{d a}-\frac {1}{a d}-\frac {9 \left (\tan ^{4}\left (d x +c \right )\right )}{4 a d}-\frac {9 x \tan \left (d x +c \right )}{4 a}-\frac {9 x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}-\frac {9 x \left (\tan ^{5}\left (d x +c \right )\right )}{4 a}-\frac {15 \left (\tan ^{2}\left (d x +c \right )\right )}{4 a d}-\frac {3 i \tan \left (d x +c \right )}{2 d a}}{\tan \left (d x +c \right ) a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(175\)

input

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

output

-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)
 

3.1.64.5 Fricas [A] (verification not implemented)

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

input

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

output

-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))
 

3.1.64.6 Sympy [A] (verification not implemented)

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

input

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

output

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)
 

3.1.64.7 Maxima [F(-2)]

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

input

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

output

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

3.1.64.8 Giac [A] (verification not implemented)

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

input

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

output

-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
 

3.1.64.9 Mupad [B] (verification not implemented)

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

input

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

output

(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)