\(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [63]

Optimal result

Integrand size = 22, antiderivative size = 71 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \] Output:

-3/4*I*x/a^2+ln(sin(d*x+c))/a^2/d+3/4/a^2/d/(1+I*tan(d*x+c))+1/4/d/(a+I*a* 
tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2}{(a+i a \tan (c+d x))^2}-\frac {7 \log (i-\tan (c+d x))-8 \log (\tan (c+d x))+\log (i+\tan (c+d x))+\frac {6 i}{-i+\tan (c+d x)}}{a^2}}{8 d} \] Input:

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

Output:

(2/(a + I*a*Tan[c + d*x])^2 - (7*Log[I - Tan[c + d*x]] - 8*Log[Tan[c + d*x 
]] + Log[I + Tan[c + d*x]] + (6*I)/(-I + Tan[c + d*x]))/a^2)/(8*d)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4042, 27, 3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {4 a^2-3 i a^2 \tan (c+d x)}{\tan (c+d x)}dx}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {\frac {4 a^2 \int \cot (c+d x)dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {-4 a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {\frac {4 a^2 \log (-\sin (c+d x))}{d}-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\)

Input:

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

Output:

(((-3*I)*a^2*x + (4*a^2*Log[-Sin[c + d*x]])/d)/(2*a^2) + 3/(2*d*(1 + I*Tan 
[c + d*x])))/(2*a^2) + 1/(4*d*(a + I*a*Tan[c + d*x])^2)
 

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 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]
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01

method	result	size
risch	\(-\frac {7 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {2 i c}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(72\)
derivativedivides	\(-\frac {3 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(90\)
default	\(-\frac {3 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(90\)
norman	\(\frac {\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {5 i \tan \left (d x +c \right )}{4 d a}-\frac {3 i \tan \left (d x +c \right )^{3}}{4 d a}-\frac {3 i x \tan \left (d x +c \right )^{2}}{2 a}-\frac {3 i x \tan \left (d x +c \right )^{4}}{4 a}+\frac {\tan \left (d x +c \right )^{2}}{2 a d}}{a \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}\)	\(144\)

Input:

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

Output:

-7/4*I*x/a^2+1/2/a^2/d*exp(-2*I*(d*x+c))+1/16/a^2/d*exp(-4*I*(d*x+c))-2*I/ 
a^2/d*c+1/a^2/d*ln(exp(2*I*(d*x+c))-1)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \] Input:

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

Output:

1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d* 
x + 2*I*c) - 1) + 8*e^(2*I*d*x + 2*I*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} - 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \] Input:

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

Output:

Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(- 
4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((- 
7*I*exp(4*I*c) - 4*I*exp(2*I*c) - I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a**2)), 
 True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)
 

Maxima [F(-2)]

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

integrate(cot(d*x+c)/(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.
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{8 \, a^{2} d} - \frac {7 \, \log \left (\tan \left (d x + c\right ) - i\right )}{8 \, a^{2} d} + \frac {\log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {-3 i \, \tan \left (d x + c\right ) - 4}{4 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \] Input:

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

Output:

-1/8*log(tan(d*x + c) + I)/(a^2*d) - 7/8*log(tan(d*x + c) - I)/(a^2*d) + l 
og(abs(tan(d*x + c)))/(a^2*d) + 1/4*(-3*I*tan(d*x + c) - 4)/(a^2*d*(tan(d* 
x + c) - I)^2)
 

Mupad [B] (verification not implemented)

Time = 1.00 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\cot \left (d x +c \right )}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

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

Output:

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