\(\int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [474]

Optimal result

Integrand size = 19, antiderivative size = 107 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 a b x}{\left (a^2+b^2\right )^2}+\frac {\log (\sin (c+d x))}{a^2 d}-\frac {b^2 \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right )^2 d}+\frac {b^2}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \] Output:

-2*a*b*x/(a^2+b^2)^2+ln(sin(d*x+c))/a^2/d-b^2*(3*a^2+b^2)*ln(a*cos(d*x+c)+ 
b*sin(d*x+c))/a^2/(a^2+b^2)^2/d+b^2/a/(a^2+b^2)/d/(a+b*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.44 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {-\frac {a (a-i b) \log (i-\tan (c+d x))}{2 (a+i b)}+\frac {\left (a^2+b^2\right ) \log (\tan (c+d x))}{a}-\frac {a (a+i b) \log (i+\tan (c+d x))}{2 (a-i b)}-\frac {b^2 \left (3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{a \left (a^2+b^2\right )}+\frac {b^2}{a+b \tan (c+d x)}}{a \left (a^2+b^2\right ) d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.23, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4052, 3042, 4134, 3042, 25, 3956, 4013}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a^2-b \tan (c+d x) a+b^2+b^2 \tan (c+d x)^2}{\tan (c+d x) (a+b \tan (c+d x))}dx}{a \left (a^2+b^2\right )}+\frac {b^2}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 4134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

\(\displaystyle \frac {-\frac {b^2 \left (3 a^2+b^2\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a \left (a^2+b^2\right )}+\frac {\left (a^2+b^2\right ) \log (-\sin (c+d x))}{a d}-\frac {2 a^2 b x}{a^2+b^2}}{a \left (a^2+b^2\right )}+\frac {b^2}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 4013

\(\displaystyle \frac {b^2}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\frac {\left (a^2+b^2\right ) \log (-\sin (c+d x))}{a d}-\frac {b^2 \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac {2 a^2 b x}{a^2+b^2}}{a \left (a^2+b^2\right )}\)

Input:

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

Output:

((-2*a^2*b*x)/(a^2 + b^2) + ((a^2 + b^2)*Log[-Sin[c + d*x]])/(a*d) - (b^2* 
(3*a^2 + b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a*(a^2 + b^2)*d))/(a* 
(a^2 + b^2)) + b^2/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4013

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si 
n[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] && EqQ[a*c + b*d, 0]
 

rule 4052

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] 
 && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4134

Int[((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^ 
2)/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])), x_Symbol] :> Simp[(a*(A*c - c*C + B*d) + b*(B*c - A*d + C*d))*(x/ 
((a^2 + b^2)*(c^2 + d^2))), x] + (Simp[(A*b^2 - a*b*B + a^2*C)/((b*c - a*d) 
*(a^2 + b^2))   Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] - Sim 
p[(c^2*C - B*c*d + A*d^2)/((b*c - a*d)*(c^2 + d^2))   Int[(d - c*Tan[e + f* 
x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] 
&& NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {b^{2}}{a \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{2} \left (3 a^{2}+b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(126\)
default	\(\frac {\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {b^{2}}{a \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{2} \left (3 a^{2}+b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(126\)
parallelrisch	\(\frac {-6 b^{2} \left (a +b \tan \left (d x +c \right )\right ) \left (a^{2}+\frac {b^{2}}{3}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )-a^{2} \left (a -b \right ) \left (a +b \right ) \left (a +b \tan \left (d x +c \right )\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+2 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right ) \ln \left (\tan \left (d x +c \right )\right )+\left (-4 b^{2} a^{3} x d -2 a^{2} b^{3}-2 b^{5}\right ) \tan \left (d x +c \right )-4 b \,a^{4} x d}{2 \left (a +b \tan \left (d x +c \right )\right ) \left (a^{2}+b^{2}\right )^{2} d \,a^{2}}\)	\(165\)
norman	\(\frac {-\frac {b^{3} \tan \left (d x +c \right )}{d \,a^{2} \left (a^{2}+b^{2}\right )}-\frac {2 b \,a^{2} x}{a^{4}+2 b^{2} a^{2}+b^{4}}-\frac {2 b^{2} a x \tan \left (d x +c \right )}{a^{4}+2 b^{2} a^{2}+b^{4}}}{a +b \tan \left (d x +c \right )}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}-\frac {b^{2} \left (3 a^{2}+b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) a^{2} d}\)	\(200\)
risch	\(-\frac {i x}{2 i a b -a^{2}+b^{2}}+\frac {6 i b^{2} x}{a^{4}+2 b^{2} a^{2}+b^{4}}+\frac {6 i b^{2} c}{d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}+\frac {2 i b^{4} x}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) a^{2}}+\frac {2 i b^{4} c}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) a^{2} d}-\frac {2 i x}{a^{2}}-\frac {2 i c}{a^{2} d}+\frac {2 i b^{3}}{\left (-i a +b \right ) d a \left (i a +b \right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) b^{2}}{d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(339\)

Input:

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

Output:

1/d*(1/a^2*ln(tan(d*x+c))+1/(a^2+b^2)^2*(1/2*(-a^2+b^2)*ln(1+tan(d*x+c)^2) 
-2*a*b*arctan(tan(d*x+c)))+b^2/a/(a^2+b^2)/(a+b*tan(d*x+c))-b^2*(3*a^2+b^2 
)/a^2/(a^2+b^2)^2*ln(a+b*tan(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (107) = 214\).

Time = 0.10 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.20 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 \, a^{4} b d x - 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (3 \, a^{3} b^{2} + a b^{4} + {\left (3 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, a^{3} b^{2} d x + a^{2} b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}} \] Input:

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

Output:

-1/2*(4*a^4*b*d*x - 2*a*b^4 - (a^5 + 2*a^3*b^2 + a*b^4 + (a^4*b + 2*a^2*b^ 
3 + b^5)*tan(d*x + c))*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) + (3*a^3*b 
^2 + a*b^4 + (3*a^2*b^3 + b^5)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a 
*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + 2*(2*a^3*b^2*d*x + a^2*b^3) 
*tan(d*x + c))/((a^6*b + 2*a^4*b^3 + a^2*b^5)*d*tan(d*x + c) + (a^7 + 2*a^ 
5*b^2 + a^3*b^4)*d)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.63 (sec) , antiderivative size = 2961, normalized size of antiderivative = 27.67 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*x*cot(c)/tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-log 
(tan(c + d*x)**2 + 1)/(2*d) + log(tan(c + d*x))/d)/a**2, Eq(b, 0)), ((log( 
tan(c + d*x)**2 + 1)/(2*d) - log(tan(c + d*x))/d - 1/(2*d*tan(c + d*x)**2) 
)/b**2, Eq(a, 0)), (3*I*d*x*tan(c + d*x)**2/(4*a**2*d*tan(c + d*x)**2 + 8* 
I*a**2*d*tan(c + d*x) - 4*a**2*d) - 6*d*x*tan(c + d*x)/(4*a**2*d*tan(c + d 
*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) - 3*I*d*x/(4*a**2*d*tan(c + d 
*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) - 2*log(tan(c + d*x)**2 + 1)* 
tan(c + d*x)**2/(4*a**2*d*tan(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a* 
*2*d) - 4*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(4*a**2*d*tan(c + d*x)** 
2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) + 2*log(tan(c + d*x)**2 + 1)/(4*a* 
*2*d*tan(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) + 4*log(tan(c + 
 d*x))*tan(c + d*x)**2/(4*a**2*d*tan(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) 
 - 4*a**2*d) + 8*I*log(tan(c + d*x))*tan(c + d*x)/(4*a**2*d*tan(c + d*x)** 
2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) - 4*log(tan(c + d*x))/(4*a**2*d*ta 
n(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) + 3*I*tan(c + d*x)/(4* 
a**2*d*tan(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d) - 4/(4*a**2*d 
*tan(c + d*x)**2 + 8*I*a**2*d*tan(c + d*x) - 4*a**2*d), Eq(b, -I*a)), (-3* 
I*d*x*tan(c + d*x)**2/(4*a**2*d*tan(c + d*x)**2 - 8*I*a**2*d*tan(c + d*x) 
- 4*a**2*d) - 6*d*x*tan(c + d*x)/(4*a**2*d*tan(c + d*x)**2 - 8*I*a**2*d*ta 
n(c + d*x) - 4*a**2*d) + 3*I*d*x/(4*a**2*d*tan(c + d*x)**2 - 8*I*a**2*d...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, b^{2}}{a^{4} + a^{2} b^{2} + {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )} + \frac {2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}}}{2 \, d} \] Input:

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

Output:

-1/2*(4*(d*x + c)*a*b/(a^4 + 2*a^2*b^2 + b^4) - 2*b^2/(a^4 + a^2*b^2 + (a^ 
3*b + a*b^3)*tan(d*x + c)) + 2*(3*a^2*b^2 + b^4)*log(b*tan(d*x + c) + a)/( 
a^6 + 2*a^4*b^2 + a^2*b^4) + (a^2 - b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2* 
a^2*b^2 + b^4) - 2*log(tan(d*x + c))/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.72 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (d x + c\right )} a b}{a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac {{\left (3 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d + 2 \, a^{4} b^{3} d + a^{2} b^{5} d} + \frac {\log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {a^{3} b^{2} + a b^{4}}{{\left (a^{2} + b^{2}\right )}^{2} {\left (b \tan \left (d x + c\right ) + a\right )} a^{2} d} \] Input:

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

Output:

-2*(d*x + c)*a*b/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/2*(a^2 - b^2)*log(tan(d 
*x + c)^2 + 1)/(a^4*d + 2*a^2*b^2*d + b^4*d) - (3*a^2*b^3 + b^5)*log(abs(b 
*tan(d*x + c) + a))/(a^6*b*d + 2*a^4*b^3*d + a^2*b^5*d) + log(abs(tan(d*x 
+ c)))/(a^2*d) + (a^3*b^2 + a*b^4)/((a^2 + b^2)^2*(b*tan(d*x + c) + a)*a^2 
*d)
 

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.42 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{2\,d\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}+\frac {b^2}{a\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {b^2\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2+b^2\right )}{a^2\,d\,{\left (a^2+b^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 541, normalized size of antiderivative = 5.06 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**5 + cos(c + d*x)*log(tan( 
(c + d*x)/2)**2 + 1)*a**3*b**2 - 3*cos(c + d*x)*log(tan((c + d*x)/2)**2*a 
- 2*tan((c + d*x)/2)*b - a)*a**3*b**2 - cos(c + d*x)*log(tan((c + d*x)/2)* 
*2*a - 2*tan((c + d*x)/2)*b - a)*a*b**4 + cos(c + d*x)*log(tan((c + d*x)/2 
))*a**5 + 2*cos(c + d*x)*log(tan((c + d*x)/2))*a**3*b**2 + cos(c + d*x)*lo 
g(tan((c + d*x)/2))*a*b**4 - 2*cos(c + d*x)*a**4*b*d*x + cos(c + d*x)*a**3 
*b**2 + cos(c + d*x)*a*b**4 - log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a* 
*4*b + log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**2*b**3 - 3*log(tan((c 
+ d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)*a**2*b**3 - log(ta 
n((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)*b**5 + log(ta 
n((c + d*x)/2))*sin(c + d*x)*a**4*b + 2*log(tan((c + d*x)/2))*sin(c + d*x) 
*a**2*b**3 + log(tan((c + d*x)/2))*sin(c + d*x)*b**5 - 2*sin(c + d*x)*a**3 
*b**2*d*x)/(a**2*d*(cos(c + d*x)*a**5 + 2*cos(c + d*x)*a**3*b**2 + cos(c + 
 d*x)*a*b**4 + sin(c + d*x)*a**4*b + 2*sin(c + d*x)*a**2*b**3 + sin(c + d* 
x)*b**5))