\(\int \cot ^{-1}(c+d \cot (a+b x)) \, dx\) [25]

Optimal result

Integrand size = 11, antiderivative size = 198 \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=x \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{2} i x \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{2} i x \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {\operatorname {PolyLog}\left (2,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b} \] Output:

x*arccot(c+d*cot(b*x+a))-1/2*I*x*ln(1-(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+ 
d))+1/2*I*x*ln(1-(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))-1/4*polylog(2 
,(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+d))/b+1/4*polylog(2,(c+I*(1+d))*exp(2 
*I*a+2*I*b*x)/(c+I*(1-d)))/b
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(701\) vs. \(2(198)=396\).

Time = 4.19 (sec) , antiderivative size = 701, normalized size of antiderivative = 3.54 \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=x \left (\cot ^{-1}(c+d \cot (a+b x))+\frac {\left (4 a \sqrt {-d^2} \arctan \left (c+\frac {\left (1+c^2\right ) \tan (a+b x)}{d}\right )+i d \log (1+i \tan (a+b x)) \log \left (\frac {-c d+\sqrt {-d^2}-\left (1+c^2\right ) \tan (a+b x)}{-i-i c^2-c d+\sqrt {-d^2}}\right )-i d \log (1-i \tan (a+b x)) \log \left (\frac {-c d+\sqrt {-d^2}-\left (1+c^2\right ) \tan (a+b x)}{i+i c^2-c d+\sqrt {-d^2}}\right )+i d \log (1-i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+\left (1+c^2\right ) \tan (a+b x)}{-i-i c^2+c d+\sqrt {-d^2}}\right )-i d \log (1+i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+\left (1+c^2\right ) \tan (a+b x)}{i+i c^2+c d+\sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {\left (1+c^2\right ) (1-i \tan (a+b x))}{1+c^2+i c d-i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {\left (1+c^2\right ) (1-i \tan (a+b x))}{1+c^2+i c d+i \sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {\left (1+c^2\right ) (1+i \tan (a+b x))}{1+c^2-i c d-i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {\left (1+c^2\right ) (1+i \tan (a+b x))}{1+c^2-i c d+i \sqrt {-d^2}}\right )\right ) \left (\left (-c d+\sqrt {-d^2}\right ) \cos (a+b x)-\left (1+c^2\right ) \sin (a+b x)\right ) \left (\left (c d+\sqrt {-d^2}\right ) \cos (a+b x)+\left (1+c^2\right ) \sin (a+b x)\right )}{\left (1+c^2\right ) \sqrt {-d^2} (2 a-i \log (1-i \tan (a+b x))+i \log (1+i \tan (a+b x))) \left (1+c^2+d^2+\left (-1-c^2+d^2\right ) \cos (2 (a+b x))+2 c d \sin (2 (a+b x))\right )}\right ) \] Input:

Integrate[ArcCot[c + d*Cot[a + b*x]],x]
 

Output:

x*(ArcCot[c + d*Cot[a + b*x]] + ((4*a*Sqrt[-d^2]*ArcTan[c + ((1 + c^2)*Tan 
[a + b*x])/d] + I*d*Log[1 + I*Tan[a + b*x]]*Log[(-(c*d) + Sqrt[-d^2] - (1 
+ c^2)*Tan[a + b*x])/(-I - I*c^2 - c*d + Sqrt[-d^2])] - I*d*Log[1 - I*Tan[ 
a + b*x]]*Log[(-(c*d) + Sqrt[-d^2] - (1 + c^2)*Tan[a + b*x])/(I + I*c^2 - 
c*d + Sqrt[-d^2])] + I*d*Log[1 - I*Tan[a + b*x]]*Log[(c*d + Sqrt[-d^2] + ( 
1 + c^2)*Tan[a + b*x])/(-I - I*c^2 + c*d + Sqrt[-d^2])] - I*d*Log[1 + I*Ta 
n[a + b*x]]*Log[(c*d + Sqrt[-d^2] + (1 + c^2)*Tan[a + b*x])/(I + I*c^2 + c 
*d + Sqrt[-d^2])] - I*d*PolyLog[2, ((1 + c^2)*(1 - I*Tan[a + b*x]))/(1 + c 
^2 + I*c*d - I*Sqrt[-d^2])] + I*d*PolyLog[2, ((1 + c^2)*(1 - I*Tan[a + b*x 
]))/(1 + c^2 + I*c*d + I*Sqrt[-d^2])] - I*d*PolyLog[2, ((1 + c^2)*(1 + I*T 
an[a + b*x]))/(1 + c^2 - I*c*d - I*Sqrt[-d^2])] + I*d*PolyLog[2, ((1 + c^2 
)*(1 + I*Tan[a + b*x]))/(1 + c^2 - I*c*d + I*Sqrt[-d^2])])*((-(c*d) + Sqrt 
[-d^2])*Cos[a + b*x] - (1 + c^2)*Sin[a + b*x])*((c*d + Sqrt[-d^2])*Cos[a + 
 b*x] + (1 + c^2)*Sin[a + b*x]))/((1 + c^2)*Sqrt[-d^2]*(2*a - I*Log[1 - I* 
Tan[a + b*x]] + I*Log[1 + I*Tan[a + b*x]])*(1 + c^2 + d^2 + (-1 - c^2 + d^ 
2)*Cos[2*(a + b*x)] + 2*c*d*Sin[2*(a + b*x)])))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5693, 2620, 2715, 2838}

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.

\(\displaystyle \int \cot ^{-1}(d \cot (a+b x)+c) \, dx\)

\(\Big \downarrow \) 5693

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

\(\Big \downarrow \) 2620

\(\displaystyle -b (i c-d+1) \left (\frac {x \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{2 b (c-i (1-d))}-\frac {\int \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )dx}{2 b (c-i (1-d))}\right )+b (-i c+d+1) \left (\frac {\int \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )dx}{2 b (c+i (d+1))}-\frac {x \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{2 b (c+i (d+1))}\right )+x \cot ^{-1}(d \cot (a+b x)+c)\)

\(\Big \downarrow \) 2715

\(\displaystyle -b (i c-d+1) \left (\frac {i \int e^{-2 i a-2 i b x} \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )de^{2 i a+2 i b x}}{4 b^2 (c-i (1-d))}+\frac {x \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{2 b (c-i (1-d))}\right )+b (-i c+d+1) \left (-\frac {i \int e^{-2 i a-2 i b x} \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )de^{2 i a+2 i b x}}{4 b^2 (c+i (d+1))}-\frac {x \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{2 b (c+i (d+1))}\right )+x \cot ^{-1}(d \cot (a+b x)+c)\)

\(\Big \downarrow \) 2838

\(\displaystyle -b (i c-d+1) \left (\frac {x \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{2 b (c-i (1-d))}-\frac {i \operatorname {PolyLog}\left (2,\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b^2 (c-i (1-d))}\right )+b (-i c+d+1) \left (\frac {i \operatorname {PolyLog}\left (2,\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2 (c+i (d+1))}-\frac {x \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{2 b (c+i (d+1))}\right )+x \cot ^{-1}(d \cot (a+b x)+c)\)

Input:

Int[ArcCot[c + d*Cot[a + b*x]],x]
 

Output:

x*ArcCot[c + d*Cot[a + b*x]] - b*(1 + I*c - d)*((x*Log[1 - ((1 + I*c - d)* 
E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/(2*b*(c - I*(1 - d))) - ((I/4)*Po 
lyLog[2, ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/(b^2*(c - 
 I*(1 - d)))) + b*(1 - I*c + d)*(-1/2*(x*Log[1 - ((c + I*(1 + d))*E^((2*I) 
*a + (2*I)*b*x))/(c + I*(1 - d))])/(b*(c + I*(1 + d))) + ((I/4)*PolyLog[2, 
 ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))])/(b^2*(c + I*( 
1 + d))))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 5693

Int[ArcCot[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcC 
ot[c + d*Cot[a + b*x]], x] + (-Simp[b*(1 + I*c - d)   Int[x*(E^(2*I*a + 2*I 
*b*x)/(1 + I*c + d - (1 + I*c - d)*E^(2*I*a + 2*I*b*x))), x], x] + Simp[b*( 
1 - I*c + d)   Int[x*(E^(2*I*a + 2*I*b*x)/(1 - I*c - d - (1 - I*c + d)*E^(2 
*I*a + 2*I*b*x))), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[(c - I*d)^2, -1 
]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1145 vs. \(2 (168 ) = 336\).

Time = 5.12 (sec) , antiderivative size = 1146, normalized size of antiderivative = 5.79

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1146\)
default	\(\text {Expression too large to display}\)	\(1146\)
risch	\(\text {Expression too large to display}\)	\(4964\)

Input:

int(arccot(c+d*cot(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b/d*(-d*(1/2*Pi-arccot(cot(b*x+a)))*arccot(c+d*cot(b*x+a))-d^2*(-1/d*arc 
tan(d*((c+d*cot(b*x+a))/d-c/d)+c)*arctan(-(c+d*cot(b*x+a))/d+c/d)-1/d^2*(- 
1/2*I*d*arctan(d*((c+d*cot(b*x+a))/d-c/d)+c)*ln(1-(I*d+I+c)*(1+I*(d*((c+d* 
cot(b*x+a))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(-I*d+I-c))- 
1/2*d*arctan(d*((c+d*cot(b*x+a))/d-c/d)+c)^2-1/4*d*polylog(2,(I*d+I+c)*(1+ 
I*(d*((c+d*cot(b*x+a))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/( 
-I*d+I-c))+1/2*I*d^2*ln(1-(c-I*d+I)*(1+I*(d*((c+d*cot(b*x+a))/d-c/d)+c))^2 
/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I-c))*arctan(d*((c+d*cot(b*x+a) 
)/d-c/d)+c)/(1+I*c+d)+1/2*I*d*ln(1-(c-I*d+I)*(1+I*(d*((c+d*cot(b*x+a))/d-c 
/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I-c))*arctan(d*((c+d*c 
ot(b*x+a))/d-c/d)+c)/(1+I*c+d)+1/2*I*d/(c-I*d-I)*ln(1-(c-I*d+I)*(1+I*(d*(( 
c+d*cot(b*x+a))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I-c 
))*c*arctan(d*((c+d*cot(b*x+a))/d-c/d)+c)+1/2*d^2*arctan(d*((c+d*cot(b*x+a 
))/d-c/d)+c)^2/(1+I*c+d)+1/4*d^2*polylog(2,(c-I*d+I)*(1+I*(d*((c+d*cot(b*x 
+a))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I-c))/(1+I*c+d 
)+1/2*d*arctan(d*((c+d*cot(b*x+a))/d-c/d)+c)^2/(1+I*c+d)+1/2*d/(c-I*d-I)*c 
*arctan(d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1/4*d*polylog(2,(c-I*d+I)*(1+I*(d* 
((c+d*cot(b*x+a))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I 
-c))/(1+I*c+d)+1/4*d/(c-I*d-I)*polylog(2,(c-I*d+I)*(1+I*(d*((c+d*cot(b*x+a 
))/d-c/d)+c))^2/((d*((c+d*cot(b*x+a))/d-c/d)+c)^2+1)/(I*d+I-c))*c)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 965 vs. \(2 (140) = 280\).

Time = 0.28 (sec) , antiderivative size = 965, normalized size of antiderivative = 4.87 \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\text {Too large to display} \] Input:

integrate(arccot(c+d*cot(b*x+a)),x, algorithm="fricas")
 

Output:

1/8*(8*b*x*arccot(d*cot(b*x + a) + c) + 2*I*a*log(1/2*c^2 + I*c*d - 1/2*d^ 
2 - 1/2*(c^2 + d^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I* 
d + I)*sin(2*b*x + 2*a) + 1/2) - 2*I*a*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2 
*(c^2 + d^2 - 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)* 
sin(2*b*x + 2*a) + 1/2) - 2*I*a*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 
+ d^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2* 
b*x + 2*a) - 1/2) + 2*I*a*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 
- 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 
2*a) - 1/2) - 2*(I*b*x + I*a)*log((c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*c 
os(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) + 2*d + 1) 
/(c^2 + d^2 + 2*d + 1)) - 2*(-I*b*x - I*a)*log((c^2 + d^2 - (c^2 - 2*I*c*d 
 - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a 
) + 2*d + 1)/(c^2 + d^2 + 2*d + 1)) - 2*(-I*b*x - I*a)*log((c^2 + d^2 - (c 
^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*si 
n(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1)) - 2*(I*b*x + I*a)*log((c^ 
2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I* 
d^2 + I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1)) - dilog(-(c^2 
+ d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d 
^2 - I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) - dilog(-(c 
^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d ...
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\text {Timed out} \] Input:

integrate(acot(c+d*cot(b*x+a)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (140) = 280\).

Time = 0.20 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.69 \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\frac {d {\left (\frac {8 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (c^{2} + 1\right )} \tan \left (b x + a\right )}{d}\right )}{d} - \frac {8 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (c^{2} + 1\right )} \tan \left (b x + a\right )}{d}\right ) - 4 \, \arctan \left (\frac {c d + {\left (c^{2} + 1\right )} \tan \left (b x + a\right )}{d}\right ) \arctan \left (\frac {c d + {\left (c^{2} + d + 1\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} + 2 \, d + 1}, -\frac {c d \tan \left (b x + a\right ) - c^{2} - d - 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) + 4 \, \arctan \left (\frac {c d + {\left (c^{2} + 1\right )} \tan \left (b x + a\right )}{d}\right ) \arctan \left (-\frac {c d + {\left (c^{2} - d + 1\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} - 2 \, d + 1}, -\frac {c d \tan \left (b x + a\right ) - c^{2} + d - 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) - {\left (\log \left (\frac {{\left (c^{2} + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - \log \left (\frac {{\left (c^{2} + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + 1}{c^{2} + d^{2} - 2 \, d + 1}\right )\right )} \log \left ({\left (c^{2} + 1\right )} d^{2} + 2 \, {\left (c^{3} + c\right )} d \tan \left (b x + a\right ) + {\left (c^{4} + 2 \, c^{2} + 1\right )} \tan \left (b x + a\right )^{2}\right ) - 2 \, {\rm Li}_2\left (\frac {{\left (i \, c - 1\right )} \tan \left (b x + a\right ) + i \, d}{c + i \, d + i}\right ) + 2 \, {\rm Li}_2\left (\frac {{\left (i \, c + 1\right )} \tan \left (b x + a\right ) + i \, d}{c + i \, d - i}\right ) + 2 \, {\rm Li}_2\left (-\frac {{\left (i \, c - 1\right )} \tan \left (b x + a\right ) + i \, d}{c - i \, d + i}\right ) - 2 \, {\rm Li}_2\left (-\frac {{\left (i \, c + 1\right )} \tan \left (b x + a\right ) + i \, d}{c - i \, d - i}\right )}{d}\right )} + 8 \, {\left (b x + a\right )} \operatorname {arccot}\left (c + \frac {d}{\tan \left (b x + a\right )}\right ) - 8 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (c^{2} + 1\right )} \tan \left (b x + a\right )}{d}\right )}{8 \, b} \] Input:

integrate(arccot(c+d*cot(b*x+a)),x, algorithm="maxima")
 

Output:

1/8*(d*(8*(b*x + a)*arctan((c*d + (c^2 + 1)*tan(b*x + a))/d)/d - (8*(b*x + 
 a)*arctan((c*d + (c^2 + 1)*tan(b*x + a))/d) - 4*arctan((c*d + (c^2 + 1)*t 
an(b*x + a))/d)*arctan2((c*d + (c^2 + d + 1)*tan(b*x + a))/(c^2 + d^2 + 2* 
d + 1), -(c*d*tan(b*x + a) - c^2 - d - 1)/(c^2 + d^2 + 2*d + 1)) + 4*arcta 
n((c*d + (c^2 + 1)*tan(b*x + a))/d)*arctan2(-(c*d + (c^2 - d + 1)*tan(b*x 
+ a))/(c^2 + d^2 - 2*d + 1), -(c*d*tan(b*x + a) - c^2 + d - 1)/(c^2 + d^2 
- 2*d + 1)) - (log(((c^2 + 1)*tan(b*x + a)^2 + c^2 + 1)/(c^2 + d^2 + 2*d + 
 1)) - log(((c^2 + 1)*tan(b*x + a)^2 + c^2 + 1)/(c^2 + d^2 - 2*d + 1)))*lo 
g((c^2 + 1)*d^2 + 2*(c^3 + c)*d*tan(b*x + a) + (c^4 + 2*c^2 + 1)*tan(b*x + 
 a)^2) - 2*dilog(((I*c - 1)*tan(b*x + a) + I*d)/(c + I*d + I)) + 2*dilog(( 
(I*c + 1)*tan(b*x + a) + I*d)/(c + I*d - I)) + 2*dilog(-((I*c - 1)*tan(b*x 
 + a) + I*d)/(c - I*d + I)) - 2*dilog(-((I*c + 1)*tan(b*x + a) + I*d)/(c - 
 I*d - I)))/d) + 8*(b*x + a)*arccot(c + d/tan(b*x + a)) - 8*(b*x + a)*arct 
an((c*d + (c^2 + 1)*tan(b*x + a))/d))/b
 

Giac [F]

\[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\int { \operatorname {arccot}\left (d \cot \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(arccot(c+d*cot(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arccot(d*cot(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\int \mathrm {acot}\left (c+d\,\mathrm {cot}\left (a+b\,x\right )\right ) \,d x \] Input:

int(acot(c + d*cot(a + b*x)),x)
 

Output:

int(acot(c + d*cot(a + b*x)), x)
 

Reduce [F]

\[ \int \cot ^{-1}(c+d \cot (a+b x)) \, dx=\int \mathit {acot} \left (\cot \left (b x +a \right ) d +c \right )d x \] Input:

int(acot(c+d*cot(b*x+a)),x)
 

Output:

int(acot(cot(a + b*x)*d + c),x)