\(\int \frac {\cot ^{-1}(a+b x)}{\sqrt {(1+a^2) c+2 a b c x+b^2 c x^2}} \, dx\) [46]

Optimal result

Integrand size = 33, antiderivative size = 216 \[ \int \frac {\cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}} \] Output:

-2*I*(1+(b*x+a)^2)^(1/2)*arccot(b*x+a)*arctan((1+I*(b*x+a))^(1/2)/(1-I*(b* 
x+a))^(1/2))/b/(c+c*(b*x+a)^2)^(1/2)-I*(1+(b*x+a)^2)^(1/2)*polylog(2,-I*(1 
+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b/(c+c*(b*x+a)^2)^(1/2)+I*(1+(b*x+a 
)^2)^(1/2)*polylog(2,I*(1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b/(c+c*(b* 
x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.64 \[ \int \frac {\cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {\left (1+(a+b x)^2\right ) \left (\cot ^{-1}(a+b x) \left (\log \left (1-e^{i \cot ^{-1}(a+b x)}\right )-\log \left (1+e^{i \cot ^{-1}(a+b x)}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i \cot ^{-1}(a+b x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \cot ^{-1}(a+b x)}\right )\right )}{b (a+b x) \sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )} \sqrt {1+\frac {1}{(a+b x)^2}}} \] Input:

Integrate[ArcCot[a + b*x]/Sqrt[(1 + a^2)*c + 2*a*b*c*x + b^2*c*x^2],x]
 

Output:

-(((1 + (a + b*x)^2)*(ArcCot[a + b*x]*(Log[1 - E^(I*ArcCot[a + b*x])] - Lo 
g[1 + E^(I*ArcCot[a + b*x])]) + I*PolyLog[2, -E^(I*ArcCot[a + b*x])] - I*P 
olyLog[2, E^(I*ArcCot[a + b*x])]))/(b*(a + b*x)*Sqrt[c*(1 + a^2 + 2*a*b*x 
+ b^2*x^2)]*Sqrt[1 + (a + b*x)^(-2)]))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.72, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5579, 5426, 5422}

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.

\(\displaystyle \int \frac {\cot ^{-1}(a+b x)}{\sqrt {\left (a^2+1\right ) c+2 a b c x+b^2 c x^2}} \, dx\)

\(\Big \downarrow \) 5579

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

\(\Big \downarrow \) 5426

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

\(\Big \downarrow \) 5422

\(\displaystyle \frac {\sqrt {(a+b x)^2+1} \left (-2 i \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)-i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )+i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )\right )}{b \sqrt {c (a+b x)^2+c}}\)

Input:

Int[ArcCot[a + b*x]/Sqrt[(1 + a^2)*c + 2*a*b*c*x + b^2*c*x^2],x]
 

Output:

(Sqrt[1 + (a + b*x)^2]*((-2*I)*ArcCot[a + b*x]*ArcTan[Sqrt[1 + I*(a + b*x) 
]/Sqrt[1 - I*(a + b*x)]] - I*PolyLog[2, ((-I)*Sqrt[1 + I*(a + b*x)])/Sqrt[ 
1 - I*(a + b*x)]] + I*PolyLog[2, (I*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + 
 b*x)]]))/(b*Sqrt[c + c*(a + b*x)^2])
 

Defintions of rubi rules used

rule 5422

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] 
 :> Simp[-2*I*(a + b*ArcCot[c*x])*(ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/ 
(c*Sqrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1 - I* 
c*x])]/(c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - 
I*c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && 
 GtQ[d, 0]
 

rule 5426

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]   Int[(a + b*ArcCot[c*x])^ 
p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] & 
& IGtQ[p, 0] &&  !GtQ[d, 0]
 

rule 5579

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((A_.) + (B_.)*(x_) + ( 
C_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/d   Subst[Int[(C/d^2 + (C/d^2)*x^2) 
^q*(a + b*ArcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B, C, 
p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
 

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {i \left (i \operatorname {arccot}\left (b x +a \right ) \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {arccot}\left (b x +a \right ) \ln \left (\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}+1\right )+\operatorname {polylog}\left (2, \frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\operatorname {polylog}\left (2, -\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\)	\(156\)

Input:

int(arccot(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(1/2),x,method=_RETURNVE 
RBOSE)
 

Output:

I*(I*arccot(b*x+a)*ln(1-(I+a+b*x)/(1+(b*x+a)^2)^(1/2))-I*arccot(b*x+a)*ln( 
(I+a+b*x)/(1+(b*x+a)^2)^(1/2)+1)+polylog(2,(I+a+b*x)/(1+(b*x+a)^2)^(1/2))- 
polylog(2,-(I+a+b*x)/(1+(b*x+a)^2)^(1/2)))*(c*(b*x+a-I)*(I+a+b*x))^(1/2)/( 
b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b/c
 

Fricas [F]

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

integrate(arccot(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(1/2),x, algorithm 
="fricas")
 

Output:

integral(arccot(b*x + a)/sqrt(b^2*c*x^2 + 2*a*b*c*x + (a^2 + 1)*c), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\text {Timed out} \] Input:

integrate(acot(b*x+a)/((a**2+1)*c+2*a*b*c*x+b**2*c*x**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(arccot(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(1/2),x, algorithm 
="maxima")
 

Output:

integrate(arccot(b*x + a)/sqrt(b^2*c*x^2 + 2*a*b*c*x + (a^2 + 1)*c), x)
 

Giac [F]

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

integrate(arccot(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(1/2),x, algorithm 
="giac")
 

Output:

integrate(arccot(b*x + a)/sqrt(b^2*c*x^2 + 2*a*b*c*x + (a^2 + 1)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int \frac {\mathrm {acot}\left (a+b\,x\right )}{\sqrt {c\,b^2\,x^2+2\,a\,c\,b\,x+c\,\left (a^2+1\right )}} \,d x \] Input:

int(acot(a + b*x)/(c*(a^2 + 1) + b^2*c*x^2 + 2*a*b*c*x)^(1/2),x)
 

Output:

int(acot(a + b*x)/(c*(a^2 + 1) + b^2*c*x^2 + 2*a*b*c*x)^(1/2), x)
 

Reduce [F]

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

int(acot(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(1/2),x)
 

Output:

int(acot(a + b*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)/sqrt(c)