\(\int \frac {\coth ^{-1}(x)^2}{(1-x^2)^2} \, dx\) [20]

Optimal result

Integrand size = 14, antiderivative size = 62 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {\text {arctanh}(x)}{4} \] Output:

x/(-4*x^2+4)-arccoth(x)/(-2*x^2+2)+x*arccoth(x)^2/(-2*x^2+2)+1/6*arccoth(x 
)^3+1/4*arctanh(x)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {-6 x+12 \coth ^{-1}(x)-12 x \coth ^{-1}(x)^2+4 \left (-1+x^2\right ) \coth ^{-1}(x)^3-3 \left (-1+x^2\right ) \log (1-x)+3 \left (-1+x^2\right ) \log (1+x)}{24 \left (-1+x^2\right )} \] Input:

Integrate[ArcCoth[x]^2/(1 - x^2)^2,x]
 

Output:

(-6*x + 12*ArcCoth[x] - 12*x*ArcCoth[x]^2 + 4*(-1 + x^2)*ArcCoth[x]^3 - 3* 
(-1 + x^2)*Log[1 - x] + 3*(-1 + x^2)*Log[1 + x])/(24*(-1 + x^2))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6519, 6557, 215, 219}

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.

\(\displaystyle \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx\)

\(\Big \downarrow \) 6519

\(\displaystyle -\int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^2}dx+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3\)

\(\Big \downarrow \) 6557

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-x^2\right )^2}dx+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-x^2}dx+\frac {x}{2 \left (1-x^2\right )}\right )+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\text {arctanh}(x)}{2}+\frac {x}{2 \left (1-x^2\right )}\right )+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3\)

Input:

Int[ArcCoth[x]^2/(1 - x^2)^2,x]
 

Output:

-1/2*ArcCoth[x]/(1 - x^2) + (x*ArcCoth[x]^2)/(2*(1 - x^2)) + ArcCoth[x]^3/ 
6 + (x/(2*(1 - x^2)) + ArcTanh[x]/2)/2
 

Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6519

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Sy 
mbol] :> Simp[x*((a + b*ArcCoth[c*x])^p/(2*d*(d + e*x^2))), x] + (Simp[(a + 
 b*ArcCoth[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x] - Simp[b*c*(p/2)   Int[x*( 
(a + b*ArcCoth[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
 

rule 6557

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q 
_.), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcCoth[c*x])^p/(2*e*(q 
+ 1))), x] + Simp[b*(p/(2*c*(q + 1)))   Int[(d + e*x^2)^q*(a + b*ArcCoth[c* 
x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && 
 GtQ[p, 0] && NeQ[q, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(157\) vs. \(2(50)=100\).

Time = 1.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.55

method	result	size
risch	\(\frac {\ln \left (x +1\right )^{3}}{48}-\frac {\left (x^{2} \ln \left (x -1\right )+2 x -\ln \left (x -1\right )\right ) \ln \left (x +1\right )^{2}}{16 \left (x^{2}-1\right )}+\frac {\left (x^{2} \ln \left (x -1\right )^{2}+4 \ln \left (x -1\right ) x -\ln \left (x -1\right )^{2}+4\right ) \ln \left (x +1\right )}{16 \left (x -1\right ) \left (x +1\right )}+\frac {-x^{2} \ln \left (x -1\right )^{3}+6 \ln \left (x +1\right ) x^{2}-6 x^{2} \ln \left (x -1\right )-6 x \ln \left (x -1\right )^{2}+\ln \left (x -1\right )^{3}-6 \ln \left (x +1\right )-6 \ln \left (x -1\right )-12 x}{48 \left (x -1\right ) \left (x +1\right )}\)	\(158\)
default	\(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x +1\right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x +1\right )}{4}-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x -1\right )}{4}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (\frac {x -1}{x +1}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right ) \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {x +1}{x -1}-1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{x +1}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{x +1}}}\right ) \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {x +1}{x -1}-1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right )^{3} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (x +1\right )}{8 x -8}-\frac {x +1}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8 x +8}+\frac {x -1}{16 x +16}\)	\(402\)
parts	\(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x +1\right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x +1\right )}{4}-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x -1\right )}{4}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (\frac {x -1}{x +1}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right ) \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {x +1}{x -1}-1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{x +1}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{x +1}}}\right ) \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{\left (x -1\right ) \left (\frac {x +1}{x -1}-1\right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {x +1}{x -1}-1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (x +1\right )}{x -1}\right )^{3} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (x +1\right )}{8 x -8}-\frac {x +1}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8 x +8}+\frac {x -1}{16 x +16}\)	\(402\)

Input:

int(arccoth(x)^2/(-x^2+1)^2,x,method=_RETURNVERBOSE)
 

Output:

1/48*ln(x+1)^3-1/16*(x^2*ln(x-1)+2*x-ln(x-1))/(x^2-1)*ln(x+1)^2+1/16*(x^2* 
ln(x-1)^2+4*ln(x-1)*x-ln(x-1)^2+4)/(x-1)/(x+1)*ln(x+1)+1/48*(-x^2*ln(x-1)^ 
3+6*ln(x+1)*x^2-6*x^2*ln(x-1)-6*x*ln(x-1)^2+ln(x-1)^3-6*ln(x+1)-6*ln(x-1)- 
12*x)/(x-1)/(x+1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{3} - 6 \, x \log \left (\frac {x + 1}{x - 1}\right )^{2} + 6 \, {\left (x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \] Input:

integrate(arccoth(x)^2/(-x^2+1)^2,x, algorithm="fricas")
 

Output:

1/48*((x^2 - 1)*log((x + 1)/(x - 1))^3 - 6*x*log((x + 1)/(x - 1))^2 + 6*(x 
^2 + 1)*log((x + 1)/(x - 1)) - 12*x)/(x^2 - 1)
 

Sympy [F]

\[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\int \frac {\operatorname {acoth}^{2}{\left (x \right )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \] Input:

integrate(acoth(x)**2/(-x**2+1)**2,x)
 

Output:

Integral(acoth(x)**2/((x - 1)**2*(x + 1)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (46) = 92\).

Time = 0.03 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.76 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\left (x\right )^{2} - \frac {{\left ({\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4\right )} \operatorname {arcoth}\left (x\right )}{8 \, {\left (x^{2} - 1\right )}} + \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{3} - 3 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} \log \left (x - 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{3} + 3 \, {\left ({\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} + 2 \, x^{2} - 2\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \] Input:

integrate(arccoth(x)^2/(-x^2+1)^2,x, algorithm="maxima")
 

Output:

-1/4*(2*x/(x^2 - 1) - log(x + 1) + log(x - 1))*arccoth(x)^2 - 1/8*((x^2 - 
1)*log(x + 1)^2 - 2*(x^2 - 1)*log(x + 1)*log(x - 1) + (x^2 - 1)*log(x - 1) 
^2 - 4)*arccoth(x)/(x^2 - 1) + 1/48*((x^2 - 1)*log(x + 1)^3 - 3*(x^2 - 1)* 
log(x + 1)^2*log(x - 1) - (x^2 - 1)*log(x - 1)^3 + 3*((x^2 - 1)*log(x - 1) 
^2 + 2*x^2 - 2)*log(x + 1) - 6*(x^2 - 1)*log(x - 1) - 12*x)/(x^2 - 1)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2}}{16 \, {\left (x + 1\right )}} - \frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )}{8 \, {\left (x + 1\right )}} - \frac {x - 1}{8 \, {\left (x + 1\right )}} \] Input:

integrate(arccoth(x)^2/(-x^2+1)^2,x, algorithm="giac")
 

Output:

-1/16*(x - 1)*log((x + 1)/(x - 1))^2/(x + 1) - 1/8*(x - 1)*log((x + 1)/(x 
- 1))/(x + 1) - 1/8*(x - 1)/(x + 1)
 

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.24 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\ln \left (\frac {1}{x}+1\right )}^3}{48}-\frac {{\ln \left (1-\frac {1}{x}\right )}^3}{48}-\frac {x}{4\,\left (x^2-1\right )}+\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\frac {3\,x}{32}-\frac {1}{8}}{x^2-1}-\frac {\frac {x}{8}+\frac {1}{8}}{x^2-1}-\frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}+\frac {x}{32\,\left (x^2-1\right )}+\ln \left (\frac {1}{x}+1\right )\,\left (\frac {\frac {x}{4}+\frac {1}{16}}{x^2-1}-\frac {1}{16\,\left (x^2-1\right )}\right )\right )+{\ln \left (1-\frac {1}{x}\right )}^2\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{16}-\frac {x}{8\,\left (x^2-1\right )}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )}-\frac {x\,{\ln \left (\frac {1}{x}+1\right )}^2}{8\,\left (x^2-1\right )}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \] Input:

int(acoth(x)^2/(x^2 - 1)^2,x)
 

Output:

log(1/x + 1)^3/48 - (atan(x*1i)*1i)/4 - log(1 - 1/x)^3/48 - x/(4*(x^2 - 1) 
) + log(1 - 1/x)*(((3*x)/32 - 1/8)/(x^2 - 1) - (x/8 + 1/8)/(x^2 - 1) - log 
(1/x + 1)^2/16 + x/(32*(x^2 - 1)) + log(1/x + 1)*((x/4 + 1/16)/(x^2 - 1) - 
 1/(16*(x^2 - 1)))) + log(1 - 1/x)^2*(log(1/x + 1)/16 - x/(8*(x^2 - 1))) + 
 log(1/x + 1)/(4*(x^2 - 1)) - (x*log(1/x + 1)^2)/(8*(x^2 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {-4 \mathit {acoth} \left (x \right )^{3} x^{2}+4 \mathit {acoth} \left (x \right )^{3}-12 \mathit {acoth} \left (x \right )^{2} x -12 \mathit {acoth} \left (x \right ) x^{2}+3 \,\mathrm {log}\left (x -1\right ) x^{2}-3 \,\mathrm {log}\left (x -1\right )-3 \,\mathrm {log}\left (x +1\right ) x^{2}+3 \,\mathrm {log}\left (x +1\right )-6 x}{24 x^{2}-24} \] Input:

int(acoth(x)^2/(-x^2+1)^2,x)
 

Output:

( - 4*acoth(x)**3*x**2 + 4*acoth(x)**3 - 12*acoth(x)**2*x - 12*acoth(x)*x* 
*2 + 3*log(x - 1)*x**2 - 3*log(x - 1) - 3*log(x + 1)*x**2 + 3*log(x + 1) - 
 6*x)/(24*(x**2 - 1))