\(\int \frac {\coth ^{-1}(x)^2}{(1-x^2)^2} \, dx\) [55]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6104, 6142, 205, 212} \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {\text {arctanh}(x)}{4}+\frac {x}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3 \]

[In]

Int[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6104

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

Rule 6142

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] + Dist[b*(p/(2*c*(q + 1))), Int[(d + e*x^2)^q*(a + b*ArcCot
h[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]

Rubi steps \begin{align*} \text {integral}& = \frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3-\int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx \\ & = -\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 {1}{2} \int \frac {1}{\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 {1}{4} \int \frac {1}{1-x^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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {-6 x+12 \coth ^{-1}(x)-12 x \coth ^{-1}(x)^2+4 \left (-1+x^2\right ) \coth ^{-1}(x)^3-3 \left (-1+x^2\right ) \log (1-x)+3 \left (-1+x^2\right ) \log (1+x)}{24 \left (-1+x^2\right )} \]

[In]

Integrate[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

(-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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(158\) vs. \(2(52)=104\).

Time = 1.03 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.56

method	result	size
risch	\(-\frac {\ln \left (x -1\right )^{3}}{48}+\frac {\left (x^{2} \ln \left (1+x \right )-2 x -\ln \left (1+x \right )\right ) \ln \left (x -1\right )^{2}}{16 x^{2}-16}-\frac {\left (x^{2} \ln \left (1+x \right )^{2}-4 \ln \left (1+x \right ) x -\ln \left (1+x \right )^{2}+4\right ) \ln \left (x -1\right )}{16 \left (x -1\right ) \left (1+x \right )}+\frac {x^{2} \ln \left (1+x \right )^{3}+6 x^{2} \ln \left (1+x \right )-6 \ln \left (x -1\right ) x^{2}-6 x \ln \left (1+x \right )^{2}-\ln \left (1+x \right )^{3}+6 \ln \left (1+x \right )+6 \ln \left (x -1\right )-12 x}{48 \left (x -1\right ) \left (1+x \right )}\)	\(159\)
default	\(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (1+x \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}{1+x}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{3} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (1+x \right )}{8 x -8}-\frac {1+x}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8+8 x}+\frac {x -1}{16+16 x}\)	\(402\)
parts	\(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (1+x \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}{1+x}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{3} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (1+x \right )}{8 x -8}-\frac {1+x}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8+8 x}+\frac {x -1}{16+16 x}\)	\(402\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{3} - 6 \, x \log \left (\frac {x + 1}{x - 1}\right )^{2} + 6 \, {\left (x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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 = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.76 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\left (x\right )^{2} - \frac {{\left ({\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4\right )} \operatorname {arcoth}\left (x\right )}{8 \, {\left (x^{2} - 1\right )}} + \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{3} - 3 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} \log \left (x - 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{3} + 3 \, {\left ({\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} + 2 \, x^{2} - 2\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

-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)

none

Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2}}{16 \, {\left (x + 1\right )}} - \frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )}{8 \, {\left (x + 1\right )}} - \frac {x - 1}{8 \, {\left (x + 1\right )}} \]

[In]

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

[Out]

-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 = 5.81 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.24 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\ln \left (\frac {1}{x}+1\right )}^3}{48}-\frac {{\ln \left (1-\frac {1}{x}\right )}^3}{48}-\frac {x}{4\,\left (x^2-1\right )}+\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\frac {3\,x}{32}-\frac {1}{8}}{x^2-1}-\frac {\frac {x}{8}+\frac {1}{8}}{x^2-1}-\frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}+\frac {x}{32\,\left (x^2-1\right )}+\ln \left (\frac {1}{x}+1\right )\,\left (\frac {\frac {x}{4}+\frac {1}{16}}{x^2-1}-\frac {1}{16\,\left (x^2-1\right )}\right )\right )+{\ln \left (1-\frac {1}{x}\right )}^2\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{16}-\frac {x}{8\,\left (x^2-1\right )}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )}-\frac {x\,{\ln \left (\frac {1}{x}+1\right )}^2}{8\,\left (x^2-1\right )}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

[In]

int(acoth(x)^2/(x^2 - 1)^2,x)

[Out]

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*l
og(1/x + 1)^2)/(8*(x^2 - 1))