\(\int \frac {\coth ^{-1}(x)}{(1-x^2)^2} \, dx\) [59]

Optimal result

Integrand size = 12, antiderivative size = 38 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \coth ^{-1}(x)^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6104, 267} \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \coth ^{-1}(x)^2 \]

[In]

Int[ArcCoth[x]/(1 - x^2)^2,x]

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=\frac {1-2 x \coth ^{-1}(x)+\left (-1+x^2\right ) \coth ^{-1}(x)^2}{4 \left (-1+x^2\right )} \]

[In]

Integrate[ArcCoth[x]/(1 - x^2)^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(79\) vs. \(2(32)=64\).

Time = 0.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.11

method	result	size
risch	\(\frac {\ln \left (x -1\right )^{2}}{16}-\frac {\left (x^{2} \ln \left (1+x \right )-2 x -\ln \left (1+x \right )\right ) \ln \left (x -1\right )}{8 \left (x^{2}-1\right )}+\frac {x^{2} \ln \left (1+x \right )^{2}-4 \ln \left (1+x \right ) x -\ln \left (1+x \right )^{2}+4}{16 \left (x -1\right ) \left (1+x \right )}\)	\(80\)
default	\(-\frac {\operatorname {arccoth}\left (x \right )}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right ) \ln \left (1+x \right )}{4}-\frac {\operatorname {arccoth}\left (x \right )}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right ) \ln \left (x -1\right )}{4}-\frac {\ln \left (1+x \right )^{2}}{16}+\frac {\left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{2}\right )}{8}+\frac {\ln \left (x -1\right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{8}-\frac {\ln \left (x -1\right )^{2}}{16}+\frac {1}{8 x -8}-\frac {1}{8 \left (1+x \right )}\)	\(99\)
parts	\(-\frac {\operatorname {arccoth}\left (x \right )}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right ) \ln \left (1+x \right )}{4}-\frac {\operatorname {arccoth}\left (x \right )}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right ) \ln \left (x -1\right )}{4}-\frac {\ln \left (1+x \right )^{2}}{16}+\frac {\left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{2}\right )}{8}+\frac {\ln \left (x -1\right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{8}-\frac {\ln \left (x -1\right )^{2}}{16}+\frac {1}{8 x -8}-\frac {1}{8 \left (1+x \right )}\)	\(99\)

[In]

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

[Out]

1/16*ln(x-1)^2-1/8*(x^2*ln(1+x)-2*x-ln(1+x))/(x^2-1)*ln(x-1)+1/16*(x^2*ln(1+x)^2-4*ln(1+x)*x-ln(1+x)^2+4)/(x-1
)/(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2} - 4 \, x \log \left (\frac {x + 1}{x - 1}\right ) + 4}{16 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

1/16*((x^2 - 1)*log((x + 1)/(x - 1))^2 - 4*x*log((x + 1)/(x - 1)) + 4)/(x^2 - 1)

Sympy [F]

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

[In]

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

[Out]

Integral(acoth(x)/((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. 76 vs. \(2 (28) = 56\).

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.00 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\left (x\right ) - \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4}{16 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

-1/4*(2*x/(x^2 - 1) - log(x + 1) + log(x - 1))*arccoth(x) - 1/16*((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)/(x^2 - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (28) = 56\).

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.11 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=-\frac {{\left (x - 1\right )} \log \left (-\frac {\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} + 1}{\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} - 1}\right )}{8 \, {\left (x + 1\right )}} - \frac {x - 1}{8 \, {\left (x + 1\right )}} \]

[In]

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

[Out]

-1/8*(x - 1)*log(-(((x + 1)/(x - 1) - 1)/((x + 1)/(x - 1) + 1) + 1)/(((x + 1)/(x - 1) - 1)/((x + 1)/(x - 1) +
1) - 1))/(x + 1) - 1/8*(x - 1)/(x + 1)

Mupad [B] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.13 \[ \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx=\frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}-\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{8}-\frac {x}{4\,\left (x^2-1\right )}\right )+\frac {{\ln \left (1-\frac {1}{x}\right )}^2}{16}+\frac {1}{4\,\left (x^2-1\right )}-\frac {x\,\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )} \]

[In]

int(acoth(x)/(x^2 - 1)^2,x)

[Out]

log(1/x + 1)^2/16 - log(1 - 1/x)*(log(1/x + 1)/8 - x/(4*(x^2 - 1))) + log(1 - 1/x)^2/16 + 1/(4*(x^2 - 1)) - (x
*log(1/x + 1))/(4*(x^2 - 1))