3.59 \(\int \frac {\coth ^{-1}(x)}{(1-x^2)^2} \, dx\)

Optimal. Leaf size=38 \[ -\frac {1}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \coth ^{-1}(x)^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5957, 261} \[ -\frac {1}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \coth ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/(1 - x^2)^2,x]

[Out]

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

Rule 261

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 5957

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*} \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx &=\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]  time = 0.03, size = 28, normalized size = 0.74 \[ \frac {\left (x^2-1\right ) \coth ^{-1}(x)^2-2 x \coth ^{-1}(x)+1}{4 \left (x^2-1\right )} \]

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 42, normalized size = 1.11 \[ \frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2} - 4 \, x \log \left (\frac {x + 1}{x - 1}\right ) + 4}{16 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\relax (x)}{{\left (x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(x)/(x^2 - 1)^2, x)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 99, normalized size = 2.61 \[ -\frac {\mathrm {arccoth}\relax (x )}{4 \left (-1+x \right )}-\frac {\mathrm {arccoth}\relax (x ) \ln \left (-1+x \right )}{4}-\frac {\mathrm {arccoth}\relax (x )}{4 \left (1+x \right )}+\frac {\mathrm {arccoth}\relax (x ) \ln \left (1+x \right )}{4}-\frac {\ln \left (1+x \right )^{2}}{16}+\frac {\left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (-\frac {x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (-1+x \right )^{2}}{16}+\frac {\ln \left (-1+x \right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{8}+\frac {1}{-8+8 x}-\frac {1}{8 \left (1+x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.30, size = 76, normalized size = 2.00 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\relax (x) - \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4}{16 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 1.22, size = 81, normalized size = 2.13 \[ \frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}-\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{8}-\frac {x}{4\,\left (x^2-1\right )}\right )+\frac {{\ln \left (1-\frac {1}{x}\right )}^2}{16}+\frac {1}{4\,\left (x^2-1\right )}-\frac {x\,\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}{\relax (x )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________