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

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Rubi [A]  time = 0.0174931, antiderivative size = 38, normalized size of antiderivative = 1., 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 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]

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]

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 = 0.0323973, 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))

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Maple [B]  time = 0.043, size = 99, normalized size = 2.6 \begin{align*} -{\frac{{\rm arccoth} \left (x\right )}{-4+4\,x}}-{\frac{{\rm arccoth} \left (x\right )\ln \left ( -1+x \right ) }{4}}-{\frac{{\rm arccoth} \left (x\right )}{4+4\,x}}+{\frac{{\rm arccoth} \left (x\right )\ln \left ( 1+x \right ) }{4}}+{\frac{1}{8} \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 ) }-{\frac{ \left ( \ln \left ( 1+x \right ) \right ) ^{2}}{16}}-{\frac{ \left ( \ln \left ( -1+x \right ) \right ) ^{2}}{16}}+{\frac{\ln \left ( -1+x \right ) }{8}\ln \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) }+{\frac{1}{-8+8\,x}}-{\frac{1}{8+8\,x}} \end{align*}

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

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Maxima [B]  time = 0.967018, size = 103, normalized size = 2.71 \begin{align*} -\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 )}} \end{align*}

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)

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Fricas [A]  time = 1.58369, size = 111, normalized size = 2.92 \begin{align*} \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 )}} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}{\left (x \right )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \end{align*}

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (x\right )}{{\left (x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

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)