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

Optimal. Leaf size=36 \[ -\frac{x}{4 \left (1-x^2\right )}+\frac{\coth ^{-1}(x)}{2 \left (1-x^2\right )}-\frac{1}{4} \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0307194, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5995, 199, 206} \[ -\frac{x}{4 \left (1-x^2\right )}+\frac{\coth ^{-1}(x)}{2 \left (1-x^2\right )}-\frac{1}{4} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCoth[x])/(1 - x^2)^2,x]

[Out]

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

Rule 5995

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]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.027431, size = 44, normalized size = 1.22 \[ \frac{x}{4 \left (x^2-1\right )}-\frac{\coth ^{-1}(x)}{2 \left (x^2-1\right )}+\frac{1}{8} \log (1-x)-\frac{1}{8} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*(-1 + x^2)) - ArcCoth[x]/(2*(-1 + x^2)) + Log[1 - x]/8 - Log[1 + x]/8

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Maple [A]  time = 0.027, size = 39, normalized size = 1.1 \begin{align*} -{\frac{{\rm arccoth} \left (x\right )}{2\,{x}^{2}-2}}+{\frac{1}{-8+8\,x}}+{\frac{\ln \left ( -1+x \right ) }{8}}+{\frac{1}{8+8\,x}}-{\frac{\ln \left ( 1+x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.946376, size = 46, normalized size = 1.28 \begin{align*} \frac{x}{4 \,{\left (x^{2} - 1\right )}} - \frac{\operatorname{arcoth}\left (x\right )}{2 \,{\left (x^{2} - 1\right )}} - \frac{1}{8} \, \log \left (x + 1\right ) + \frac{1}{8} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58823, size = 74, normalized size = 2.06 \begin{align*} -\frac{{\left (x^{2} + 1\right )} \log \left (\frac{x + 1}{x - 1}\right ) - 2 \, x}{8 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.819492, size = 31, normalized size = 0.86 \begin{align*} - \frac{x^{2} \operatorname{acoth}{\left (x \right )}}{4 x^{2} - 4} + \frac{x}{4 x^{2} - 4} - \frac{\operatorname{acoth}{\left (x \right )}}{4 x^{2} - 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*acoth(x)/(4*x**2 - 4) + x/(4*x**2 - 4) - acoth(x)/(4*x**2 - 4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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(x*arccoth(x)/(-x^2+1)^2,x, algorithm="giac")

[Out]

integrate(x*arccoth(x)/(x^2 - 1)^2, x)