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]

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

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.39, size = 29, normalized size = 0.81 \[ -\frac {{\left (x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 2 \, x}{8 \, {\left (x^{2} - 1\right )}} \]

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

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)

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maple [A]  time = 0.04, size = 39, normalized size = 1.08 \[ -\frac {\mathrm {arccoth}\relax (x )}{2 \left (x^{2}-1\right )}+\frac {1}{-8+8 x}+\frac {\ln \left (-1+x \right )}{8}+\frac {1}{8+8 x}-\frac {\ln \left (1+x \right )}{8} \]

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 = 0.30, size = 34, normalized size = 0.94 \[ \frac {x}{4 \, {\left (x^{2} - 1\right )}} - \frac {\operatorname {arcoth}\relax (x)}{2 \, {\left (x^{2} - 1\right )}} - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \]

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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mupad [B]  time = 1.16, size = 21, normalized size = 0.58 \[ \frac {\frac {x}{4}-\frac {\mathrm {acoth}\relax (x)}{2}}{x^2-1}-\frac {\mathrm {acoth}\relax (x)}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*acoth(x))/(x^2 - 1)^2,x)

[Out]

(x/4 - acoth(x)/2)/(x^2 - 1) - acoth(x)/4

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sympy [A]  time = 0.56, size = 31, normalized size = 0.86 \[ - \frac {x^{2} \operatorname {acoth}{\relax (x )}}{4 x^{2} - 4} + \frac {x}{4 x^{2} - 4} - \frac {\operatorname {acoth}{\relax (x )}}{4 x^{2} - 4} \]

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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