3.60 \(\int \frac {x \coth ^{-1}(x)}{(1-x^2)^3} \, dx\)

Optimal. Leaf size=50 \[ -\frac {3 x}{32 \left (1-x^2\right )}-\frac {x}{16 \left (1-x^2\right )^2}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{32} \tanh ^{-1}(x) \]

[Out]

-1/16*x/(-x^2+1)^2-3/32*x/(-x^2+1)+1/4*arccoth(x)/(-x^2+1)^2-3/32*arctanh(x)

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5995, 199, 206} \[ -\frac {3 x}{32 \left (1-x^2\right )}-\frac {x}{16 \left (1-x^2\right )^2}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{32} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCoth[x])/(1 - x^2)^3,x]

[Out]

-x/(16*(1 - x^2)^2) - (3*x)/(32*(1 - x^2)) + ArcCoth[x]/(4*(1 - x^2)^2) - (3*ArcTanh[x])/32

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 )^3} \, dx &=\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {1}{4} \int \frac {1}{\left (1-x^2\right )^3} \, dx\\ &=-\frac {x}{16 \left (1-x^2\right )^2}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{16} \int \frac {1}{\left (1-x^2\right )^2} \, dx\\ &=-\frac {x}{16 \left (1-x^2\right )^2}-\frac {3 x}{32 \left (1-x^2\right )}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{32} \int \frac {1}{1-x^2} \, dx\\ &=-\frac {x}{16 \left (1-x^2\right )^2}-\frac {3 x}{32 \left (1-x^2\right )}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{32} \tanh ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 50, normalized size = 1.00 \[ \frac {1}{64} \left (\frac {6 x}{x^2-1}-\frac {4 x}{\left (x^2-1\right )^2}+\frac {16 \coth ^{-1}(x)}{\left (x^2-1\right )^2}+3 \log (1-x)-3 \log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcCoth[x])/(1 - x^2)^3,x]

[Out]

((-4*x)/(-1 + x^2)^2 + (6*x)/(-1 + x^2) + (16*ArcCoth[x])/(-1 + x^2)^2 + 3*Log[1 - x] - 3*Log[1 + x])/64

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fricas [A]  time = 0.61, size = 47, normalized size = 0.94 \[ \frac {6 \, x^{3} - {\left (3 \, x^{4} - 6 \, x^{2} - 5\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 10 \, x}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="fricas")

[Out]

1/64*(6*x^3 - (3*x^4 - 6*x^2 - 5)*log((x + 1)/(x - 1)) - 10*x)/(x^4 - 2*x^2 + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x \operatorname {arcoth}\relax (x)}{{\left (x^{2} - 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="giac")

[Out]

integrate(-x*arccoth(x)/(x^2 - 1)^3, x)

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maple [A]  time = 0.04, size = 53, normalized size = 1.06 \[ \frac {\mathrm {arccoth}\relax (x )}{4 \left (x^{2}-1\right )^{2}}-\frac {1}{64 \left (-1+x \right )^{2}}+\frac {3}{64 \left (-1+x \right )}+\frac {3 \ln \left (-1+x \right )}{64}+\frac {1}{64 \left (1+x \right )^{2}}+\frac {3}{64 \left (1+x \right )}-\frac {3 \ln \left (1+x \right )}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(x)/(-x^2+1)^3,x)

[Out]

1/4/(x^2-1)^2*arccoth(x)-1/64/(-1+x)^2+3/64/(-1+x)+3/64*ln(-1+x)+1/64/(1+x)^2+3/64/(1+x)-3/64*ln(1+x)

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maxima [A]  time = 0.30, size = 47, normalized size = 0.94 \[ \frac {3 \, x^{3} - 5 \, x}{32 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} + \frac {\operatorname {arcoth}\relax (x)}{4 \, {\left (x^{2} - 1\right )}^{2}} - \frac {3}{64} \, \log \left (x + 1\right ) + \frac {3}{64} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="maxima")

[Out]

1/32*(3*x^3 - 5*x)/(x^4 - 2*x^2 + 1) + 1/4*arccoth(x)/(x^2 - 1)^2 - 3/64*log(x + 1) + 3/64*log(x - 1)

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mupad [B]  time = 1.21, size = 34, normalized size = 0.68 \[ \frac {3\,\ln \left (x-1\right )}{64}-\frac {3\,\ln \left (x+1\right )}{64}+\frac {\frac {\mathrm {acoth}\relax (x)}{4}-\frac {5\,x}{32}+\frac {3\,x^3}{32}}{{\left (x^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*acoth(x))/(x^2 - 1)^3,x)

[Out]

(3*log(x - 1))/64 - (3*log(x + 1))/64 + (acoth(x)/4 - (5*x)/32 + (3*x^3)/32)/(x^2 - 1)^2

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sympy [B]  time = 0.91, size = 88, normalized size = 1.76 \[ - \frac {3 x^{4} \operatorname {acoth}{\relax (x )}}{32 x^{4} - 64 x^{2} + 32} + \frac {3 x^{3}}{32 x^{4} - 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {acoth}{\relax (x )}}{32 x^{4} - 64 x^{2} + 32} - \frac {5 x}{32 x^{4} - 64 x^{2} + 32} + \frac {5 \operatorname {acoth}{\relax (x )}}{32 x^{4} - 64 x^{2} + 32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(x)/(-x**2+1)**3,x)

[Out]

-3*x**4*acoth(x)/(32*x**4 - 64*x**2 + 32) + 3*x**3/(32*x**4 - 64*x**2 + 32) + 6*x**2*acoth(x)/(32*x**4 - 64*x*
*2 + 32) - 5*x/(32*x**4 - 64*x**2 + 32) + 5*acoth(x)/(32*x**4 - 64*x**2 + 32)

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