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

Optimal. Leaf size=62 \[ \frac{x}{4 \left (1-x^2\right )}+\frac{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac{\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac{1}{4} \tanh ^{-1}(x)+\frac{1}{6} \coth ^{-1}(x)^3 \]

[Out]

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

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Rubi [A]  time = 0.0506488, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5957, 5995, 199, 206} \[ \frac{x}{4 \left (1-x^2\right )}+\frac{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac{\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac{1}{4} \tanh ^{-1}(x)+\frac{1}{6} \coth ^{-1}(x)^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

x/(4*(1 - x^2)) - ArcCoth[x]/(2*(1 - x^2)) + (x*ArcCoth[x]^2)/(2*(1 - x^2)) + ArcCoth[x]^3/6 + ArcTanh[x]/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 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{\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx &=\frac{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac{1}{6} \coth ^{-1}(x)^3-\int \frac{x \coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx\\ &=-\frac{\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac{1}{6} \coth ^{-1}(x)^3+\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{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac{1}{6} \coth ^{-1}(x)^3+\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{x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac{1}{6} \coth ^{-1}(x)^3+\frac{1}{4} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0831476, size = 61, normalized size = 0.98 \[ \frac{-3 \left (x^2-1\right ) \log (1-x)+3 \left (x^2-1\right ) \log (x+1)+4 \left (x^2-1\right ) \coth ^{-1}(x)^3-6 x-12 x \coth ^{-1}(x)^2+12 \coth ^{-1}(x)}{24 \left (x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

(-6*x + 12*ArcCoth[x] - 12*x*ArcCoth[x]^2 + 4*(-1 + x^2)*ArcCoth[x]^3 - 3*(-1 + x^2)*Log[1 - x] + 3*(-1 + x^2)
*Log[1 + x])/(24*(-1 + x^2))

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Maple [C]  time = 0.494, size = 707, normalized size = 11.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arccoth(x)^2/(-1+x)-1/4*arccoth(x)^2*ln(-1+x)-1/4*arccoth(x)^2/(1+x)+1/4*arccoth(x)^2*ln(1+x)+1/4*arccoth
(x)^2*ln((-1+x)/(1+x))+1/24*(6*I*csgn(I*(1+x)/(-1+x))^2*csgn(I/((-1+x)/(1+x))^(1/2))*arccoth(x)^2*Pi-6*I*arcco
th(x)^2*Pi*csgn(I/((-1+x)/(1+x))^(1/2))*csgn(I*(1+x)/(-1+x))^2*x^2+3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x))^3*
x^2+3*I*csgn(I/((1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^2*arccoth(x)^2*Pi+3*I*arccoth(x)^2*Pi*c
sgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x))*csgn(I/((1+x)/(-1+x)-1))*x^2-3*I*arccoth(x)^2*Pi*csg
n(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^2*csgn(I*(1+x)/(-1+x))*x^2+3*I*arccoth(x)^2*Pi*csgn(I/((-1+x)/(1+x))^(1/2))
^2*csgn(I*(1+x)/(-1+x))*x^2-3*I*csgn(I*(1+x)/(-1+x))^3*arccoth(x)^2*Pi-3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x)
/((1+x)/(-1+x)-1))^2*csgn(I/((1+x)/(-1+x)-1))*x^2-3*I*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^3*arccoth(x)^2*Pi+
3*I*csgn(I*(1+x)/(-1+x))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^2*arccoth(x)^2*Pi+3*I*arccoth(x)^2*Pi*csgn(I*(1
+x)/(-1+x)/((1+x)/(-1+x)-1))^3*x^2-3*I*csgn(I*(1+x)/(-1+x))*csgn(I/((-1+x)/(1+x))^(1/2))^2*arccoth(x)^2*Pi-3*I
*csgn(I/((1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))*arccoth(x)^2*Pi+4*x^2*arc
coth(x)^3-4*arccoth(x)^3+6*arccoth(x)*x^2+6*arccoth(x)-6*x)/(-1+x)/(1+x)

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Maxima [B]  time = 1.02543, size = 231, normalized size = 3.73 \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 )^{2} - \frac{{\left ({\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\right )} \operatorname{arcoth}\left (x\right )}{8 \,{\left (x^{2} - 1\right )}} + \frac{{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{3} - 3 \,{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} \log \left (x - 1\right ) -{\left (x^{2} - 1\right )} \log \left (x - 1\right )^{3} + 3 \,{\left ({\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} + 2 \, x^{2} - 2\right )} \log \left (x + 1\right ) - 6 \,{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 12 \, x}{48 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(x^2 - 1) - log(x + 1) + log(x - 1))*arccoth(x)^2 - 1/8*((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)*arccoth(x)/(x^2 - 1) + 1/48*((x^2 - 1)*log(x + 1)^3 - 3*(x^2 - 1
)*log(x + 1)^2*log(x - 1) - (x^2 - 1)*log(x - 1)^3 + 3*((x^2 - 1)*log(x - 1)^2 + 2*x^2 - 2)*log(x + 1) - 6*(x^
2 - 1)*log(x - 1) - 12*x)/(x^2 - 1)

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Fricas [A]  time = 1.56907, size = 165, normalized size = 2.66 \begin{align*} \frac{{\left (x^{2} - 1\right )} \log \left (\frac{x + 1}{x - 1}\right )^{3} - 6 \, x \log \left (\frac{x + 1}{x - 1}\right )^{2} + 6 \,{\left (x^{2} + 1\right )} \log \left (\frac{x + 1}{x - 1}\right ) - 12 \, x}{48 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((x^2 - 1)*log((x + 1)/(x - 1))^3 - 6*x*log((x + 1)/(x - 1))^2 + 6*(x^2 + 1)*log((x + 1)/(x - 1)) - 12*x)
/(x^2 - 1)

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

[Out]

Integral(acoth(x)**2/((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 )^{2}}{{\left (x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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