3.1.59 \(\int \frac {\coth ^{-1}(x)}{(1-x^2)^2} \, dx\) [59]

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

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Rubi [A]
time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 = {6104, 267} \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/(1 - x^2)^2,x]

[Out]

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

Rule 267

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]

Rule 6104

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]

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*}

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Mathematica [A]
time = 0.01, size = 28, normalized size = 0.74 \begin {gather*} \frac {1-2 x \coth ^{-1}(x)+\left (-1+x^2\right ) \coth ^{-1}(x)^2}{4 \left (-1+x^2\right )} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(98\) vs. \(2(32)=64\).
time = 0.11, size = 99, normalized size = 2.61

method	result	size
risch	\(\frac {\ln \left (1+x \right )^{2}}{16}-\frac {\left (\ln \left (-1+x \right ) x^{2}+2 x -\ln \left (-1+x \right )\right ) \ln \left (1+x \right )}{8 \left (x^{2}-1\right )}+\frac {x^{2} \ln \left (-1+x \right )^{2}+4 \ln \left (-1+x \right ) x -\ln \left (-1+x \right )^{2}+4}{16 \left (-1+x \right ) \left (1+x \right )}\)	\(80\)
default	\(-\frac {\mathrm {arccoth}\left (x \right )}{4 \left (1+x \right )}+\frac {\mathrm {arccoth}\left (x \right ) \ln \left (1+x \right )}{4}-\frac {\mathrm {arccoth}\left (x \right )}{4 \left (-1+x \right )}-\frac {\mathrm {arccoth}\left (x \right ) \ln \left (-1+x \right )}{4}+\frac {\left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{2}\right )}{8}-\frac {\ln \left (1+x \right )^{2}}{16}+\frac {\ln \left (-1+x \right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{8}-\frac {\ln \left (-1+x \right )^{2}}{16}-\frac {1}{8 \left (1+x \right )}+\frac {1}{-8+8 x}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (28) = 56\).
time = 0.25, size = 76, normalized size = 2.00 \begin {gather*} -\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 {gather*}

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 = 0.36, size = 42, normalized size = 1.11 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acoth}{\left (x \right )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (28) = 56\).
time = 0.42, size = 80, normalized size = 2.11 \begin {gather*} -\frac {{\left (x - 1\right )} \log \left (-\frac {\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} + 1}{\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} - 1}\right )}{8 \, {\left (x + 1\right )}} - \frac {x - 1}{8 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.22, size = 81, normalized size = 2.13 \begin {gather*} \frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}-\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{8}-\frac {x}{4\,\left (x^2-1\right )}\right )+\frac {{\ln \left (1-\frac {1}{x}\right )}^2}{16}+\frac {1}{4\,\left (x^2-1\right )}-\frac {x\,\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(x)/(x^2 - 1)^2,x)

[Out]

log(1/x + 1)^2/16 - log(1 - 1/x)*(log(1/x + 1)/8 - x/(4*(x^2 - 1))) + log(1 - 1/x)^2/16 + 1/(4*(x^2 - 1)) - (x
*log(1/x + 1))/(4*(x^2 - 1))

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