3.1.61 \(\int \frac {\coth ^{-1}(x)}{(1-x^2)^3} \, dx\) [61]

Optimal. Leaf size=67 \[ -\frac {1}{16 \left (1-x^2\right )^2}-\frac {3}{16 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3 x \coth ^{-1}(x)}{8 \left (1-x^2\right )}+\frac {3}{16} \coth ^{-1}(x)^2 \]

[Out]

-1/16/(-x^2+1)^2-3/16/(-x^2+1)+1/4*x*arccoth(x)/(-x^2+1)^2+3/8*x*arccoth(x)/(-x^2+1)+3/16*arccoth(x)^2

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Rubi [A]
time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6108, 6104, 267} \begin {gather*} -\frac {3}{16 \left (1-x^2\right )}-\frac {1}{16 \left (1-x^2\right )^2}+\frac {3 x \coth ^{-1}(x)}{8 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3}{16} \coth ^{-1}(x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/(1 - x^2)^3,x]

[Out]

-1/16*1/(1 - x^2)^2 - 3/(16*(1 - x^2)) + (x*ArcCoth[x])/(4*(1 - x^2)^2) + (3*x*ArcCoth[x])/(8*(1 - x^2)) + (3*
ArcCoth[x]^2)/16

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]

Rule 6108

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1
)/(4*c*d*(q + 1)^2)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcCoth[c*x]), x], x]
- Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcCoth[c*x])/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^
2*d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx &=-\frac {1}{16 \left (1-x^2\right )^2}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3}{4} \int \frac {\coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx\\ &=-\frac {1}{16 \left (1-x^2\right )^2}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3 x \coth ^{-1}(x)}{8 \left (1-x^2\right )}+\frac {3}{16} \coth ^{-1}(x)^2-\frac {3}{8} \int \frac {x}{\left (1-x^2\right )^2} \, dx\\ &=-\frac {1}{16 \left (1-x^2\right )^2}-\frac {3}{16 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3 x \coth ^{-1}(x)}{8 \left (1-x^2\right )}+\frac {3}{16} \coth ^{-1}(x)^2\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.64 \begin {gather*} -\frac {4-3 x^2+2 x \left (-5+3 x^2\right ) \coth ^{-1}(x)-3 \left (-1+x^2\right )^2 \coth ^{-1}(x)^2}{16 \left (-1+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[x]/(1 - x^2)^3,x]

[Out]

-1/16*(4 - 3*x^2 + 2*x*(-5 + 3*x^2)*ArcCoth[x] - 3*(-1 + x^2)^2*ArcCoth[x]^2)/(-1 + x^2)^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(130\) vs. \(2(57)=114\).
time = 0.12, size = 131, normalized size = 1.96

method	result	size
risch	\(\frac {3 \ln \left (1+x \right )^{2}}{64}-\frac {\left (3 \ln \left (-1+x \right ) x^{4}+6 x^{3}-6 \ln \left (-1+x \right ) x^{2}-10 x +3 \ln \left (-1+x \right )\right ) \ln \left (1+x \right )}{32 \left (x^{2}-1\right )^{2}}+\frac {3 x^{4} \ln \left (-1+x \right )^{2}+12 x^{3} \ln \left (-1+x \right )-6 x^{2} \ln \left (-1+x \right )^{2}+12 x^{2}-20 \ln \left (-1+x \right ) x +3 \ln \left (-1+x \right )^{2}-16}{64 \left (1+x \right ) \left (-1+x \right ) \left (x^{2}-1\right )}\)	\(128\)
default	\(-\frac {\mathrm {arccoth}\left (x \right )}{16 \left (1+x \right )^{2}}-\frac {3 \,\mathrm {arccoth}\left (x \right )}{16 \left (1+x \right )}+\frac {3 \,\mathrm {arccoth}\left (x \right ) \ln \left (1+x \right )}{16}+\frac {\mathrm {arccoth}\left (x \right )}{16 \left (-1+x \right )^{2}}-\frac {3 \,\mathrm {arccoth}\left (x \right )}{16 \left (-1+x \right )}-\frac {3 \,\mathrm {arccoth}\left (x \right ) \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (-1+x \right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{32}-\frac {3 \ln \left (-1+x \right )^{2}}{64}+\frac {3 \left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{2}\right )}{32}-\frac {3 \ln \left (1+x \right )^{2}}{64}-\frac {1}{64 \left (-1+x \right )^{2}}+\frac {7}{64 \left (-1+x \right )}-\frac {1}{64 \left (1+x \right )^{2}}-\frac {7}{64 \left (1+x \right )}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(x)/(-x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-1/16*arccoth(x)/(1+x)^2-3/16*arccoth(x)/(1+x)+3/16*arccoth(x)*ln(1+x)+1/16*arccoth(x)/(-1+x)^2-3/16*arccoth(x
)/(-1+x)-3/16*arccoth(x)*ln(-1+x)+3/32*ln(-1+x)*ln(1/2+1/2*x)-3/64*ln(-1+x)^2+3/32*(ln(1+x)-ln(1/2+1/2*x))*ln(
1/2-1/2*x)-3/64*ln(1+x)^2-1/64/(-1+x)^2+7/64/(-1+x)-1/64/(1+x)^2-7/64/(1+x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (49) = 98\).
time = 0.26, size = 118, normalized size = 1.76 \begin {gather*} -\frac {1}{16} \, {\left (\frac {2 \, {\left (3 \, x^{3} - 5 \, x\right )}}{x^{4} - 2 \, x^{2} + 1} - 3 \, \log \left (x + 1\right ) + 3 \, \log \left (x - 1\right )\right )} \operatorname {arcoth}\left (x\right ) - \frac {3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right )^{2} - 6 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + 3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right )^{2} - 12 \, x^{2} + 16}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(3*x^3 - 5*x)/(x^4 - 2*x^2 + 1) - 3*log(x + 1) + 3*log(x - 1))*arccoth(x) - 1/64*(3*(x^4 - 2*x^2 + 1)
*log(x + 1)^2 - 6*(x^4 - 2*x^2 + 1)*log(x + 1)*log(x - 1) + 3*(x^4 - 2*x^2 + 1)*log(x - 1)^2 - 12*x^2 + 16)/(x
^4 - 2*x^2 + 1)

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Fricas [A]
time = 0.35, size = 66, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2} + 12 \, x^{2} - 4 \, {\left (3 \, x^{3} - 5 \, x\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 16}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(3*(x^4 - 2*x^2 + 1)*log((x + 1)/(x - 1))^2 + 12*x^2 - 4*(3*x^3 - 5*x)*log((x + 1)/(x - 1)) - 16)/(x^4 -
2*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 )}}{x^{6} - 3 x^{4} + 3 x^{2} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(acoth(x)/(x**6 - 3*x**4 + 3*x**2 - 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-arccoth(x)/(x^2 - 1)^3, x)

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Mupad [B]
time = 1.32, size = 112, normalized size = 1.67 \begin {gather*} \frac {3\,{\ln \left (\frac {1}{x}+1\right )}^2}{64}-\ln \left (1-\frac {1}{x}\right )\,\left (\frac {3\,\ln \left (\frac {1}{x}+1\right )}{32}+\frac {\frac {5\,x}{16}-\frac {3\,x^3}{16}}{x^4-2\,x^2+1}\right )+\frac {3\,{\ln \left (1-\frac {1}{x}\right )}^2}{64}+\frac {\frac {3\,x^2}{16}-\frac {1}{4}}{x^4-2\,x^2+1}+\frac {\ln \left (\frac {1}{x}+1\right )\,\left (\frac {5\,x}{16}-\frac {3\,x^3}{16}\right )}{x^4-2\,x^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-acoth(x)/(x^2 - 1)^3,x)

[Out]

(3*log(1/x + 1)^2)/64 - log(1 - 1/x)*((3*log(1/x + 1))/32 + ((5*x)/16 - (3*x^3)/16)/(x^4 - 2*x^2 + 1)) + (3*lo
g(1 - 1/x)^2)/64 + ((3*x^2)/16 - 1/4)/(x^4 - 2*x^2 + 1) + (log(1/x + 1)*((5*x)/16 - (3*x^3)/16))/(x^4 - 2*x^2
+ 1)

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