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

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

[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 = 0.04, 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 = {5961, 5957, 261} \[ -\frac {3}{16 \left (1-x^2\right )}-\frac {1}{16 \left (1-x^2\right )^2}+\frac {3 x \coth ^{-1}(x)}{8 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)}{4 \left (1-x^2\right )^2}+\frac {3}{16} \coth ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/(1 - x^2)^3,x]

[Out]

-1/(16*(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 261

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 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 5961

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 = 0.06, size = 43, normalized size = 0.64 \[ -\frac {-3 x^2+2 \left (3 x^2-5\right ) x \coth ^{-1}(x)-3 \left (x^2-1\right )^2 \coth ^{-1}(x)^2+4}{16 \left (x^2-1\right )^2} \]

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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fricas [A]  time = 0.40, size = 66, normalized size = 0.99 \[ \frac {3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2} + 12 \, x^{2} - 4 \, {\left (3 \, x^{3} - 5 \, x\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 16}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

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

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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maple [B]  time = 0.07, size = 131, normalized size = 1.96 \[ \frac {\mathrm {arccoth}\relax (x )}{16 \left (-1+x \right )^{2}}-\frac {3 \,\mathrm {arccoth}\relax (x )}{16 \left (-1+x \right )}-\frac {3 \,\mathrm {arccoth}\relax (x ) \ln \left (-1+x \right )}{16}-\frac {\mathrm {arccoth}\relax (x )}{16 \left (1+x \right )^{2}}-\frac {3 \,\mathrm {arccoth}\relax (x )}{16 \left (1+x \right )}+\frac {3 \,\mathrm {arccoth}\relax (x ) \ln \left (1+x \right )}{16}-\frac {3 \ln \left (-1+x \right )^{2}}{64}+\frac {3 \ln \left (-1+x \right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{32}-\frac {3 \ln \left (1+x \right )^{2}}{64}+\frac {3 \left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (-\frac {x}{2}+\frac {1}{2}\right )}{32}-\frac {1}{64 \left (-1+x \right )^{2}}+\frac {7}{64 \left (-1+x \right )}-\frac {1}{64 \left (1+x \right )^{2}}-\frac {7}{64 \left (1+x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)-3/64*ln(-1+x)^2+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*x+1/2)-1/64/(-1+x)^2+7/64/(-1+x)-1/64/(1+x)^2-7/64/(1+x)

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maxima [B]  time = 0.31, size = 118, normalized size = 1.76 \[ -\frac {1}{16} \, {\left (\frac {2 \, {\left (3 \, x^{3} - 5 \, x\right )}}{x^{4} - 2 \, x^{2} + 1} - 3 \, \log \left (x + 1\right ) + 3 \, \log \left (x - 1\right )\right )} \operatorname {arcoth}\relax (x) - \frac {3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right )^{2} - 6 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + 3 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right )^{2} - 12 \, x^{2} + 16}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {acoth}{\relax (x )}}{x^{6} - 3 x^{4} + 3 x^{2} - 1}\, dx \]

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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