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

Optimal. Leaf size=61 \[ -\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac {1}{2} a^2 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{2 x^2}-\frac {a \coth ^{-1}(a x)}{x} \]

[Out]

-a*arccoth(a*x)/x+1/2*a^2*arccoth(a*x)^2-1/2*arccoth(a*x)^2/x^2+a^2*ln(x)-1/2*a^2*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.10, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5917, 5983, 266, 36, 29, 31, 5949} \[ -\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac {1}{2} a^2 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{2 x^2}-\frac {a \coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^2/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5949

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

Rule 5983

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcCoth[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcCoth[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.93 \[ -\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+\frac {\left (a^2 x^2-1\right ) \coth ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)-\frac {a \coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]^2/x^3,x]

[Out]

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

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fricas [A]  time = 0.55, size = 79, normalized size = 1.30 \[ -\frac {4 \, a^{2} x^{2} \log \left (a^{2} x^{2} - 1\right ) - 8 \, a^{2} x^{2} \log \relax (x) + 4 \, a x \log \left (\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)^2/x^3, x)

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maple [B]  time = 0.06, size = 164, normalized size = 2.69 \[ -\frac {\mathrm {arccoth}\left (a x \right )^{2}}{2 x^{2}}-\frac {a \,\mathrm {arccoth}\left (a x \right )}{x}-\frac {a^{2} \mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{2}+\frac {a^{2} \mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {a^{2} \ln \left (a x -1\right )^{2}}{8}+\frac {a^{2} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4}+a^{2} \ln \left (a x \right )-\frac {a^{2} \ln \left (a x -1\right )}{2}-\frac {a^{2} \ln \left (a x +1\right )}{2}-\frac {a^{2} \ln \left (a x +1\right )^{2}}{8}+\frac {a^{2} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4}-\frac {a^{2} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^2/x^3,x)

[Out]

-1/2*arccoth(a*x)^2/x^2-a*arccoth(a*x)/x-1/2*a^2*arccoth(a*x)*ln(a*x-1)+1/2*a^2*arccoth(a*x)*ln(a*x+1)-1/8*a^2
*ln(a*x-1)^2+1/4*a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+a^2*ln(a*x)-1/2*a^2*ln(a*x-1)-1/2*a^2*ln(a*x+1)-1/8*a^2*ln(a*x+
1)^2+1/4*a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/4*a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)

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maxima [A]  time = 0.31, size = 96, normalized size = 1.57 \[ \frac {1}{8} \, {\left (2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right ) + 8 \, \log \relax (x)\right )} a^{2} + \frac {1}{2} \, {\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac {2}{x}\right )} a \operatorname {arcoth}\left (a x\right ) - \frac {\operatorname {arcoth}\left (a x\right )^{2}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.45, size = 145, normalized size = 2.38 \[ a^2\,\ln \relax (x)+{\ln \left (\frac {1}{a\,x}+1\right )}^2\,\left (\frac {a^2}{8}-\frac {1}{8\,x^2}\right )+{\ln \left (1-\frac {1}{a\,x}\right )}^2\,\left (\frac {a^2}{8}-\frac {1}{8\,x^2}\right )-\frac {a^2\,\ln \left (a^2\,x^2-1\right )}{2}+\ln \left (1-\frac {1}{a\,x}\right )\,\left (\frac {4\,a\,x-2}{16\,x^2}+\frac {4\,a\,x+2}{16\,x^2}-\ln \left (\frac {1}{a\,x}+1\right )\,\left (\frac {a^2}{4}-\frac {1}{4\,x^2}\right )\right )-\frac {a\,\ln \left (\frac {1}{a\,x}+1\right )}{2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)^2/x^3,x)

[Out]

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

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sympy [A]  time = 1.58, size = 56, normalized size = 0.92 \[ a^{2} \log {\relax (x )} - a^{2} \log {\left (a x + 1 \right )} + \frac {a^{2} \operatorname {acoth}^{2}{\left (a x \right )}}{2} + a^{2} \operatorname {acoth}{\left (a x \right )} - \frac {a \operatorname {acoth}{\left (a x \right )}}{x} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**2/x**3,x)

[Out]

a**2*log(x) - a**2*log(a*x + 1) + a**2*acoth(a*x)**2/2 + a**2*acoth(a*x) - a*acoth(a*x)/x - acoth(a*x)**2/(2*x
**2)

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