3.15 \(\int x^2 \coth ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=103 \[ -\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a^3}-\frac {\tanh ^{-1}(a x)}{3 a^3}+\frac {\coth ^{-1}(a x)^2}{3 a^3}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{3 a^3}+\frac {x}{3 a^2}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2+\frac {x^2 \coth ^{-1}(a x)}{3 a} \]

[Out]

1/3*x/a^2+1/3*x^2*arccoth(a*x)/a+1/3*arccoth(a*x)^2/a^3+1/3*x^3*arccoth(a*x)^2-1/3*arctanh(a*x)/a^3-2/3*arccot
h(a*x)*ln(2/(-a*x+1))/a^3-1/3*polylog(2,1-2/(-a*x+1))/a^3

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Rubi [A]  time = 0.15, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5917, 5981, 321, 206, 5985, 5919, 2402, 2315} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{3 a^3}+\frac {x}{3 a^2}-\frac {\tanh ^{-1}(a x)}{3 a^3}+\frac {\coth ^{-1}(a x)^2}{3 a^3}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{3 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2+\frac {x^2 \coth ^{-1}(a x)}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(3*a^2) + (x^2*ArcCoth[a*x])/(3*a) + ArcCoth[a*x]^2/(3*a^3) + (x^3*ArcCoth[a*x]^2)/3 - ArcTanh[a*x]/(3*a^3)
- (2*ArcCoth[a*x]*Log[2/(1 - a*x)])/(3*a^3) - PolyLog[2, 1 - 2/(1 - a*x)]/(3*a^3)

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])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5919

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

Rule 5981

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

Rule 5985

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

Rubi steps

\begin {align*} \int x^2 \coth ^{-1}(a x)^2 \, dx &=\frac {1}{3} x^3 \coth ^{-1}(a x)^2-\frac {1}{3} (2 a) \int \frac {x^3 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(a x)^2+\frac {2 \int x \coth ^{-1}(a x) \, dx}{3 a}-\frac {2 \int \frac {x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a}\\ &=\frac {x^2 \coth ^{-1}(a x)}{3 a}+\frac {\coth ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2-\frac {1}{3} \int \frac {x^2}{1-a^2 x^2} \, dx-\frac {2 \int \frac {\coth ^{-1}(a x)}{1-a x} \, dx}{3 a^2}\\ &=\frac {x}{3 a^2}+\frac {x^2 \coth ^{-1}(a x)}{3 a}+\frac {\coth ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a^3}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{3 a^2}+\frac {2 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{3 a^2}\\ &=\frac {x}{3 a^2}+\frac {x^2 \coth ^{-1}(a x)}{3 a}+\frac {\coth ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)}{3 a^3}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a^3}-\frac {2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{3 a^3}\\ &=\frac {x}{3 a^2}+\frac {x^2 \coth ^{-1}(a x)}{3 a}+\frac {\coth ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)}{3 a^3}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a^3}-\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 66, normalized size = 0.64 \[ \frac {\left (a^3 x^3-1\right ) \coth ^{-1}(a x)^2+\coth ^{-1}(a x) \left (a^2 x^2-2 \log \left (1-e^{-2 \coth ^{-1}(a x)}\right )-1\right )+\text {Li}_2\left (e^{-2 \coth ^{-1}(a x)}\right )+a x}{3 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x + (-1 + a^3*x^3)*ArcCoth[a*x]^2 + ArcCoth[a*x]*(-1 + a^2*x^2 - 2*Log[1 - E^(-2*ArcCoth[a*x])]) + PolyLog[
2, E^(-2*ArcCoth[a*x])])/(3*a^3)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arcoth}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arccoth(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccoth(a*x)^2, x)

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maple [A]  time = 0.06, size = 176, normalized size = 1.71 \[ \frac {x^{3} \mathrm {arccoth}\left (a x \right )^{2}}{3}+\frac {x^{2} \mathrm {arccoth}\left (a x \right )}{3 a}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{3 a^{3}}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{3 a^{3}}+\frac {x}{3 a^{2}}+\frac {\ln \left (a x -1\right )}{6 a^{3}}-\frac {\ln \left (a x +1\right )}{6 a^{3}}+\frac {\ln \left (a x -1\right )^{2}}{12 a^{3}}-\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{3 a^{3}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{6 a^{3}}-\frac {\ln \left (a x +1\right )^{2}}{12 a^{3}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{6 a^{3}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{6 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(a*x)^2+1/3*x^2*arccoth(a*x)/a+1/3/a^3*arccoth(a*x)*ln(a*x-1)+1/3/a^3*arccoth(a*x)*ln(a*x+1)+1/
3*x/a^2+1/6/a^3*ln(a*x-1)-1/6/a^3*ln(a*x+1)+1/12/a^3*ln(a*x-1)^2-1/3/a^3*dilog(1/2+1/2*a*x)-1/6/a^3*ln(a*x-1)*
ln(1/2+1/2*a*x)-1/12/a^3*ln(a*x+1)^2-1/6/a^3*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/6/a^3*ln(-1/2*a*x+1/2)*ln(a*x+
1)

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maxima [A]  time = 0.31, size = 134, normalized size = 1.30 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{12} \, a^{2} {\left (\frac {4 \, a x - \log \left (a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2} + 2 \, \log \left (a x - 1\right )}{a^{5}} - \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{5}} - \frac {2 \, \log \left (a x + 1\right )}{a^{5}}\right )} + \frac {1}{3} \, a {\left (\frac {x^{2}}{a^{2}} + \frac {\log \left (a^{2} x^{2} - 1\right )}{a^{4}}\right )} \operatorname {arcoth}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(a*x)^2 + 1/12*a^2*((4*a*x - log(a*x + 1)^2 + 2*log(a*x + 1)*log(a*x - 1) + log(a*x - 1)^2 + 2*
log(a*x - 1))/a^5 - 4*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a^5 - 2*log(a*x + 1)/a^5) + 1/
3*a*(x^2/a^2 + log(a^2*x^2 - 1)/a^4)*arccoth(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {acoth}\left (a\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*acoth(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {acoth}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*acoth(a*x)**2, x)

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