3.13 \(\int x^4 \coth ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=127 \[ -\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{5 a^5}-\frac {3 \tanh ^{-1}(a x)}{10 a^5}+\frac {\coth ^{-1}(a x)^2}{5 a^5}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{5 a^5}+\frac {3 x}{10 a^4}+\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {x^3}{30 a^2}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2+\frac {x^4 \coth ^{-1}(a x)}{10 a} \]

[Out]

3/10*x/a^4+1/30*x^3/a^2+1/5*x^2*arccoth(a*x)/a^3+1/10*x^4*arccoth(a*x)/a+1/5*arccoth(a*x)^2/a^5+1/5*x^5*arccot
h(a*x)^2-3/10*arctanh(a*x)/a^5-2/5*arccoth(a*x)*ln(2/(-a*x+1))/a^5-1/5*polylog(2,1-2/(-a*x+1))/a^5

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Rubi [A]  time = 0.23, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {5917, 5981, 302, 206, 321, 5985, 5919, 2402, 2315} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{5 a^5}+\frac {x^3}{30 a^2}+\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {3 x}{10 a^4}-\frac {3 \tanh ^{-1}(a x)}{10 a^5}+\frac {\coth ^{-1}(a x)^2}{5 a^5}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{5 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2+\frac {x^4 \coth ^{-1}(a x)}{10 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcCoth[a*x]^2,x]

[Out]

(3*x)/(10*a^4) + x^3/(30*a^2) + (x^2*ArcCoth[a*x])/(5*a^3) + (x^4*ArcCoth[a*x])/(10*a) + ArcCoth[a*x]^2/(5*a^5
) + (x^5*ArcCoth[a*x]^2)/5 - (3*ArcTanh[a*x])/(10*a^5) - (2*ArcCoth[a*x]*Log[2/(1 - a*x)])/(5*a^5) - PolyLog[2
, 1 - 2/(1 - a*x)]/(5*a^5)

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^4 \coth ^{-1}(a x)^2 \, dx &=\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {1}{5} (2 a) \int \frac {x^5 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \coth ^{-1}(a x)^2+\frac {2 \int x^3 \coth ^{-1}(a x) \, dx}{5 a}-\frac {2 \int \frac {x^3 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{5 a}\\ &=\frac {x^4 \coth ^{-1}(a x)}{10 a}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {1}{10} \int \frac {x^4}{1-a^2 x^2} \, dx+\frac {2 \int x \coth ^{-1}(a x) \, dx}{5 a^3}-\frac {2 \int \frac {x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{5 a^3}\\ &=\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {x^4 \coth ^{-1}(a x)}{10 a}+\frac {\coth ^{-1}(a x)^2}{5 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {1}{10} \int \left (-\frac {1}{a^4}-\frac {x^2}{a^2}+\frac {1}{a^4 \left (1-a^2 x^2\right )}\right ) \, dx-\frac {2 \int \frac {\coth ^{-1}(a x)}{1-a x} \, dx}{5 a^4}-\frac {\int \frac {x^2}{1-a^2 x^2} \, dx}{5 a^2}\\ &=\frac {3 x}{10 a^4}+\frac {x^3}{30 a^2}+\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {x^4 \coth ^{-1}(a x)}{10 a}+\frac {\coth ^{-1}(a x)^2}{5 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{5 a^5}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{10 a^4}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{5 a^4}+\frac {2 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{5 a^4}\\ &=\frac {3 x}{10 a^4}+\frac {x^3}{30 a^2}+\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {x^4 \coth ^{-1}(a x)}{10 a}+\frac {\coth ^{-1}(a x)^2}{5 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {3 \tanh ^{-1}(a x)}{10 a^5}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{5 a^5}-\frac {2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{5 a^5}\\ &=\frac {3 x}{10 a^4}+\frac {x^3}{30 a^2}+\frac {x^2 \coth ^{-1}(a x)}{5 a^3}+\frac {x^4 \coth ^{-1}(a x)}{10 a}+\frac {\coth ^{-1}(a x)^2}{5 a^5}+\frac {1}{5} x^5 \coth ^{-1}(a x)^2-\frac {3 \tanh ^{-1}(a x)}{10 a^5}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{5 a^5}-\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{5 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 87, normalized size = 0.69 \[ \frac {6 \left (a^5 x^5-1\right ) \coth ^{-1}(a x)^2+a x \left (a^2 x^2+9\right )+3 \coth ^{-1}(a x) \left (a^4 x^4+2 a^2 x^2-4 \log \left (1-e^{-2 \coth ^{-1}(a x)}\right )-3\right )+6 \text {Li}_2\left (e^{-2 \coth ^{-1}(a x)}\right )}{30 a^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*ArcCoth[a*x]^2,x]

[Out]

(a*x*(9 + a^2*x^2) + 6*(-1 + a^5*x^5)*ArcCoth[a*x]^2 + 3*ArcCoth[a*x]*(-3 + 2*a^2*x^2 + a^4*x^4 - 4*Log[1 - E^
(-2*ArcCoth[a*x])]) + 6*PolyLog[2, E^(-2*ArcCoth[a*x])])/(30*a^5)

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} \operatorname {arcoth}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccoth(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^4*arccoth(a*x)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \operatorname {arcoth}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccoth(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^4*arccoth(a*x)^2, x)

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maple [A]  time = 0.06, size = 196, normalized size = 1.54 \[ \frac {x^{5} \mathrm {arccoth}\left (a x \right )^{2}}{5}+\frac {x^{4} \mathrm {arccoth}\left (a x \right )}{10 a}+\frac {x^{2} \mathrm {arccoth}\left (a x \right )}{5 a^{3}}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{5 a^{5}}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{5 a^{5}}+\frac {x^{3}}{30 a^{2}}+\frac {3 x}{10 a^{4}}+\frac {3 \ln \left (a x -1\right )}{20 a^{5}}-\frac {3 \ln \left (a x +1\right )}{20 a^{5}}+\frac {\ln \left (a x -1\right )^{2}}{20 a^{5}}-\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{5 a^{5}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{10 a^{5}}-\frac {\ln \left (a x +1\right )^{2}}{20 a^{5}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{10 a^{5}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{10 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arccoth(a*x)^2,x)

[Out]

1/5*x^5*arccoth(a*x)^2+1/10*x^4*arccoth(a*x)/a+1/5*x^2*arccoth(a*x)/a^3+1/5/a^5*arccoth(a*x)*ln(a*x-1)+1/5/a^5
*arccoth(a*x)*ln(a*x+1)+1/30*x^3/a^2+3/10*x/a^4+3/20/a^5*ln(a*x-1)-3/20*ln(a*x+1)/a^5+1/20/a^5*ln(a*x-1)^2-1/5
/a^5*dilog(1/2+1/2*a*x)-1/10/a^5*ln(a*x-1)*ln(1/2+1/2*a*x)-1/20/a^5*ln(a*x+1)^2-1/10/a^5*ln(-1/2*a*x+1/2)*ln(1
/2+1/2*a*x)+1/10/a^5*ln(-1/2*a*x+1/2)*ln(a*x+1)

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maxima [A]  time = 0.32, size = 155, normalized size = 1.22 \[ \frac {1}{5} \, x^{5} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{60} \, a^{2} {\left (\frac {2 \, a^{3} x^{3} + 18 \, a x - 3 \, \log \left (a x + 1\right )^{2} + 6 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + 3 \, \log \left (a x - 1\right )^{2} + 9 \, \log \left (a x - 1\right )}{a^{7}} - \frac {12 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{7}} - \frac {9 \, \log \left (a x + 1\right )}{a^{7}}\right )} + \frac {1}{10} \, a {\left (\frac {a^{2} x^{4} + 2 \, x^{2}}{a^{4}} + \frac {2 \, \log \left (a^{2} x^{2} - 1\right )}{a^{6}}\right )} \operatorname {arcoth}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccoth(a*x)^2,x, algorithm="maxima")

[Out]

1/5*x^5*arccoth(a*x)^2 + 1/60*a^2*((2*a^3*x^3 + 18*a*x - 3*log(a*x + 1)^2 + 6*log(a*x + 1)*log(a*x - 1) + 3*lo
g(a*x - 1)^2 + 9*log(a*x - 1))/a^7 - 12*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a^7 - 9*log(
a*x + 1)/a^7) + 1/10*a*((a^2*x^4 + 2*x^2)/a^4 + 2*log(a^2*x^2 - 1)/a^6)*arccoth(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,{\mathrm {acoth}\left (a\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*acoth(a*x)^2,x)

[Out]

int(x^4*acoth(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \operatorname {acoth}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*acoth(a*x)**2,x)

[Out]

Integral(x**4*acoth(a*x)**2, x)

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