3.25 \(\int x^3 \coth ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=139 \[ -\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^4}-\frac {\tanh ^{-1}(a x)}{4 a^4}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {\coth ^{-1}(a x)^2}{a^4}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{a^4}+\frac {x}{4 a^3}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3+\frac {x^3 \coth ^{-1}(a x)^2}{4 a} \]

[Out]

1/4*x/a^3+1/4*x^2*arccoth(a*x)/a^2+arccoth(a*x)^2/a^4+3/4*x*arccoth(a*x)^2/a^3+1/4*x^3*arccoth(a*x)^2/a-1/4*ar
ccoth(a*x)^3/a^4+1/4*x^4*arccoth(a*x)^3-1/4*arctanh(a*x)/a^4-2*arccoth(a*x)*ln(2/(-a*x+1))/a^4-polylog(2,1-2/(
-a*x+1))/a^4

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Rubi [A]  time = 0.42, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5917, 5981, 321, 206, 5985, 5919, 2402, 2315, 5911, 5949} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^4}+\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {x}{4 a^3}-\frac {\tanh ^{-1}(a x)}{4 a^4}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {\coth ^{-1}(a x)^2}{a^4}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3+\frac {x^3 \coth ^{-1}(a x)^2}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5911

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcCoth[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 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 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^3 \coth ^{-1}(a x)^3 \, dx &=\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {1}{4} (3 a) \int \frac {x^4 \coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \coth ^{-1}(a x)^3+\frac {3 \int x^2 \coth ^{-1}(a x)^2 \, dx}{4 a}-\frac {3 \int \frac {x^2 \coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{4 a}\\ &=\frac {x^3 \coth ^{-1}(a x)^2}{4 a}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {1}{2} \int \frac {x^3 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx+\frac {3 \int \coth ^{-1}(a x)^2 \, dx}{4 a^3}-\frac {3 \int \frac {\coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{4 a^3}\\ &=\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^3 \coth ^{-1}(a x)^2}{4 a}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3+\frac {\int x \coth ^{-1}(a x) \, dx}{2 a^2}-\frac {\int \frac {x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^2}-\frac {3 \int \frac {x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^2}\\ &=\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {\coth ^{-1}(a x)^2}{a^4}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^3 \coth ^{-1}(a x)^2}{4 a}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {\int \frac {\coth ^{-1}(a x)}{1-a x} \, dx}{2 a^3}-\frac {3 \int \frac {\coth ^{-1}(a x)}{1-a x} \, dx}{2 a^3}-\frac {\int \frac {x^2}{1-a^2 x^2} \, dx}{4 a}\\ &=\frac {x}{4 a^3}+\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {\coth ^{-1}(a x)^2}{a^4}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^3 \coth ^{-1}(a x)^2}{4 a}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{4 a^3}+\frac {\int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}+\frac {3 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}\\ &=\frac {x}{4 a^3}+\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {\coth ^{-1}(a x)^2}{a^4}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^3 \coth ^{-1}(a x)^2}{4 a}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)}{4 a^4}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {\operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{2 a^4}-\frac {3 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{2 a^4}\\ &=\frac {x}{4 a^3}+\frac {x^2 \coth ^{-1}(a x)}{4 a^2}+\frac {\coth ^{-1}(a x)^2}{a^4}+\frac {3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac {x^3 \coth ^{-1}(a x)^2}{4 a}-\frac {\coth ^{-1}(a x)^3}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)}{4 a^4}-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^4}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 88, normalized size = 0.63 \[ \frac {\left (a^4 x^4-1\right ) \coth ^{-1}(a x)^3+\left (a^3 x^3+3 a x-4\right ) \coth ^{-1}(a x)^2+\coth ^{-1}(a x) \left (a^2 x^2-8 \log \left (1-e^{-2 \coth ^{-1}(a x)}\right )-1\right )+4 \text {Li}_2\left (e^{-2 \coth ^{-1}(a x)}\right )+a x}{4 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x + (-4 + 3*a*x + a^3*x^3)*ArcCoth[a*x]^2 + (-1 + a^4*x^4)*ArcCoth[a*x]^3 + ArcCoth[a*x]*(-1 + a^2*x^2 - 8*
Log[1 - E^(-2*ArcCoth[a*x])]) + 4*PolyLog[2, E^(-2*ArcCoth[a*x])])/(4*a^4)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \operatorname {arcoth}\left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arccoth(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccoth(a*x)^3, x)

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maple [C]  time = 1.59, size = 684, normalized size = 4.92 \[ -\frac {\mathrm {arccoth}\left (a x \right )}{4 a^{4}}+\frac {\mathrm {arccoth}\left (a x \right )^{2}}{a^{4}}+\frac {3 \mathrm {arccoth}\left (a x \right )^{2} \ln \left (a x -1\right )}{8 a^{4}}-\frac {3 \mathrm {arccoth}\left (a x \right )^{2} \ln \left (a x +1\right )}{8 a^{4}}-\frac {2 \,\mathrm {arccoth}\left (a x \right ) \ln \left (1+\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a^{4}}-\frac {\sqrt {\frac {a x -1}{a x +1}}}{4 a^{4} \left (\sqrt {\frac {a x -1}{a x +1}}+1\right )}-\frac {3 \mathrm {arccoth}\left (a x \right )^{2} \ln \left (\frac {a x -1}{a x +1}\right )}{8 a^{4}}-\frac {\sqrt {\frac {a x -1}{a x +1}}}{4 a^{4} \left (-1+\sqrt {\frac {a x -1}{a x +1}}\right )}+\frac {2 \dilog \left (\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a^{4}}-\frac {\mathrm {arccoth}\left (a x \right )^{3}}{4 a^{4}}+\frac {x^{4} \mathrm {arccoth}\left (a x \right )^{3}}{4}+\frac {3 x \mathrm {arccoth}\left (a x \right )^{2}}{4 a^{3}}+\frac {x^{3} \mathrm {arccoth}\left (a x \right )^{2}}{4 a}+\frac {x^{2} \mathrm {arccoth}\left (a x \right )}{4 a^{2}}+\frac {3 i \pi \,\mathrm {csgn}\left (\frac {i}{\frac {a x +1}{a x -1}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\left (a x -1\right ) \left (\frac {a x +1}{a x -1}-1\right )}\right )^{2} \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}}-\frac {3 i \pi \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\left (a x -1\right ) \left (\frac {a x +1}{a x -1}-1\right )}\right )^{3} \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}}+\frac {3 i \pi \,\mathrm {csgn}\left (\frac {i}{\sqrt {\frac {a x -1}{a x +1}}}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{a x -1}\right )^{2} \mathrm {arccoth}\left (a x \right )^{2}}{8 a^{4}}-\frac {2 \dilog \left (1+\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a^{4}}+\frac {3 i \pi \,\mathrm {csgn}\left (\frac {i \left (a x +1\right )}{a x -1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\left (a x -1\right ) \left (\frac {a x +1}{a x -1}-1\right )}\right )^{2} \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}}-\frac {3 i \pi \mathrm {csgn}\left (\frac {i}{\sqrt {\frac {a x -1}{a x +1}}}\right )^{2} \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{a x -1}\right ) \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}}-\frac {3 i \pi \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{a x -1}\right )^{3} \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}}-\frac {3 i \pi \,\mathrm {csgn}\left (\frac {i \left (a x +1\right )}{a x -1}\right ) \mathrm {csgn}\left (\frac {i}{\frac {a x +1}{a x -1}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\left (a x -1\right ) \left (\frac {a x +1}{a x -1}-1\right )}\right ) \mathrm {arccoth}\left (a x \right )^{2}}{16 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^4*arccoth(a*x)+arccoth(a*x)^2/a^4+3/8/a^4*arccoth(a*x)^2*ln(a*x-1)-3/8/a^4*arccoth(a*x)^2*ln(a*x+1)-2/a
^4*arccoth(a*x)*ln(1+1/((a*x-1)/(a*x+1))^(1/2))-1/4/a^4/(((a*x-1)/(a*x+1))^(1/2)+1)*((a*x-1)/(a*x+1))^(1/2)-3/
8/a^4*arccoth(a*x)^2*ln((a*x-1)/(a*x+1))-1/4/a^4/(-1+((a*x-1)/(a*x+1))^(1/2))*((a*x-1)/(a*x+1))^(1/2)+2/a^4*di
log(1/((a*x-1)/(a*x+1))^(1/2))-1/4*arccoth(a*x)^3/a^4+1/4*x^4*arccoth(a*x)^3+3/4*x*arccoth(a*x)^2/a^3+1/4*x^3*
arccoth(a*x)^2/a+1/4*x^2*arccoth(a*x)/a^2+3/16*I/a^4*Pi*csgn(I/((a*x+1)/(a*x-1)-1))*csgn(I*(a*x+1)/(a*x-1)/((a
*x+1)/(a*x-1)-1))^2*arccoth(a*x)^2-3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))^3*arccoth(a*x)^2+
3/8*I/a^4*Pi*csgn(I/((a*x-1)/(a*x+1))^(1/2))*csgn(I*(a*x+1)/(a*x-1))^2*arccoth(a*x)^2-2/a^4*dilog(1+1/((a*x-1)
/(a*x+1))^(1/2))+3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1))*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))^2*arccoth(a
*x)^2-3/16*I/a^4*Pi*csgn(I/((a*x-1)/(a*x+1))^(1/2))^2*csgn(I*(a*x+1)/(a*x-1))*arccoth(a*x)^2-3/16*I/a^4*Pi*csg
n(I*(a*x+1)/(a*x-1))^3*arccoth(a*x)^2-3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1))*csgn(I/((a*x+1)/(a*x-1)-1))*csgn(I
*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))*arccoth(a*x)^2

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maxima [B]  time = 0.33, size = 262, normalized size = 1.88 \[ \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (a x\right )^{3} + \frac {1}{8} \, a {\left (\frac {2 \, {\left (a^{2} x^{3} + 3 \, x\right )}}{a^{4}} - \frac {3 \, \log \left (a x + 1\right )}{a^{5}} + \frac {3 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{32} \, a {\left (\frac {\frac {{\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right )^{2} - \log \left (a x + 1\right )^{3} + \log \left (a x - 1\right )^{3} + 8 \, a x - {\left (3 \, \log \left (a x - 1\right )^{2} - 16 \, \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) + 8 \, \log \left (a x - 1\right )^{2} + 4 \, \log \left (a x - 1\right )}{a} - \frac {32 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a} - \frac {4 \, \log \left (a x + 1\right )}{a}}{a^{4}} + \frac {2 \, {\left (4 \, a^{2} x^{2} - 2 \, {\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right ) + 3 \, \log \left (a x + 1\right )^{2} + 3 \, \log \left (a x - 1\right )^{2} + 16 \, \log \left (a x - 1\right )\right )} \operatorname {arcoth}\left (a x\right )}{a^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccoth(a*x)^3 + 1/8*a*(2*(a^2*x^3 + 3*x)/a^4 - 3*log(a*x + 1)/a^5 + 3*log(a*x - 1)/a^5)*arccoth(a*x)^
2 + 1/32*a*((((3*log(a*x - 1) - 8)*log(a*x + 1)^2 - log(a*x + 1)^3 + log(a*x - 1)^3 + 8*a*x - (3*log(a*x - 1)^
2 - 16*log(a*x - 1))*log(a*x + 1) + 8*log(a*x - 1)^2 + 4*log(a*x - 1))/a - 32*(log(a*x - 1)*log(1/2*a*x + 1/2)
+ dilog(-1/2*a*x + 1/2))/a - 4*log(a*x + 1)/a)/a^4 + 2*(4*a^2*x^2 - 2*(3*log(a*x - 1) - 8)*log(a*x + 1) + 3*l
og(a*x + 1)^2 + 3*log(a*x - 1)^2 + 16*log(a*x - 1))*arccoth(a*x)/a^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\mathrm {acoth}\left (a\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*acoth(a*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {acoth}^{3}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*acoth(a*x)**3, x)

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