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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 103, 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, 325, 206, 5989, 5933, 2447} \[ -\frac {1}{3} a^3 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {a^2}{3 x}+\frac {1}{3} a^3 \tanh ^{-1}(a x)+\frac {1}{3} a^3 \coth ^{-1}(a x)^2+\frac {2}{3} a^3 \log \left (2-\frac {2}{a x+1}\right ) \coth ^{-1}(a x)-\frac {a \coth ^{-1}(a x)}{3 x^2}-\frac {\coth ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a^2/(3*x) - (a*ArcCoth[a*x])/(3*x^2) + (a^3*ArcCoth[a*x]^2)/3 - ArcCoth[a*x]^2/(3*x^3) + (a^3*ArcTanh[a*x])/3
+ (2*a^3*ArcCoth[a*x]*Log[2 - 2/(1 + a*x)])/3 - (a^3*PolyLog[2, -1 + 2/(1 + a*x)])/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 325

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 5933

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

Rule 5989

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 87, normalized size = 0.84 \[ \frac {-a^3 x^3 \text {Li}_2\left (-e^{-2 \coth ^{-1}(a x)}\right )+\left (a^3 x^3-1\right ) \coth ^{-1}(a x)^2-a^2 x^2+a x \coth ^{-1}(a x) \left (a^2 x^2+2 a^2 x^2 \log \left (e^{-2 \coth ^{-1}(a x)}+1\right )-1\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (a x\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(a*x)^2/x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)^2/x^4, x)

________________________________________________________________________________________

maple [B]  time = 0.07, size = 224, normalized size = 2.17 \[ -\frac {\mathrm {arccoth}\left (a x \right )^{2}}{3 x^{3}}-\frac {a \,\mathrm {arccoth}\left (a x \right )}{3 x^{2}}+\frac {2 a^{3} \mathrm {arccoth}\left (a x \right ) \ln \left (a x \right )}{3}-\frac {a^{3} \mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{3}-\frac {a^{3} \mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {a^{2}}{3 x}-\frac {a^{3} \ln \left (a x -1\right )}{6}+\frac {a^{3} \ln \left (a x +1\right )}{6}-\frac {a^{3} \ln \left (a x -1\right )^{2}}{12}+\frac {a^{3} \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{3}+\frac {a^{3} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{6}+\frac {a^{3} \ln \left (a x +1\right )^{2}}{12}+\frac {a^{3} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{6}-\frac {a^{3} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{6}-\frac {a^{3} \dilog \left (a x \right )}{3}-\frac {a^{3} \dilog \left (a x +1\right )}{3}-\frac {a^{3} \ln \left (a x \right ) \ln \left (a x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 176, normalized size = 1.71 \[ \frac {1}{12} \, {\left (4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )} a - 4 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )} a + 4 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )} a + 2 \, a \log \left (a x + 1\right ) - 2 \, a \log \left (a x - 1\right ) + \frac {a x \log \left (a x + 1\right )^{2} - 2 \, a x \log \left (a x + 1\right ) \log \left (a x - 1\right ) - a x \log \left (a x - 1\right )^{2} - 4}{x}\right )} a^{2} - \frac {1}{3} \, {\left (a^{2} \log \left (a^{2} x^{2} - 1\right ) - a^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} a \operatorname {arcoth}\left (a x\right ) - \frac {\operatorname {arcoth}\left (a x\right )^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acoth(a*x)^2/x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a*x)**2/x**4, x)

________________________________________________________________________________________