3.22 \(\int \frac {\coth ^{-1}(a x)^2}{x^5} \, dx\)

Optimal. Leaf size=90 \[ \frac {2}{3} a^4 \log (x)+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {a^3 \coth ^{-1}(a x)}{2 x}-\frac {a^2}{12 x^2}-\frac {1}{3} a^4 \log \left (1-a^2 x^2\right )-\frac {\coth ^{-1}(a x)^2}{4 x^4}-\frac {a \coth ^{-1}(a x)}{6 x^3} \]

[Out]

-1/12*a^2/x^2-1/6*a*arccoth(a*x)/x^3-1/2*a^3*arccoth(a*x)/x+1/4*a^4*arccoth(a*x)^2-1/4*arccoth(a*x)^2/x^4+2/3*
a^4*ln(x)-1/3*a^4*ln(-a^2*x^2+1)

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5917, 5983, 266, 44, 36, 29, 31, 5949} \[ -\frac {a^2}{12 x^2}-\frac {1}{3} a^4 \log \left (1-a^2 x^2\right )+\frac {2}{3} a^4 \log (x)+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {a^3 \coth ^{-1}(a x)}{2 x}-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {\coth ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^2/x^5,x]

[Out]

-a^2/(12*x^2) - (a*ArcCoth[a*x])/(6*x^3) - (a^3*ArcCoth[a*x])/(2*x) + (a^4*ArcCoth[a*x]^2)/4 - ArcCoth[a*x]^2/
(4*x^4) + (2*a^4*Log[x])/3 - (a^4*Log[1 - a^2*x^2])/3

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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^5} \, dx &=-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{2} a \int \frac {\coth ^{-1}(a x)}{x^4 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{2} a \int \frac {\coth ^{-1}(a x)}{x^4} \, dx+\frac {1}{2} a^3 \int \frac {\coth ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{6} a^2 \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx+\frac {1}{2} a^3 \int \frac {\coth ^{-1}(a x)}{x^2} \, dx+\frac {1}{2} a^5 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {a^3 \coth ^{-1}(a x)}{2 x}+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{12} a^2 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} a^4 \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {a^3 \coth ^{-1}(a x)}{2 x}+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{12} a^2 \operatorname {Subst}\left (\int \left (\frac {1}{x^2}+\frac {a^2}{x}-\frac {a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right )+\frac {1}{4} a^4 \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a^2}{12 x^2}-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {a^3 \coth ^{-1}(a x)}{2 x}+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {1}{6} a^4 \log (x)-\frac {1}{12} a^4 \log \left (1-a^2 x^2\right )+\frac {1}{4} a^4 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} a^6 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a^2}{12 x^2}-\frac {a \coth ^{-1}(a x)}{6 x^3}-\frac {a^3 \coth ^{-1}(a x)}{2 x}+\frac {1}{4} a^4 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{4 x^4}+\frac {2}{3} a^4 \log (x)-\frac {1}{3} a^4 \log \left (1-a^2 x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 82, normalized size = 0.91 \[ \frac {\left (a^4 x^4-1\right ) \coth ^{-1}(a x)^2}{4 x^4}+\frac {2}{3} a^4 \log (x)-\frac {a^2}{12 x^2}-\frac {a \left (3 a^2 x^2+1\right ) \coth ^{-1}(a x)}{6 x^3}-\frac {1}{3} a^4 \log \left (1-a^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]^2/x^5,x]

[Out]

-1/12*a^2/x^2 - (a*(1 + 3*a^2*x^2)*ArcCoth[a*x])/(6*x^3) + ((-1 + a^4*x^4)*ArcCoth[a*x]^2)/(4*x^4) + (2*a^4*Lo
g[x])/3 - (a^4*Log[1 - a^2*x^2])/3

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 97, normalized size = 1.08 \[ -\frac {16 \, a^{4} x^{4} \log \left (a^{2} x^{2} - 1\right ) - 32 \, a^{4} x^{4} \log \relax (x) + 4 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (3 \, a^{3} x^{3} + a x\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{48 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^5,x, algorithm="fricas")

[Out]

-1/48*(16*a^4*x^4*log(a^2*x^2 - 1) - 32*a^4*x^4*log(x) + 4*a^2*x^2 - 3*(a^4*x^4 - 1)*log((a*x + 1)/(a*x - 1))^
2 + 4*(3*a^3*x^3 + a*x)*log((a*x + 1)/(a*x - 1)))/x^4

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (a x\right )^{2}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^5,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)^2/x^5, x)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 185, normalized size = 2.06 \[ -\frac {\mathrm {arccoth}\left (a x \right )^{2}}{4 x^{4}}-\frac {a \,\mathrm {arccoth}\left (a x \right )}{6 x^{3}}-\frac {a^{3} \mathrm {arccoth}\left (a x \right )}{2 x}-\frac {a^{4} \mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{4}+\frac {a^{4} \mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {a^{4} \ln \left (a x -1\right )^{2}}{16}+\frac {a^{4} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8}-\frac {a^{4} \ln \left (a x +1\right )^{2}}{16}-\frac {a^{4} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8}+\frac {a^{4} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8}-\frac {a^{2}}{12 x^{2}}+\frac {2 a^{4} \ln \left (a x \right )}{3}-\frac {a^{4} \ln \left (a x -1\right )}{3}-\frac {a^{4} \ln \left (a x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^2/x^5,x)

[Out]

-1/4*arccoth(a*x)^2/x^4-1/6*a*arccoth(a*x)/x^3-1/2*a^3*arccoth(a*x)/x-1/4*a^4*arccoth(a*x)*ln(a*x-1)+1/4*a^4*a
rccoth(a*x)*ln(a*x+1)-1/16*a^4*ln(a*x-1)^2+1/8*a^4*ln(a*x-1)*ln(1/2+1/2*a*x)-1/16*a^4*ln(a*x+1)^2-1/8*a^4*ln(-
1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/8*a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/12*a^2/x^2+2/3*a^4*ln(a*x)-1/3*a^4*ln(a*x-1)
-1/3*a^4*ln(a*x+1)

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 154, normalized size = 1.71 \[ \frac {1}{48} \, {\left (32 \, a^{2} \log \relax (x) - \frac {3 \, a^{2} x^{2} \log \left (a x + 1\right )^{2} + 3 \, a^{2} x^{2} \log \left (a x - 1\right )^{2} + 16 \, a^{2} x^{2} \log \left (a x - 1\right ) - 2 \, {\left (3 \, a^{2} x^{2} \log \left (a x - 1\right ) - 8 \, a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 4}{x^{2}}\right )} a^{2} + \frac {1}{12} \, {\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} + 1\right )}}{x^{3}}\right )} a \operatorname {arcoth}\left (a x\right ) - \frac {\operatorname {arcoth}\left (a x\right )^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^5,x, algorithm="maxima")

[Out]

1/48*(32*a^2*log(x) - (3*a^2*x^2*log(a*x + 1)^2 + 3*a^2*x^2*log(a*x - 1)^2 + 16*a^2*x^2*log(a*x - 1) - 2*(3*a^
2*x^2*log(a*x - 1) - 8*a^2*x^2)*log(a*x + 1) + 4)/x^2)*a^2 + 1/12*(3*a^3*log(a*x + 1) - 3*a^3*log(a*x - 1) - 2
*(3*a^2*x^2 + 1)/x^3)*a*arccoth(a*x) - 1/4*arccoth(a*x)^2/x^4

________________________________________________________________________________________

mupad [B]  time = 1.55, size = 196, normalized size = 2.18 \[ \frac {2\,a^4\,\ln \relax (x)}{3}+{\ln \left (\frac {1}{a\,x}+1\right )}^2\,\left (\frac {a^4}{16}-\frac {1}{16\,x^4}\right )+{\ln \left (1-\frac {1}{a\,x}\right )}^2\,\left (\frac {a^4}{16}-\frac {1}{16\,x^4}\right )+\ln \left (1-\frac {1}{a\,x}\right )\,\left (\frac {24\,a^3\,x^3-12\,a^2\,x^2+8\,a\,x-6}{192\,x^4}+\frac {24\,a^3\,x^3+12\,a^2\,x^2+8\,a\,x+6}{192\,x^4}-\ln \left (\frac {1}{a\,x}+1\right )\,\left (\frac {a^4}{8}-\frac {1}{8\,x^4}\right )\right )-\frac {a^4\,\ln \left (a^2\,x^2-1\right )}{3}-\frac {a^2}{12\,x^2}-\frac {a\,\ln \left (\frac {1}{a\,x}+1\right )\,\left (\frac {a^2\,x^2}{4}+\frac {1}{12}\right )}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)^2/x^5,x)

[Out]

(2*a^4*log(x))/3 + log(1/(a*x) + 1)^2*(a^4/16 - 1/(16*x^4)) + log(1 - 1/(a*x))^2*(a^4/16 - 1/(16*x^4)) + log(1
- 1/(a*x))*((8*a*x - 12*a^2*x^2 + 24*a^3*x^3 - 6)/(192*x^4) + (8*a*x + 12*a^2*x^2 + 24*a^3*x^3 + 6)/(192*x^4)
- log(1/(a*x) + 1)*(a^4/8 - 1/(8*x^4))) - (a^4*log(a^2*x^2 - 1))/3 - a^2/(12*x^2) - (a*log(1/(a*x) + 1)*((a^2
*x^2)/4 + 1/12))/x^3

________________________________________________________________________________________

sympy [A]  time = 2.64, size = 90, normalized size = 1.00 \[ \frac {2 a^{4} \log {\relax (x )}}{3} - \frac {2 a^{4} \log {\left (a x + 1 \right )}}{3} + \frac {a^{4} \operatorname {acoth}^{2}{\left (a x \right )}}{4} + \frac {2 a^{4} \operatorname {acoth}{\left (a x \right )}}{3} - \frac {a^{3} \operatorname {acoth}{\left (a x \right )}}{2 x} - \frac {a^{2}}{12 x^{2}} - \frac {a \operatorname {acoth}{\left (a x \right )}}{6 x^{3}} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**2/x**5,x)

[Out]

2*a**4*log(x)/3 - 2*a**4*log(a*x + 1)/3 + a**4*acoth(a*x)**2/4 + 2*a**4*acoth(a*x)/3 - a**3*acoth(a*x)/(2*x) -
a**2/(12*x**2) - a*acoth(a*x)/(6*x**3) - acoth(a*x)**2/(4*x**4)

________________________________________________________________________________________