3.33 \(\int \frac{\coth ^{-1}(a x)^3}{x^5} \, dx\)

Optimal. Leaf size=141 \[ -a^4 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a^2 \coth ^{-1}(a x)}{4 x^2}-\frac{a^3}{4 x}+\frac{1}{4} a^4 \tanh ^{-1}(a x)+\frac{1}{4} a^4 \coth ^{-1}(a x)^3+a^4 \coth ^{-1}(a x)^2-\frac{3 a^3 \coth ^{-1}(a x)^2}{4 x}+2 a^4 \log \left (2-\frac{2}{a x+1}\right ) \coth ^{-1}(a x)-\frac{a \coth ^{-1}(a x)^2}{4 x^3}-\frac{\coth ^{-1}(a x)^3}{4 x^4} \]

[Out]

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

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Rubi [A]  time = 0.463885, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5917, 5983, 325, 206, 5989, 5933, 2447, 5949} \[ -a^4 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a^2 \coth ^{-1}(a x)}{4 x^2}-\frac{a^3}{4 x}+\frac{1}{4} a^4 \tanh ^{-1}(a x)+\frac{1}{4} a^4 \coth ^{-1}(a x)^3+a^4 \coth ^{-1}(a x)^2-\frac{3 a^3 \coth ^{-1}(a x)^2}{4 x}+2 a^4 \log \left (2-\frac{2}{a x+1}\right ) \coth ^{-1}(a x)-\frac{a \coth ^{-1}(a x)^2}{4 x^3}-\frac{\coth ^{-1}(a x)^3}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^3/x^5,x]

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.230209, size = 118, normalized size = 0.84 \[ \frac{-4 a^4 x^4 \text{PolyLog}\left (2,-e^{-2 \coth ^{-1}(a x)}\right )-a^3 x^3+a x \left (4 a^3 x^3-3 a^2 x^2-1\right ) \coth ^{-1}(a x)^2+\left (a^4 x^4-1\right ) \coth ^{-1}(a x)^3+a^2 x^2 \coth ^{-1}(a x) \left (a^2 x^2+8 a^2 x^2 \log \left (e^{-2 \coth ^{-1}(a x)}+1\right )-1\right )}{4 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a*x]^3/x^5,x]

[Out]

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

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Maple [C]  time = 0.536, size = 661, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^3/x^5,x)

[Out]

1/4*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+I/((a*x-1)/(
a*x+1))^(1/2))+3/8*a^4*arccoth(a*x)^2*ln((a*x-1)/(a*x+1))+2*a^4*arccoth(a*x)*ln(1-I/((a*x-1)/(a*x+1))^(1/2))+1
/4*a^4*arccoth(a*x)+1/4*a^4*arccoth(a*x)^3+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*csgn(I*(a*x+1)/(a*x-1))^3*arccoth(a*x)^2-1/4*a^2*arccoth(a*x)/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+3/16*I*a^4*Pi*csgn(I*(a*x+1)/(a*x-
1)/((a*x+1)/(a*x-1)-1))*csgn(I*(a*x+1)/(a*x-1))*csgn(I/((a*x+1)/(a*x-1)-1))*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)^2-3/16*I*a^4*Pi*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-
1))^2*csgn(I/((a*x+1)/(a*x-1)-1))*arccoth(a*x)^2-1/4*arccoth(a*x)^3/x^4-1/4*a*arccoth(a*x)^2/x^3-3/4*a^3*arcco
th(a*x)^2/x-1/4*a^3/x-a^4*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))^2*csgn(I*(a
*x+1)/(a*x-1))*arccoth(a*x)^2+2*a^4*dilog(1+I/((a*x-1)/(a*x+1))^(1/2))+2*a^4*dilog(1-I/((a*x-1)/(a*x+1))^(1/2)
)

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Maxima [B]  time = 1.01002, size = 462, normalized size = 3.28 \begin{align*} \frac{1}{8} \,{\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 )^{2} + \frac{1}{32} \,{\left ({\left (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 - 32 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )} a + 32 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )} a + 4 \, a \log \left (a x + 1\right ) - 4 \, a \log \left (a x - 1\right ) + \frac{a x \log \left (a x + 1\right )^{3} - a x \log \left (a x - 1\right )^{3} - 8 \, a x \log \left (a x - 1\right )^{2} -{\left (3 \, a x \log \left (a x - 1\right ) - 8 \, a x\right )} \log \left (a x + 1\right )^{2} +{\left (3 \, a x \log \left (a x - 1\right )^{2} - 16 \, a x \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) - 8}{x}\right )} a^{2} + 2 \,{\left (32 \, a^{2} \log \left (x\right ) - \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 \operatorname{arcoth}\left (a x\right )\right )} a - \frac{\operatorname{arcoth}\left (a x\right )^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3/x^5,x, algorithm="maxima")

[Out]

1/8*(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)^2 + 1/32*((32*(log(a*x -
1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))*a - 32*(log(a*x + 1)*log(x) + dilog(-a*x))*a + 32*(log(-a*x + 1
)*log(x) + dilog(a*x))*a + 4*a*log(a*x + 1) - 4*a*log(a*x - 1) + (a*x*log(a*x + 1)^3 - a*x*log(a*x - 1)^3 - 8*
a*x*log(a*x - 1)^2 - (3*a*x*log(a*x - 1) - 8*a*x)*log(a*x + 1)^2 + (3*a*x*log(a*x - 1)^2 - 16*a*x*log(a*x - 1)
)*log(a*x + 1) - 8)/x)*a^2 + 2*(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*arccoth(a*x))*a - 1/4*arcco
th(a*x)^3/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (a x\right )^{3}}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3/x^5,x, algorithm="fricas")

[Out]

integral(arccoth(a*x)^3/x^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}^{3}{\left (a x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**3/x**5,x)

[Out]

Integral(acoth(a*x)**3/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )^{3}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3/x^5,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)^3/x^5, x)