3.265 \(\int \frac{\tanh ^{-1}(a x)}{x^3 (1-a^2 x^2)^2} \, dx\)

Optimal. Leaf size=123 \[ -a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a^3 x}{4 \left (1-a^2 x^2\right )}+\frac{a^2 \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+a^2 \tanh ^{-1}(a x)^2+\frac{1}{4} a^2 \tanh ^{-1}(a x)+2 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.358372, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6030, 5982, 5916, 325, 206, 5988, 5932, 2447, 5994, 199} \[ -a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a^3 x}{4 \left (1-a^2 x^2\right )}+\frac{a^2 \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+a^2 \tanh ^{-1}(a x)^2+\frac{1}{4} a^2 \tanh ^{-1}(a x)+2 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^2),x]

[Out]

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

Rule 6030

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int
[x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTanh
[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[
m, 0] && NeQ[p, -1]

Rule 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[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 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 5988

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[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 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[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 5994

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rubi steps

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

Mathematica [A]  time = 0.429802, size = 83, normalized size = 0.67 \[ \frac{1}{8} a^2 \left (-8 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \left (-\frac{2}{a^2 x^2}+8 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+\cosh \left (2 \tanh ^{-1}(a x)\right )+2\right )-\frac{4}{a x}+8 \tanh ^{-1}(a x)^2-\sinh \left (2 \tanh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^2),x]

[Out]

(a^2*(-4/(a*x) + 8*ArcTanh[a*x]^2 + 2*ArcTanh[a*x]*(2 - 2/(a^2*x^2) + Cosh[2*ArcTanh[a*x]] + 8*Log[1 - E^(-2*A
rcTanh[a*x])]) - 8*PolyLog[2, E^(-2*ArcTanh[a*x])] - Sinh[2*ArcTanh[a*x]]))/8

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Maple [B]  time = 0.066, size = 265, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{4\,ax-4}}-{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) -{\frac{{\it Artanh} \left ( ax \right ) }{2\,{x}^{2}}}+2\,{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) +{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{4\,ax+4}}-{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) -{a}^{2}{\it dilog} \left ( ax \right ) -{a}^{2}{\it dilog} \left ( ax+1 \right ) -{a}^{2}\ln \left ( ax \right ) \ln \left ( ax+1 \right ) -{\frac{{a}^{2} \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{4}}+{a}^{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) +{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{2}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{2} \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{4}}-{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{2}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{{a}^{2}}{2}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{2}}{8\,ax-8}}-{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{8}}-{\frac{a}{2\,x}}+{\frac{{a}^{2}}{8\,ax+8}}+{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/x^3/(-a^2*x^2+1)^2,x)

[Out]

-1/4*a^2*arctanh(a*x)/(a*x-1)-a^2*arctanh(a*x)*ln(a*x-1)-1/2*arctanh(a*x)/x^2+2*a^2*arctanh(a*x)*ln(a*x)+1/4*a
^2*arctanh(a*x)/(a*x+1)-a^2*arctanh(a*x)*ln(a*x+1)-a^2*dilog(a*x)-a^2*dilog(a*x+1)-a^2*ln(a*x)*ln(a*x+1)-1/4*a
^2*ln(a*x-1)^2+a^2*dilog(1/2+1/2*a*x)+1/2*a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/4*a^2*ln(a*x+1)^2-1/2*a^2*ln(-1/2*a*
x+1/2)*ln(a*x+1)+1/2*a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/8*a^2/(a*x-1)-1/8*a^2*ln(a*x-1)-1/2*a/x+1/8*a^2/(a
*x+1)+1/8*a^2*ln(a*x+1)

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Maxima [B]  time = 0.990467, size = 315, normalized size = 2.56 \begin{align*} \frac{1}{8} \,{\left (8 \,{\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 - 8 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )} a + 8 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )} a + a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac{2 \,{\left (a^{2} x^{2} -{\left (a^{3} x^{3} - a x\right )} \log \left (a x + 1\right )^{2} + 2 \,{\left (a^{3} x^{3} - a x\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) +{\left (a^{3} x^{3} - a x\right )} \log \left (a x - 1\right )^{2} - 2\right )}}{a^{2} x^{3} - x}\right )} a - \frac{1}{2} \,{\left (2 \, a^{2} \log \left (a^{2} x^{2} - 1\right ) - 2 \, a^{2} \log \left (x^{2}\right ) + \frac{2 \, a^{2} x^{2} - 1}{a^{2} x^{4} - x^{2}}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^2,x, algorithm="maxima")

[Out]

1/8*(8*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))*a - 8*(log(a*x + 1)*log(x) + dilog(-a*x))*a +
8*(log(-a*x + 1)*log(x) + dilog(a*x))*a + a*log(a*x + 1) - a*log(a*x - 1) - 2*(a^2*x^2 - (a^3*x^3 - a*x)*log(
a*x + 1)^2 + 2*(a^3*x^3 - a*x)*log(a*x + 1)*log(a*x - 1) + (a^3*x^3 - a*x)*log(a*x - 1)^2 - 2)/(a^2*x^3 - x))*
a - 1/2*(2*a^2*log(a^2*x^2 - 1) - 2*a^2*log(x^2) + (2*a^2*x^2 - 1)/(a^2*x^4 - x^2))*arctanh(a*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{artanh}\left (a x\right )}{a^{4} x^{7} - 2 \, a^{2} x^{5} + x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

integral(arctanh(a*x)/(a^4*x^7 - 2*a^2*x^5 + x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}{\left (a x \right )}}{x^{3} \left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/x**3/(-a**2*x**2+1)**2,x)

[Out]

Integral(atanh(a*x)/(x**3*(a*x - 1)**2*(a*x + 1)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^2,x, algorithm="giac")

[Out]

integrate(arctanh(a*x)/((a^2*x^2 - 1)^2*x^3), x)