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

Optimal. Leaf size=41 \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)+\frac{1}{2} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

[Out]

-(ArcTanh[a*x]/x) + (a*ArcTanh[a*x]^2)/2 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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Rubi [A]  time = 0.077544, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {5982, 5916, 266, 36, 29, 31, 5948} \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)+\frac{1}{2} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[a*x]/x) + (a*ArcTanh[a*x]^2)/2 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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 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 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 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 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[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{\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac{\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{x}+\frac{1}{2} a \tanh ^{-1}(a x)^2+a \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{x}+\frac{1}{2} a \tanh ^{-1}(a x)^2+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{\tanh ^{-1}(a x)}{x}+\frac{1}{2} a \tanh ^{-1}(a x)^2+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} a^3 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{\tanh ^{-1}(a x)}{x}+\frac{1}{2} a \tanh ^{-1}(a x)^2+a \log (x)-\frac{1}{2} a \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0402725, size = 41, normalized size = 1. \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)+\frac{1}{2} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[a*x]/x) + (a*ArcTanh[a*x]^2)/2 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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Maple [B]  time = 0.056, size = 132, normalized size = 3.2 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{x}}-{\frac{a{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2}}+{\frac{a{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}-{\frac{a \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8}}+{\frac{a\ln \left ( ax-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{a\ln \left ( ax-1 \right ) }{2}}+a\ln \left ( ax \right ) -{\frac{a\ln \left ( ax+1 \right ) }{2}}+{\frac{a\ln \left ( ax+1 \right ) }{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{a}{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{a \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.959937, size = 111, normalized size = 2.71 \begin{align*} \frac{1}{8} \,{\left (2 \,{\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right ) + 8 \, \log \left (x\right )\right )} a + \frac{1}{2} \,{\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac{2}{x}\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^2/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.1324, size = 150, normalized size = 3.66 \begin{align*} \frac{a x \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 4 \, a x \log \left (a^{2} x^{2} - 1\right ) + 8 \, a x \log \left (x\right ) - 4 \, \log \left (-\frac{a x + 1}{a x - 1}\right )}{8 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.38263, size = 37, normalized size = 0.9 \begin{align*} \begin{cases} a \log{\left (x \right )} - a \log{\left (x - \frac{1}{a} \right )} + \frac{a \operatorname{atanh}^{2}{\left (a x \right )}}{2} - a \operatorname{atanh}{\left (a x \right )} - \frac{\operatorname{atanh}{\left (a x \right )}}{x} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*log(x) - a*log(x - 1/a) + a*atanh(a*x)**2/2 - a*atanh(a*x) - atanh(a*x)/x, Ne(a, 0)), (0, True))

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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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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