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

Optimal. Leaf size=42 \[ -\frac{\log \left (1-a^2 x^2\right )}{2 a^3}+\frac{\tanh ^{-1}(a x)^2}{2 a^3}-\frac{x \tanh ^{-1}(a x)}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.0692734, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5980, 5910, 260, 5948} \[ -\frac{\log \left (1-a^2 x^2\right )}{2 a^3}+\frac{\tanh ^{-1}(a x)^2}{2 a^3}-\frac{x \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5980

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

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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

Mathematica [A]  time = 0.0370791, size = 42, normalized size = 1. \[ -\frac{\log \left (1-a^2 x^2\right )}{2 a^3}+\frac{\tanh ^{-1}(a x)^2}{2 a^3}-\frac{x \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.049, size = 145, normalized size = 3.5 \begin{align*} -{\frac{x{\it Artanh} \left ( ax \right ) }{{a}^{2}}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2\,{a}^{3}}}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2\,{a}^{3}}}-{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8\,{a}^{3}}}+{\frac{\ln \left ( ax-1 \right ) }{4\,{a}^{3}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax-1 \right ) }{2\,{a}^{3}}}-{\frac{\ln \left ( ax+1 \right ) }{2\,{a}^{3}}}-{\frac{1}{4\,{a}^{3}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{3}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.970101, size = 115, normalized size = 2.74 \begin{align*} -\frac{1}{2} \,{\left (\frac{2 \, x}{a^{2}} - \frac{\log \left (a x + 1\right )}{a^{3}} + \frac{\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname{artanh}\left (a x\right ) + \frac{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 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.09577, size = 128, normalized size = 3.05 \begin{align*} -\frac{4 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) - \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \, \log \left (a^{2} x^{2} - 1\right )}{8 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.61955, size = 41, normalized size = 0.98 \begin{align*} \begin{cases} - \frac{x \operatorname{atanh}{\left (a x \right )}}{a^{2}} - \frac{\log{\left (x - \frac{1}{a} \right )}}{a^{3}} + \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{2 a^{3}} - \frac{\operatorname{atanh}{\left (a x \right )}}{a^{3}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x*atanh(a*x)/a**2 - log(x - 1/a)/a**3 + atanh(a*x)**2/(2*a**3) - atanh(a*x)/a**3, Ne(a, 0)), (0, T
rue))

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Giac [A]  time = 1.19701, size = 80, normalized size = 1.9 \begin{align*} -\frac{x \log \left (-\frac{a x + 1}{a x - 1}\right )}{2 \, a^{2}} + \frac{\log \left (-\frac{a x + 1}{a x - 1}\right )^{2}}{8 \, a^{3}} - \frac{\log \left (a^{2} x^{2} - 1\right )}{2 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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