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

Optimal. Leaf size=57 \[ -\frac{1}{4 a^3 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{4 a^3} \]

[Out]

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

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Rubi [A]  time = 0.0628238, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {5998, 5948} \[ -\frac{1}{4 a^3 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5998

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(
q + 1))/(4*c^3*d*(q + 1)^2), x] + (Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x],
x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

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

Mathematica [A]  time = 0.0961048, size = 45, normalized size = 0.79 \[ \frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-2 a x \tanh ^{-1}(a x)+1}{4 a^3 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.053, size = 169, normalized size = 3. \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{4\,{a}^{3} \left ( ax-1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{4\,{a}^{3}}}-{\frac{{\it Artanh} \left ( ax \right ) }{4\,{a}^{3} \left ( ax+1 \right ) }}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{4\,{a}^{3}}}+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{16\,{a}^{3}}}-{\frac{\ln \left ( ax+1 \right ) }{8\,{a}^{3}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{8\,{a}^{3}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{16\,{a}^{3}}}-{\frac{\ln \left ( ax-1 \right ) }{8\,{a}^{3}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{1}{8\,{a}^{3} \left ( ax-1 \right ) }}-{\frac{1}{8\,{a}^{3} \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^3*arctanh(a*x)/(a*x-1)+1/4/a^3*arctanh(a*x)*ln(a*x-1)-1/4/a^3*arctanh(a*x)/(a*x+1)-1/4/a^3*arctanh(a*x)
*ln(a*x+1)+1/16/a^3*ln(a*x+1)^2-1/8/a^3*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/8/a^3*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/
16/a^3*ln(a*x-1)^2-1/8/a^3*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8/a^3/(a*x-1)-1/8/a^3/(a*x+1)

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Maxima [B]  time = 0.964538, size = 170, normalized size = 2.98 \begin{align*} -\frac{1}{4} \,{\left (\frac{2 \, x}{a^{4} x^{2} - 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{{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 4\right )} a}{16 \,{\left (a^{6} x^{2} - a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^4*x^2 - a^2) + log(a*x + 1)/a^3 - log(a*x - 1)/a^3)*arctanh(a*x) + 1/16*((a^2*x^2 - 1)*log(a*x +
1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 + 4)*a/(a^6*x^2 - a^4)

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Fricas [A]  time = 2.00204, size = 142, normalized size = 2.49 \begin{align*} -\frac{4 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) +{\left (a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 4}{16 \,{\left (a^{5} x^{2} - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a*x*log(-(a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4)/(a^5*x^2 - a^3)

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

[Out]

Integral(x**2*atanh(a*x)/((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{x^{2} \operatorname{artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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