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

Optimal. Leaf size=82 \[ \frac{1}{4 a^2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{4 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0679388, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5994, 5956, 261} \[ \frac{1}{4 a^2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5956

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

Rule 261

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

Rubi steps

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

Mathematica [A]  time = 0.0512578, size = 43, normalized size = 0.52 \[ \frac{\left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2-2 a x \tanh ^{-1}(a x)+1}{4 a^2-4 a^4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTanh[a*x]^2)/(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^2 - 4*a^4*x^2)

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Maple [B]  time = 0.055, size = 191, normalized size = 2.3 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{a}^{2} \left ({a}^{2}{x}^{2}-1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) }{4\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{4\,{a}^{2}}}+{\frac{{\it Artanh} \left ( ax \right ) }{4\,{a}^{2} \left ( ax+1 \right ) }}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{4\,{a}^{2}}}+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{16\,{a}^{2}}}+{\frac{1}{8\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax+1 \right ) }{8\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{16\,{a}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{8\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{1}{8\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{1}{8\,{a}^{2} \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.966322, size = 197, normalized size = 2.4 \begin{align*} \frac{{\left (\frac{2 \, x}{a^{2} x^{2} - 1} - \frac{\log \left (a x + 1\right )}{a} + \frac{\log \left (a x - 1\right )}{a}\right )} \operatorname{artanh}\left (a x\right )}{4 \, a} + \frac{{\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}{16 \,{\left (a^{4} x^{2} - a^{2}\right )}} - \frac{\operatorname{artanh}\left (a x\right )^{2}}{2 \,{\left (a^{2} x^{2} - 1\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.09665, size = 140, normalized size = 1.71 \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^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)^2/(-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^4*x^2 - a^2)

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

[Out]

Integral(x*atanh(a*x)**2/((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 \operatorname{artanh}\left (a x\right )^{2}}{{\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*arctanh(a*x)^2/(-a^2*x^2+1)^2,x, algorithm="giac")

[Out]

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