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

Optimal. Leaf size=100 \[ \frac{1}{16 a^3 \left (1-a^2 x^2\right )}-\frac{1}{16 a^3 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{\tanh ^{-1}(a x)^2}{16 a^3} \]

[Out]

-1/(16*a^3*(1 - a^2*x^2)^2) + 1/(16*a^3*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(4*a^2*(1 - a^2*x^2)^2) - (x*ArcTanh
[a*x])/(8*a^2*(1 - a^2*x^2)) - ArcTanh[a*x]^2/(16*a^3)

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Rubi [A]  time = 0.071921, antiderivative size = 100, 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 = {5998, 5956, 261} \[ \frac{1}{16 a^3 \left (1-a^2 x^2\right )}-\frac{1}{16 a^3 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{\tanh ^{-1}(a x)^2}{16 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(16*a^3*(1 - a^2*x^2)^2) + 1/(16*a^3*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(4*a^2*(1 - a^2*x^2)^2) - (x*ArcTanh
[a*x])/(8*a^2*(1 - a^2*x^2)) - ArcTanh[a*x]^2/(16*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 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^2 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx &=-\frac{1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{\int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{4 a^2}\\ &=-\frac{1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{16 a^3}+\frac{\int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx}{8 a}\\ &=-\frac{1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac{1}{16 a^3 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)^2}{16 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0840135, size = 61, normalized size = 0.61 \[ -\frac{a^2 x^2+\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2-2 \left (a^3 x^3+a x\right ) \tanh ^{-1}(a x)}{16 a^3 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x^2 - 2*(a*x + a^3*x^3)*ArcTanh[a*x] + (-1 + a^2*x^2)^2*ArcTanh[a*x]^2)/(16*a^3*(-1 + a^2*x^2)^2)

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Maple [B]  time = 0.057, size = 225, normalized size = 2.3 \begin{align*}{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{3} \left ( ax-1 \right ) ^{2}}}+{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{3} \left ( ax-1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{16\,{a}^{3}}}-{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{3} \left ( ax+1 \right ) ^{2}}}+{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{3} \left ( ax+1 \right ) }}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{16\,{a}^{3}}}+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{64\,{a}^{3}}}-{\frac{\ln \left ( ax+1 \right ) }{32\,{a}^{3}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{32\,{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}}{64\,{a}^{3}}}-{\frac{\ln \left ( ax-1 \right ) }{32\,{a}^{3}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{1}{64\,{a}^{3} \left ( ax-1 \right ) ^{2}}}-{\frac{1}{64\,{a}^{3} \left ( ax-1 \right ) }}-{\frac{1}{64\,{a}^{3} \left ( ax+1 \right ) ^{2}}}+{\frac{1}{64\,{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)^3,x)

[Out]

1/16/a^3*arctanh(a*x)/(a*x-1)^2+1/16/a^3*arctanh(a*x)/(a*x-1)+1/16/a^3*arctanh(a*x)*ln(a*x-1)-1/16/a^3*arctanh
(a*x)/(a*x+1)^2+1/16/a^3*arctanh(a*x)/(a*x+1)-1/16/a^3*arctanh(a*x)*ln(a*x+1)+1/64/a^3*ln(a*x+1)^2-1/32/a^3*ln
(-1/2*a*x+1/2)*ln(a*x+1)+1/32/a^3*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/64/a^3*ln(a*x-1)^2-1/32/a^3*ln(a*x-1)*ln(
1/2+1/2*a*x)-1/64/a^3/(a*x-1)^2-1/64/a^3/(a*x-1)-1/64/a^3/(a*x+1)^2+1/64/a^3/(a*x+1)

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Maxima [B]  time = 0.985418, size = 242, normalized size = 2.42 \begin{align*} \frac{1}{16} \,{\left (\frac{2 \,{\left (a^{2} x^{3} + x\right )}}{a^{6} x^{4} - 2 \, 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 (4 \, a^{2} x^{2} -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2}\right )} a}{64 \,{\left (a^{8} x^{4} - 2 \, 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)^3,x, algorithm="maxima")

[Out]

1/16*(2*(a^2*x^3 + x)/(a^6*x^4 - 2*a^4*x^2 + a^2) - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)*arctanh(a*x) - 1/64*(
4*a^2*x^2 - (a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*x - 1) -
(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2)*a/(a^8*x^4 - 2*a^6*x^2 + a^4)

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Fricas [A]  time = 1.93931, size = 201, normalized size = 2.01 \begin{align*} -\frac{4 \, a^{2} x^{2} +{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 4 \,{\left (a^{3} x^{3} + a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{64 \,{\left (a^{7} x^{4} - 2 \, 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)^3,x, algorithm="fricas")

[Out]

-1/64*(4*a^2*x^2 + (a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(a^3*x^3 + a*x)*log(-(a*x + 1)/(a
*x - 1)))/(a^7*x^4 - 2*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 )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, 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)**3,x)

[Out]

-Integral(x**2*atanh(a*x)/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), 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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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