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

Optimal. Leaf size=62 \[ \frac{\log \left (1-a^2 x^2\right )}{15 a^3}-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{a x^4}{20}+\frac{x^2}{15 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

[Out]

x^2/(15*a) - (a*x^4)/20 + (x^3*ArcTanh[a*x])/3 - (a^2*x^5*ArcTanh[a*x])/5 + Log[1 - a^2*x^2]/(15*a^3)

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Rubi [A]  time = 0.0874176, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6014, 5916, 266, 43} \[ \frac{\log \left (1-a^2 x^2\right )}{15 a^3}-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{a x^4}{20}+\frac{x^2}{15 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

x^2/(15*a) - (a*x^4)/20 + (x^3*ArcTanh[a*x])/3 - (a^2*x^5*ArcTanh[a*x])/5 + Log[1 - a^2*x^2]/(15*a^3)

Rule 6014

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist
[d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d +
e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[q
, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^4 \tanh ^{-1}(a x) \, dx\right )+\int x^2 \tanh ^{-1}(a x) \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{1}{3} a \int \frac{x^3}{1-a^2 x^2} \, dx+\frac{1}{5} a^3 \int \frac{x^5}{1-a^2 x^2} \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{1}{6} a \operatorname{Subst}\left (\int \frac{x}{1-a^2 x} \, dx,x,x^2\right )+\frac{1}{10} a^3 \operatorname{Subst}\left (\int \frac{x^2}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{1}{6} a \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}-\frac{1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac{1}{10} a^3 \operatorname{Subst}\left (\int \left (-\frac{1}{a^4}-\frac{x}{a^2}-\frac{1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{15 a}-\frac{a x^4}{20}+\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac{\log \left (1-a^2 x^2\right )}{15 a^3}\\ \end{align*}

Mathematica [A]  time = 0.014658, size = 62, normalized size = 1. \[ \frac{\log \left (1-a^2 x^2\right )}{15 a^3}-\frac{1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{a x^4}{20}+\frac{x^2}{15 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(15*a) - (a*x^4)/20 + (x^3*ArcTanh[a*x])/3 - (a^2*x^5*ArcTanh[a*x])/5 + Log[1 - a^2*x^2]/(15*a^3)

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Maple [A]  time = 0.03, size = 59, normalized size = 1. \begin{align*} -{\frac{{a}^{2}{x}^{5}{\it Artanh} \left ( ax \right ) }{5}}+{\frac{{x}^{3}{\it Artanh} \left ( ax \right ) }{3}}-{\frac{{x}^{4}a}{20}}+{\frac{{x}^{2}}{15\,a}}+{\frac{\ln \left ( ax-1 \right ) }{15\,{a}^{3}}}+{\frac{\ln \left ( ax+1 \right ) }{15\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a^2*x^5*arctanh(a*x)+1/3*x^3*arctanh(a*x)-1/20*x^4*a+1/15*x^2/a+1/15/a^3*ln(a*x-1)+1/15/a^3*ln(a*x+1)

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Maxima [A]  time = 0.951199, size = 88, normalized size = 1.42 \begin{align*} -\frac{1}{60} \, a{\left (\frac{3 \, a^{2} x^{4} - 4 \, x^{2}}{a^{2}} - \frac{4 \, \log \left (a x + 1\right )}{a^{4}} - \frac{4 \, \log \left (a x - 1\right )}{a^{4}}\right )} - \frac{1}{15} \,{\left (3 \, a^{2} x^{5} - 5 \, x^{3}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*a*((3*a^2*x^4 - 4*x^2)/a^2 - 4*log(a*x + 1)/a^4 - 4*log(a*x - 1)/a^4) - 1/15*(3*a^2*x^5 - 5*x^3)*arctanh
(a*x)

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Fricas [A]  time = 2.18957, size = 149, normalized size = 2.4 \begin{align*} -\frac{3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 2 \,{\left (3 \, a^{5} x^{5} - 5 \, a^{3} x^{3}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 4 \, \log \left (a^{2} x^{2} - 1\right )}{60 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^4*x^4 - 4*a^2*x^2 + 2*(3*a^5*x^5 - 5*a^3*x^3)*log(-(a*x + 1)/(a*x - 1)) - 4*log(a^2*x^2 - 1))/a^3

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Sympy [A]  time = 1.83875, size = 63, normalized size = 1.02 \begin{align*} \begin{cases} - \frac{a^{2} x^{5} \operatorname{atanh}{\left (a x \right )}}{5} - \frac{a x^{4}}{20} + \frac{x^{3} \operatorname{atanh}{\left (a x \right )}}{3} + \frac{x^{2}}{15 a} + \frac{2 \log{\left (x - \frac{1}{a} \right )}}{15 a^{3}} + \frac{2 \operatorname{atanh}{\left (a x \right )}}{15 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*(-a**2*x**2+1)*atanh(a*x),x)

[Out]

Piecewise((-a**2*x**5*atanh(a*x)/5 - a*x**4/20 + x**3*atanh(a*x)/3 + x**2/(15*a) + 2*log(x - 1/a)/(15*a**3) +
2*atanh(a*x)/(15*a**3), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.16168, size = 95, normalized size = 1.53 \begin{align*} -\frac{1}{30} \,{\left (3 \, a^{2} x^{5} - 5 \, x^{3}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{\log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{15 \, a^{3}} - \frac{3 \, a^{5} x^{4} - 4 \, a^{3} x^{2}}{60 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(3*a^2*x^5 - 5*x^3)*log(-(a*x + 1)/(a*x - 1)) + 1/15*log(abs(a^2*x^2 - 1))/a^3 - 1/60*(3*a^5*x^4 - 4*a^3
*x^2)/a^4