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

Optimal. Leaf size=64 \[ \frac{1-a^2 x^2}{6 a}+\frac{\log \left (1-a^2 x^2\right )}{3 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2}{3} x \tanh ^{-1}(a x) \]

[Out]

(1 - a^2*x^2)/(6*a) + (2*x*ArcTanh[a*x])/3 + (x*(1 - a^2*x^2)*ArcTanh[a*x])/3 + Log[1 - a^2*x^2]/(3*a)

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Rubi [A]  time = 0.0212375, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5942, 5910, 260} \[ \frac{1-a^2 x^2}{6 a}+\frac{\log \left (1-a^2 x^2\right )}{3 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2}{3} x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

(1 - a^2*x^2)/(6*a) + (2*x*ArcTanh[a*x])/3 + (x*(1 - a^2*x^2)*ArcTanh[a*x])/3 + Log[1 - a^2*x^2]/(3*a)

Rule 5942

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0089424, size = 47, normalized size = 0.73 \[ \frac{\log \left (1-a^2 x^2\right )}{3 a}-\frac{1}{3} a^2 x^3 \tanh ^{-1}(a x)-\frac{a x^2}{6}+x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x^2)/6 + x*ArcTanh[a*x] - (a^2*x^3*ArcTanh[a*x])/3 + Log[1 - a^2*x^2]/(3*a)

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Maple [A]  time = 0.026, size = 48, normalized size = 0.8 \begin{align*} -{\frac{{a}^{2}{\it Artanh} \left ( ax \right ){x}^{3}}{3}}+x{\it Artanh} \left ( ax \right ) -{\frac{a{x}^{2}}{6}}+{\frac{\ln \left ( ax-1 \right ) }{3\,a}}+{\frac{\ln \left ( ax+1 \right ) }{3\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*arctanh(a*x)*x^3+x*arctanh(a*x)-1/6*a*x^2+1/3/a*ln(a*x-1)+1/3/a*ln(a*x+1)

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Maxima [A]  time = 0.954495, size = 63, normalized size = 0.98 \begin{align*} -\frac{1}{6} \,{\left (x^{2} - \frac{2 \, \log \left (a x + 1\right )}{a^{2}} - \frac{2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a - \frac{1}{3} \,{\left (a^{2} x^{3} - 3 \, x\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.21157, size = 115, normalized size = 1.8 \begin{align*} -\frac{a^{2} x^{2} +{\left (a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 2 \, \log \left (a^{2} x^{2} - 1\right )}{6 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.995619, size = 49, normalized size = 0.77 \begin{align*} \begin{cases} - \frac{a^{2} x^{3} \operatorname{atanh}{\left (a x \right )}}{3} - \frac{a x^{2}}{6} + x \operatorname{atanh}{\left (a x \right )} + \frac{2 \log{\left (x - \frac{1}{a} \right )}}{3 a} + \frac{2 \operatorname{atanh}{\left (a x \right )}}{3 a} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**3*atanh(a*x)/3 - a*x**2/6 + x*atanh(a*x) + 2*log(x - 1/a)/(3*a) + 2*atanh(a*x)/(3*a), Ne(a
, 0)), (0, True))

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Giac [A]  time = 1.16982, size = 69, normalized size = 1.08 \begin{align*} -\frac{1}{6} \, a x^{2} - \frac{1}{6} \,{\left (a^{2} x^{3} - 3 \, x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{\log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{3 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*x^2 - 1/6*(a^2*x^3 - 3*x)*log(-(a*x + 1)/(a*x - 1)) + 1/3*log(abs(a^2*x^2 - 1))/a