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

Optimal. Leaf size=59 \[ \frac{x \sqrt{1-a^2 x^2}}{6 a}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 a^2}+\frac{\sin ^{-1}(a x)}{6 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0485783, antiderivative size = 59, 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, 195, 216} \[ \frac{x \sqrt{1-a^2 x^2}}{6 a}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 a^2}+\frac{\sin ^{-1}(a x)}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]

[Out]

(x*Sqrt[1 - a^2*x^2])/(6*a) + ArcSin[a*x]/(6*a^2) - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/(3*a^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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0420423, size = 49, normalized size = 0.83 \[ \frac{a x \sqrt{1-a^2 x^2}-2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\sin ^{-1}(a x)}{6 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.233, size = 99, normalized size = 1.7 \begin{align*}{\frac{2\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) +ax-2\,{\it Artanh} \left ( ax \right ) }{6\,{a}^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\frac{i}{6}}}{{a}^{2}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+i \right ) }-{\frac{{\frac{i}{6}}}{{a}^{2}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.43938, size = 80, normalized size = 1.36 \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{3 \, a^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} x + \frac{\arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}}}{6 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.06703, size = 163, normalized size = 2.76 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1}{\left (a x +{\left (a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} - 2 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x), x)

________________________________________________________________________________________

Giac [A]  time = 1.2134, size = 86, normalized size = 1.46 \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \log \left (-\frac{a x + 1}{a x - 1}\right )}{6 \, a^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} x + \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}}}{6 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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