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

Optimal. Leaf size=81 \[ \frac{x \left (1-a^2 x^2\right )^{3/2}}{20 a}+\frac{3 x \sqrt{1-a^2 x^2}}{40 a}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}+\frac{3 \sin ^{-1}(a x)}{40 a^2} \]

[Out]

(3*x*Sqrt[1 - a^2*x^2])/(40*a) + (x*(1 - a^2*x^2)^(3/2))/(20*a) + (3*ArcSin[a*x])/(40*a^2) - ((1 - a^2*x^2)^(5
/2)*ArcTanh[a*x])/(5*a^2)

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Rubi [A]  time = 0.0575997, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5994, 195, 216} \[ \frac{x \left (1-a^2 x^2\right )^{3/2}}{20 a}+\frac{3 x \sqrt{1-a^2 x^2}}{40 a}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}+\frac{3 \sin ^{-1}(a x)}{40 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*Sqrt[1 - a^2*x^2])/(40*a) + (x*(1 - a^2*x^2)^(3/2))/(20*a) + (3*ArcSin[a*x])/(40*a^2) - ((1 - a^2*x^2)^(5
/2)*ArcTanh[a*x])/(5*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 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}+\frac{\int \left (1-a^2 x^2\right )^{3/2} \, dx}{5 a}\\ &=\frac{x \left (1-a^2 x^2\right )^{3/2}}{20 a}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}+\frac{3 \int \sqrt{1-a^2 x^2} \, dx}{20 a}\\ &=\frac{3 x \sqrt{1-a^2 x^2}}{40 a}+\frac{x \left (1-a^2 x^2\right )^{3/2}}{20 a}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}+\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{40 a}\\ &=\frac{3 x \sqrt{1-a^2 x^2}}{40 a}+\frac{x \left (1-a^2 x^2\right )^{3/2}}{20 a}+\frac{3 \sin ^{-1}(a x)}{40 a^2}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0648409, size = 61, normalized size = 0.75 \[ \frac{a x \left (5-2 a^2 x^2\right ) \sqrt{1-a^2 x^2}-8 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+3 \sin ^{-1}(a x)}{40 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(5 - 2*a^2*x^2)*Sqrt[1 - a^2*x^2] + 3*ArcSin[a*x] - 8*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/(40*a^2)

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Maple [C]  time = 0.195, size = 120, normalized size = 1.5 \begin{align*} -{\frac{8\,{a}^{4}{x}^{4}{\it Artanh} \left ( ax \right ) +2\,{x}^{3}{a}^{3}-16\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) -5\,ax+8\,{\it Artanh} \left ( ax \right ) }{40\,{a}^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\frac{3\,i}{40}}}{{a}^{2}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+i \right ) }-{\frac{{\frac{3\,i}{40}}}{{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*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x)

[Out]

-1/40/a^2*(-(a*x-1)*(a*x+1))^(1/2)*(8*a^4*x^4*arctanh(a*x)+2*x^3*a^3-16*a^2*x^2*arctanh(a*x)-5*a*x+8*arctanh(a
*x))+3/40*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)+I)/a^2-3/40*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^2

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Maxima [A]  time = 1.43464, size = 103, normalized size = 1.27 \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} \operatorname{artanh}\left (a x\right )}{5 \, a^{2}} + \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x + 3 \, \sqrt{-a^{2} x^{2} + 1} x + \frac{3 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}}}{40 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="maxima")

[Out]

-1/5*(-a^2*x^2 + 1)^(5/2)*arctanh(a*x)/a^2 + 1/40*(2*(-a^2*x^2 + 1)^(3/2)*x + 3*sqrt(-a^2*x^2 + 1)*x + 3*arcsi
n(a^2*x/sqrt(a^2))/sqrt(a^2))/a

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Fricas [A]  time = 1.44603, size = 204, normalized size = 2.52 \begin{align*} -\frac{{\left (2 \, a^{3} x^{3} - 5 \, a x + 4 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{-a^{2} x^{2} + 1} + 6 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{40 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="fricas")

[Out]

-1/40*((2*a^3*x^3 - 5*a*x + 4*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1) + 6*arct
an((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}} \operatorname{atanh}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a**2*x**2+1)**(3/2)*atanh(a*x),x)

[Out]

Integral(x*(-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x), x)

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Giac [A]  time = 1.181, size = 116, normalized size = 1.43 \begin{align*} -\frac{{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1} \log \left (-\frac{a x + 1}{a x - 1}\right )}{10 \, a^{2}} - \frac{{\left (2 \, a^{2} x^{2} - 5\right )} \sqrt{-a^{2} x^{2} + 1} x - \frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}}}{40 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="giac")

[Out]

-1/10*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))/a^2 - 1/40*((2*a^2*x^2 - 5)*sqrt(-a^2*x^2 +
1)*x - 3*arcsin(a*x)*sgn(a)/abs(a))/a