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

Optimal. Leaf size=87 \[ -\frac{x \sqrt{1-a^2 x^2}}{6 a^3}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}+\frac{5 \sin ^{-1}(a x)}{6 a^4} \]

[Out]

-(x*Sqrt[1 - a^2*x^2])/(6*a^3) + (5*ArcSin[a*x])/(6*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^4) - (x^2*S
qrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^2)

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Rubi [A]  time = 0.124237, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6016, 321, 216, 5994} \[ -\frac{x \sqrt{1-a^2 x^2}}{6 a^3}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}+\frac{5 \sin ^{-1}(a x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Sqrt[1 - a^2*x^2])/(6*a^3) + (5*ArcSin[a*x])/(6*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^4) - (x^2*S
qrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^2)

Rule 6016

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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]

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]

Rubi steps

\begin{align*} \int \frac{x^3 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx &=-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}+\frac{2 \int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{3 a^2}+\frac{\int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{3 a}\\ &=-\frac{x \sqrt{1-a^2 x^2}}{6 a^3}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{6 a^3}+\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{3 a^3}\\ &=-\frac{x \sqrt{1-a^2 x^2}}{6 a^3}+\frac{5 \sin ^{-1}(a x)}{6 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0688944, size = 60, normalized size = 0.69 \[ -\frac{a x \sqrt{1-a^2 x^2}+2 \sqrt{1-a^2 x^2} \left (a^2 x^2+2\right ) \tanh ^{-1}(a x)-5 \sin ^{-1}(a x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x*Sqrt[1 - a^2*x^2] - 5*ArcSin[a*x] + 2*Sqrt[1 - a^2*x^2]*(2 + a^2*x^2)*ArcTanh[a*x])/(6*a^4)

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Maple [C]  time = 0.236, size = 99, normalized size = 1.1 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) +ax+4\,{\it Artanh} \left ( ax \right ) }{6\,{a}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\frac{5\,i}{6}}}{{a}^{4}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+i \right ) }-{\frac{{\frac{5\,i}{6}}}{{a}^{4}}\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^3*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/6/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(2*a^2*x^2*arctanh(a*x)+a*x+4*arctanh(a*x))+5/6*I*ln((a*x+1)/(-a^2*x^2+1)^(1
/2)+I)/a^4-5/6*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^4

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Maxima [A]  time = 1.44765, size = 151, normalized size = 1.74 \begin{align*} -\frac{1}{6} \, a{\left (\frac{\frac{\sqrt{-a^{2} x^{2} + 1} x}{a^{2}} - \frac{\arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{2}}}{a^{2}} - \frac{4 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{4}}\right )} - \frac{1}{3} \,{\left (\frac{\sqrt{-a^{2} x^{2} + 1} x^{2}}{a^{2}} + \frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*((sqrt(-a^2*x^2 + 1)*x/a^2 - arcsin(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2))/a^2 - 4*arcsin(a^2*x/sqrt(a^2))/(
sqrt(a^2)*a^4)) - 1/3*(sqrt(-a^2*x^2 + 1)*x^2/a^2 + 2*sqrt(-a^2*x^2 + 1)/a^4)*arctanh(a*x)

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Fricas [A]  time = 2.18607, size = 166, normalized size = 1.91 \begin{align*} -\frac{\sqrt{-a^{2} x^{2} + 1}{\left (a x +{\left (a^{2} x^{2} + 2\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} + 10 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*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 + 2)*log(-(a*x + 1)/(a*x - 1))) + 10*arctan((sqrt(-a^2*x^2 + 1) - 1)/
(a*x)))/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22851, size = 109, normalized size = 1.25 \begin{align*} -\frac{\sqrt{-a^{2} x^{2} + 1} x}{6 \, a^{3}} + \frac{{\left ({\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{-a^{2} x^{2} + 1}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{6 \, a^{4}} + \frac{5 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{6 \, a^{3}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(-a^2*x^2 + 1)*x/a^3 + 1/6*((-a^2*x^2 + 1)^(3/2) - 3*sqrt(-a^2*x^2 + 1))*log(-(a*x + 1)/(a*x - 1))/a^
4 + 5/6*arcsin(a*x)*sgn(a)/(a^3*abs(a))