3.434 \(\int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=70 \[ -\frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0816485, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6008, 266, 47, 63, 208} \[ -\frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^4,x]

[Out]

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

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && 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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x^4} \, dx &=-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{3} a \int \frac{\sqrt{1-a^2 x^2}}{x^3} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-\frac{1}{12} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )\\ &=-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0719546, size = 79, normalized size = 1.13 \[ -\frac{a x \sqrt{1-a^2 x^2}+a^3 x^3 \log (x)-a^3 x^3 \log \left (\sqrt{1-a^2 x^2}+1\right )+2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.236, size = 96, normalized size = 1.4 \begin{align*}{\frac{2\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) -ax-2\,{\it Artanh} \left ( ax \right ) }{6\,{x}^{3}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{a}^{3}}{6}\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{a}^{3}}{6}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x)

[Out]

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

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Maxima [A]  time = 1.43899, size = 122, normalized size = 1.74 \begin{align*} \frac{1}{6} \,{\left (a^{2} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - \sqrt{-a^{2} x^{2} + 1} a^{2} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{x^{2}}\right )} a - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.04219, size = 163, normalized size = 2.33 \begin{align*} -\frac{a^{3} x^{3} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + \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 )}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname{atanh}{\left (a x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)*(-a**2*x**2+1)**(1/2)/x**4,x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)/x**4, x)

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Giac [B]  time = 1.27278, size = 263, normalized size = 3.76 \begin{align*} \frac{1}{12} \, a^{3} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) - \frac{1}{12} \, a^{3} \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{1}{48} \,{\left (\frac{{\left (a^{4} - \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{x^{2}}\right )} a^{6} x^{3}}{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left | a \right |}} + \frac{\frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4}}{x} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{x^{3}}}{a^{2}{\left | a \right |}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{\sqrt{-a^{2} x^{2} + 1} a}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="giac")

[Out]

1/12*a^3*log(sqrt(-a^2*x^2 + 1) + 1) - 1/12*a^3*log(-sqrt(-a^2*x^2 + 1) + 1) + 1/48*((a^4 - 3*(sqrt(-a^2*x^2 +
1)*abs(a) + a)^2/x^2)*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) + (3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
*a^4/x - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/x^3)/(a^2*abs(a)))*log(-(a*x + 1)/(a*x - 1)) - 1/6*sqrt(-a^2*x^2 +
1)*a/x^2