3.457 \(\int \frac{(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=94 \[ \frac{3 a^3 \sqrt{1-a^2 x^2}}{40 x^2}-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5} \]

[Out]

(3*a^3*Sqrt[1 - a^2*x^2])/(40*x^2) - (a*(1 - a^2*x^2)^(3/2))/(20*x^4) - ((1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/(5*
x^5) - (3*a^5*ArcTanh[Sqrt[1 - a^2*x^2]])/40

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Rubi [A]  time = 0.100864, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{3 a^3 \sqrt{1-a^2 x^2}}{40 x^2}-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^6,x]

[Out]

(3*a^3*Sqrt[1 - a^2*x^2])/(40*x^2) - (a*(1 - a^2*x^2)^(3/2))/(20*x^4) - ((1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/(5*
x^5) - (3*a^5*ArcTanh[Sqrt[1 - a^2*x^2]])/40

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{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{x^6} \, dx &=-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac{1}{5} a \int \frac{\left (1-a^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac{1}{10} a \operatorname{Subst}\left (\int \frac{\left (1-a^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac{1}{40} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac{3 a^3 \sqrt{1-a^2 x^2}}{40 x^2}-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac{1}{80} \left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac{3 a^3 \sqrt{1-a^2 x^2}}{40 x^2}-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac{1}{40} \left (3 a^3\right ) \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{3 a^3 \sqrt{1-a^2 x^2}}{40 x^2}-\frac{a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac{3}{40} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0818989, size = 104, normalized size = 1.11 \[ \left (\frac{a^3}{8 x^2}-\frac{a}{20 x^4}\right ) \sqrt{1-a^2 x^2}-\frac{3}{40} a^5 \log \left (\sqrt{1-a^2 x^2}+1\right )-\frac{\sqrt{1-a^2 x^2} \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)}{5 x^5}+\frac{3}{40} a^5 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^6,x]

[Out]

(-a/(20*x^4) + a^3/(8*x^2))*Sqrt[1 - a^2*x^2] - (Sqrt[1 - a^2*x^2]*(-1 + a^2*x^2)^2*ArcTanh[a*x])/(5*x^5) + (3
*a^5*Log[x])/40 - (3*a^5*Log[1 + Sqrt[1 - a^2*x^2]])/40

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Maple [A]  time = 0.169, size = 116, normalized size = 1.2 \begin{align*} -{\frac{8\,{a}^{4}{x}^{4}{\it Artanh} \left ( ax \right ) -5\,{x}^{3}{a}^{3}-16\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) +2\,ax+8\,{\it Artanh} \left ( ax \right ) }{40\,{x}^{5}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{3\,{a}^{5}}{40}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-1 \right ) }-{\frac{3\,{a}^{5}}{40}\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45061, size = 170, normalized size = 1.81 \begin{align*} \frac{1}{40} \,{\left ({\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{4} - 3 \, a^{4} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + 3 \, \sqrt{-a^{2} x^{2} + 1} a^{4} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} a^{2}}{x^{2}} - \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}}}{x^{4}}\right )} a - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} \operatorname{artanh}\left (a x\right )}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="maxima")

[Out]

1/40*((-a^2*x^2 + 1)^(3/2)*a^4 - 3*a^4*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 3*sqrt(-a^2*x^2 + 1)*a^4
+ (-a^2*x^2 + 1)^(5/2)*a^2/x^2 - 2*(-a^2*x^2 + 1)^(5/2)/x^4)*a - 1/5*(-a^2*x^2 + 1)^(5/2)*arctanh(a*x)/x^5

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Fricas [A]  time = 1.32788, size = 204, normalized size = 2.17 \begin{align*} \frac{3 \, a^{5} x^{5} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (5 \, a^{3} x^{3} - 2 \, 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}}{40 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**(3/2)*atanh(a*x)/x**6,x)

[Out]

Timed out

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Giac [B]  time = 1.82267, size = 374, normalized size = 3.98 \begin{align*} -\frac{3}{80} \, a^{5} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{3}{80} \, a^{5} \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{1}{320} \,{\left (\frac{{\left (a^{6} - \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2}}{x^{2}} + \frac{10 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{2} x^{4}}\right )} a^{10} x^{5}}{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}{\left | a \right |}} - \frac{\frac{10 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{8}}{x} - \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} a^{4}}{x^{3}} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}}{x^{5}}}{a^{4}{\left | a \right |}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{5 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{5} - 3 \, \sqrt{-a^{2} x^{2} + 1} a^{5}}{40 \, a^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="giac")

[Out]

-3/80*a^5*log(sqrt(-a^2*x^2 + 1) + 1) + 3/80*a^5*log(-sqrt(-a^2*x^2 + 1) + 1) + 1/320*((a^6 - 5*(sqrt(-a^2*x^2
+ 1)*abs(a) + a)^2*a^2/x^2 + 10*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^2*x^4))*a^10*x^5/((sqrt(-a^2*x^2 + 1)*ab
s(a) + a)^5*abs(a)) - (10*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^8/x - 5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*a^4/x^3
+ (sqrt(-a^2*x^2 + 1)*abs(a) + a)^5/x^5)/(a^4*abs(a)))*log(-(a*x + 1)/(a*x - 1)) - 1/40*(5*(-a^2*x^2 + 1)^(3/2
)*a^5 - 3*sqrt(-a^2*x^2 + 1)*a^5)/(a^4*x^4)