3.436 \(\int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=150 \[ -\frac{a^3 \sqrt{1-a^2 x^2}}{24 x^2}-\frac{a \sqrt{1-a^2 x^2}}{20 x^4}+\frac{11}{120} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5} \]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(20*x^4) - (a^3*Sqrt[1 - a^2*x^2])/(24*x^2) - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(5*x^5)
+ (a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*x^3) + (2*a^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*x) + (11*a^5*ArcT
anh[Sqrt[1 - a^2*x^2]])/120

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Rubi [A]  time = 0.37311, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6010, 6026, 266, 51, 63, 208, 6008} \[ -\frac{a^3 \sqrt{1-a^2 x^2}}{24 x^2}-\frac{a \sqrt{1-a^2 x^2}}{20 x^4}+\frac{11}{120} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[1 - a^2*x^2])/(20*x^4) - (a^3*Sqrt[1 - a^2*x^2])/(24*x^2) - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(5*x^5)
+ (a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*x^3) + (2*a^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*x) + (11*a^5*ArcT
anh[Sqrt[1 - a^2*x^2]])/120

Rule 6010

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

Rule 6026

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcTanh[c*x])^p)/(d*f*(m + 1)), x] + (-Dist[(b*c*p)/(f*(m + 1)), Int[((
f*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/Sqrt[d + e*x^2], x], x] + Dist[(c^2*(m + 2))/(f^2*(m + 1)), 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] && LtQ[m, -1] && NeQ[m, -2]

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 51

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

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x^6} \, dx &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^5}-\frac{1}{4} \int \frac{\tanh ^{-1}(a x)}{x^6 \sqrt{1-a^2 x^2}} \, dx+\frac{1}{4} a \int \frac{1}{x^5 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}-\frac{1}{20} a \int \frac{1}{x^5 \sqrt{1-a^2 x^2}} \, dx+\frac{1}{8} a \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{5} a^2 \int \frac{\tanh ^{-1}(a x)}{x^4 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-a^2 x^2}}{16 x^4}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}-\frac{1}{40} a \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{15} a^3 \int \frac{1}{x^3 \sqrt{1-a^2 x^2}} \, dx+\frac{1}{32} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{15} \left (2 a^4\right ) \int \frac{\tanh ^{-1}(a x)}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-a^2 x^2}}{20 x^4}-\frac{3 a^3 \sqrt{1-a^2 x^2}}{32 x^2}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}-\frac{1}{160} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{30} a^3 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^2\right )+\frac{1}{64} \left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{15} \left (2 a^5\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-a^2 x^2}}{20 x^4}-\frac{a^3 \sqrt{1-a^2 x^2}}{24 x^2}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}-\frac{1}{32} \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{1}{320} \left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{60} a^5 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{15} a^5 \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}}{20 x^4}-\frac{a^3 \sqrt{1-a^2 x^2}}{24 x^2}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}-\frac{3}{32} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{1}{160} \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{1}{30} a^3 \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{1}{15} \left (2 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{a \sqrt{1-a^2 x^2}}{20 x^4}-\frac{a^3 \sqrt{1-a^2 x^2}}{24 x^2}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x^3}+\frac{2 a^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 x}+\frac{11}{120} a^5 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.121911, size = 104, normalized size = 0.69 \[ \frac{1}{120} \left (-\frac{a \sqrt{1-a^2 x^2} \left (5 a^2 x^2+6\right )}{x^4}+11 a^5 \log \left (\sqrt{1-a^2 x^2}+1\right )+\frac{8 \sqrt{1-a^2 x^2} \left (2 a^4 x^4+a^2 x^2-3\right ) \tanh ^{-1}(a x)}{x^5}-11 a^5 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*Sqrt[1 - a^2*x^2]*(6 + 5*a^2*x^2))/x^4) + (8*Sqrt[1 - a^2*x^2]*(-3 + a^2*x^2 + 2*a^4*x^4)*ArcTanh[a*x])/
x^5 - 11*a^5*Log[x] + 11*a^5*Log[1 + Sqrt[1 - a^2*x^2]])/120

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Maple [A]  time = 0.276, size = 116, normalized size = 0.8 \begin{align*}{\frac{16\,{a}^{4}{x}^{4}{\it Artanh} \left ( ax \right ) -5\,{x}^{3}{a}^{3}+8\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) -6\,ax-24\,{\it Artanh} \left ( ax \right ) }{120\,{x}^{5}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{11\,{a}^{5}}{120}\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{11\,{a}^{5}}{120}\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^6,x)

[Out]

1/120*(-(a*x-1)*(a*x+1))^(1/2)*(16*a^4*x^4*arctanh(a*x)-5*x^3*a^3+8*a^2*x^2*arctanh(a*x)-6*a*x-24*arctanh(a*x)
)/x^5+11/120*a^5*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-11/120*a^5*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-1)

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Maxima [A]  time = 1.454, size = 275, normalized size = 1.83 \begin{align*} \frac{1}{120} \,{\left (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} + 8 \,{\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^{2} - \frac{3 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{2}}{x^{2}} - \frac{6 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{x^{4}}\right )} a - \frac{1}{15} \,{\left (\frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{2}}{x^{3}} + \frac{3 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{x^{5}}\right )} \operatorname{artanh}\left (a x\right ) \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^6,x, algorithm="maxima")

[Out]

1/120*(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 + 8*(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^2 - 3*(-a^2*x^2 + 1)^(3/2)*a
^2/x^2 - 6*(-a^2*x^2 + 1)^(3/2)/x^4)*a - 1/15*(2*(-a^2*x^2 + 1)^(3/2)*a^2/x^3 + 3*(-a^2*x^2 + 1)^(3/2)/x^5)*ar
ctanh(a*x)

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Fricas [A]  time = 1.98047, size = 208, normalized size = 1.39 \begin{align*} -\frac{11 \, a^{5} x^{5} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (5 \, a^{3} x^{3} + 6 \, a x - 4 \,{\left (2 \, a^{4} x^{4} + a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{-a^{2} x^{2} + 1}}{120 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

-1/120*(11*a^5*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (5*a^3*x^3 + 6*a*x - 4*(2*a^4*x^4 + a^2*x^2 - 3)*log(-(a*
x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1))/x^5

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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^{6}}\, 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**6,x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)/x**6, x)

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Giac [B]  time = 2.2548, size = 377, normalized size = 2.51 \begin{align*} \frac{11}{240} \, a^{5} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) - \frac{11}{240} \, a^{5} \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{1}{960} \,{\left (\frac{{\left (3 \, a^{6} + \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2}}{x^{2}} - \frac{30 \,{\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{30 \,{\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{3 \,{\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} - 11 \, \sqrt{-a^{2} x^{2} + 1} a^{5}}{120 \, a^{4} x^{4}} \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^6,x, algorithm="giac")

[Out]

11/240*a^5*log(sqrt(-a^2*x^2 + 1) + 1) - 11/240*a^5*log(-sqrt(-a^2*x^2 + 1) + 1) + 1/960*((3*a^6 + 5*(sqrt(-a^
2*x^2 + 1)*abs(a) + a)^2*a^2/x^2 - 30*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^2*x^4))*a^10*x^5/((sqrt(-a^2*x^2 +
1)*abs(a) + a)^5*abs(a)) + (30*(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 - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^5/x^5)/(a^4*abs(a)))*log(-(a*x + 1)/(a*x - 1)) + 1/120*(5*(-a^2*x^2 +
1)^(3/2)*a^5 - 11*sqrt(-a^2*x^2 + 1)*a^5)/(a^4*x^4)