3.455 \(\int \frac{(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=189 \[ -i a^3 \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )+i a^3 \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{7}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x) \]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(6*x^2) + (a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/(
3*x^3) - 2*a^3*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x] + (7*a^3*ArcTanh[Sqrt[1 - a^2*x^2]])/6 - I*a^3
*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]] + I*a^3*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]]

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Rubi [A]  time = 0.31227, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6014, 6008, 266, 47, 63, 208, 5950} \[ -i a^3 \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )+i a^3 \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{7}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^4,x]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(6*x^2) + (a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/(
3*x^3) - 2*a^3*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x] + (7*a^3*ArcTanh[Sqrt[1 - a^2*x^2]])/6 - I*a^3
*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]] + I*a^3*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]]

Rule 6014

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

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]

Rule 5950

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*(a + b*ArcTanh[c*x])*
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(c*Sqrt[d]), x] + (-Simp[(I*b*PolyLog[2, -((I*Sqrt[1 - c*x])/Sqrt[1 + c*x
])])/(c*Sqrt[d]), x] + Simp[(I*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]])/(c*Sqrt[d]), x]) /; FreeQ[{a, b,
c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{x^4} \, dx &=-\left (a^2 \int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x^2} \, dx\right )+\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-a^2 \int \frac{\tanh ^{-1}(a x)}{x^2 \sqrt{1-a^2 x^2}} \, dx+a^4 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)-i a^3 \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+i a^3 \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )-a^3 \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)-i a^3 \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+i a^3 \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )-\frac{1}{12} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{1}{2} 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{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)-i a^3 \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+i a^3 \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+\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 )+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{a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x}-\frac{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-2 a^3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)+\frac{7}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-i a^3 \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )+i a^3 \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )\\ \end{align*}

Mathematica [A]  time = 1.1868, size = 199, normalized size = 1.05 \[ -\frac{\left (1-a^2 x^2\right )^{3/2} \left (\log \left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right ) \left (\sinh \left (3 \tanh ^{-1}(a x)\right )-\frac{3 a x}{\sqrt{1-a^2 x^2}}\right )+8 \tanh ^{-1}(a x)+2 \sinh \left (2 \tanh ^{-1}(a x)\right )\right )}{24 x^3}+a^3 \left (-\left (i \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-i \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a x}+i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+\log \left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^4,x]

[Out]

-(a^3*(-((Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(a*x)) + I*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] - I*ArcTanh[a*x]*L
og[1 + I/E^ArcTanh[a*x]] + Log[Tanh[ArcTanh[a*x]/2]] + I*PolyLog[2, (-I)/E^ArcTanh[a*x]] - I*PolyLog[2, I/E^Ar
cTanh[a*x]])) - ((1 - a^2*x^2)^(3/2)*(8*ArcTanh[a*x] + 2*Sinh[2*ArcTanh[a*x]] + Log[Tanh[ArcTanh[a*x]/2]]*((-3
*a*x)/Sqrt[1 - a^2*x^2] + Sinh[3*ArcTanh[a*x]])))/(24*x^3)

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Maple [A]  time = 0.223, size = 220, normalized size = 1.2 \begin{align*}{\frac{8\,{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{7\,{a}^{3}}{6}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-1 \right ) }+{\frac{7\,{a}^{3}}{6}\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-i{a}^{3}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\it Artanh} \left ( ax \right ) +i{a}^{3}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\it Artanh} \left ( ax \right ) -i{a}^{3}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) +i{a}^{3}{\it dilog} \left ( 1-{i \left ( ax+1 \right ){\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^4,x)

[Out]

1/6*(-(a*x-1)*(a*x+1))^(1/2)*(8*a^2*x^2*arctanh(a*x)-a*x-2*arctanh(a*x))/x^3-7/6*a^3*ln((a*x+1)/(-a^2*x^2+1)^(
1/2)-1)+7/6*a^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-I*a^3*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)+I*a^3*l
n(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)-I*a^3*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+I*a^3*dilog(1-I*(a*
x+1)/(-a^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{x^{4}}\,{d x} \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^4,x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*arctanh(a*x)/x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a^{2} x^{2} - 1\right )} \sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )}{x^{4}}, x\right ) \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^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}} \operatorname{atanh}{\left (a x \right )}}{x^{4}}\, dx \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**4,x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{x^{4}}\,{d x} \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^4,x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*arctanh(a*x)/x^4, x)