3.5.55 \(\int \frac {(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^4} \, dx\) [455]

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

[Out]

-1/3*(-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^3-2*a^3*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)+7/6*a^3*arcta
nh((-a^2*x^2+1)^(1/2))-I*a^3*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))+I*a^3*polylog(2,I*(-a*x+1)^(1/2)/(a*x+
1)^(1/2))-1/6*a*(-a^2*x^2+1)^(1/2)/x^2+a^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x

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Rubi [A]
time = 0.23, antiderivative size = 189, normalized size of antiderivative = 1.00, 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 = {6161, 6155, 272, 43, 65, 214, 6097} \begin {gather*} -2 a^3 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)-i a^3 \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )+i a^3 \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\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}+\frac {7}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*Sqrt[1 - a^2*x^2])/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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 214

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

Rule 272

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 6097

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]

Rule 6155

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 6161

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]))

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 \text {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 \text {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {1}{2} a^3 \text {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 \text {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )+a \text {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*}

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Mathematica [A]
time = 0.87, size = 199, normalized size = 1.05 \begin {gather*} -a^3 \left (-\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 )+i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-i \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \left (8 \tanh ^{-1}(a x)+2 \sinh \left (2 \tanh ^{-1}(a x)\right )+\log \left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right ) \left (-\frac {3 a x}{\sqrt {1-a^2 x^2}}+\sinh \left (3 \tanh ^{-1}(a x)\right )\right )\right )}{24 x^3} \end {gather*}

Antiderivative was successfully verified.

[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 = 1.42, size = 220, normalized size = 1.16

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (8 a^{2} x^{2} \arctanh \left (a x \right )-a x -2 \arctanh \left (a x \right )\right )}{6 x^{3}}+\frac {7 a^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{6}-\frac {7 a^{3} \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right )}{6}-i a^{3} \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )+i a^{3} \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )-i a^{3} \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )+i a^{3} \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )\)	\(220\)

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,method=_RETURNVERBOSE)

[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(1+(a*x+1)/(-a^2*x^2+1)
^(1/2))-7/6*a^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-1)-I*a^3*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)+I*a^3*d
ilog(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))+I*a^3*ln(1-I*(a*x+1)/(-a^2*x^
2+1)^(1/2))*arctanh(a*x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^4,x)

[Out]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^4, x)

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