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

Optimal. Leaf size=169 \[ -\frac {1}{3} a^3 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {1}{3} a^3 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {2}{3} a^3 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {a^2 \sqrt {1-a^2 x^2}}{3 x}-\frac {a \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)^2}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6008, 6010, 6026, 264, 6018} \[ -\frac {1}{3} a^3 \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {1}{3} a^3 \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {a^2 \sqrt {1-a^2 x^2}}{3 x}-\frac {a \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)^2}{3 x^3}+\frac {2}{3} a^3 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sqrt[1 - a^2*x^2])/(3*x) - (a*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*x^2) - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x
]^2)/(3*x^3) + (2*a^3*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/3 - (a^3*PolyLog[2, -(Sqrt[1 - a*x]/S
qrt[1 + a*x])])/3 + (a^3*PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]])/3

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 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 6018

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

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]

Rubi steps

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

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Mathematica [A]  time = 2.06, size = 177, normalized size = 1.05 \[ -\frac {1}{3} a^3 \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \left (\tanh ^{-1}(a x) \left (\frac {\left (\log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\log \left (e^{-\tanh ^{-1}(a x)}+1\right )\right ) \left (\sqrt {1-a^2 x^2} \sinh \left (3 \tanh ^{-1}(a x)\right )-3 a x\right )}{\sqrt {1-a^2 x^2}}+2 \sinh \left (2 \tanh ^{-1}(a x)\right )\right )-\frac {4 a^3 x^3 \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )}{\left (1-a^2 x^2\right )^{3/2}}+4 \tanh ^{-1}(a x)^2+2 \left (\cosh \left (2 \tanh ^{-1}(a x)\right )-1\right )\right )}{12 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*(a^3*PolyLog[2, -E^(-ArcTanh[a*x])]) - ((1 - a^2*x^2)^(3/2)*(4*ArcTanh[a*x]^2 + 2*(-1 + Cosh[2*ArcTanh[a*
x]]) - (4*a^3*x^3*PolyLog[2, E^(-ArcTanh[a*x])])/(1 - a^2*x^2)^(3/2) + ArcTanh[a*x]*(2*Sinh[2*ArcTanh[a*x]] +
((Log[1 - E^(-ArcTanh[a*x])] - Log[1 + E^(-ArcTanh[a*x])])*(-3*a*x + Sqrt[1 - a^2*x^2]*Sinh[3*ArcTanh[a*x]]))/
Sqrt[1 - a^2*x^2])))/(12*x^3)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.44, size = 171, normalized size = 1.01 \[ \frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a^{2} x^{2} \arctanh \left (a x \right )^{2}-a^{2} x^{2}-a x \arctanh \left (a x \right )-\arctanh \left (a x \right )^{2}\right )}{3 x^{3}}+\frac {a^{3} \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}+\frac {a^{3} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-\frac {a^{3} \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-\frac {a^{3} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-(a*x-1)*(a*x+1))^(1/2)*(a^2*x^2*arctanh(a*x)^2-a^2*x^2-a*x*arctanh(a*x)-arctanh(a*x)^2)/x^3+1/3*a^3*arct
anh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+1/3*a^3*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-1/3*a^3*arctanh(a*x)*
ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))-1/3*a^3*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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