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

Optimal. Leaf size=179 \[ \frac {3}{2} a^2 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{2} a^2 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {a \sqrt {1-a^2 x^2}}{2 x}+\frac {a^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}-3 a^2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {a^3 x}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6030, 6026, 264, 6018, 5994, 191} \[ \frac {3}{2} a^2 \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{2} a^2 \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {a^3 x}{\sqrt {1-a^2 x^2}}-\frac {a \sqrt {1-a^2 x^2}}{2 x}+\frac {a^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}-3 a^2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 191

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

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 5994

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

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]

Rule 6030

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

Rubi steps

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

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Mathematica [A]  time = 1.63, size = 182, normalized size = 1.02 \[ \frac {1}{8} a^2 \left (-\frac {8 a x}{\sqrt {1-a^2 x^2}}+\frac {8 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {a x \text {csch}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{\sqrt {1-a^2 x^2}}+12 \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-12 \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )+2 \tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )+12 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-12 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )-\tanh ^{-1}(a x) \text {csch}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )-\tanh ^{-1}(a x) \text {sech}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*((-8*a*x)/Sqrt[1 - a^2*x^2] + (8*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (a*x*Csch[ArcTanh[a*x]/2]^2)/Sqrt[1 -
a^2*x^2] - ArcTanh[a*x]*Csch[ArcTanh[a*x]/2]^2 + 12*ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] - 12*ArcTanh[a*x]*
Log[1 + E^(-ArcTanh[a*x])] + 12*PolyLog[2, -E^(-ArcTanh[a*x])] - 12*PolyLog[2, E^(-ArcTanh[a*x])] - ArcTanh[a*
x]*Sech[ArcTanh[a*x]/2]^2 + 2*Tanh[ArcTanh[a*x]/2]))/8

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )}{a^{4} x^{7} - 2 \, a^{2} x^{5} + x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*arctanh(a*x)/(a^4*x^7 - 2*a^2*x^5 + x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.45, size = 205, normalized size = 1.15 \[ -\frac {a^{2} \left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )+1\right ) a^{2} \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a x +2}-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a x +\arctanh \left (a x \right )\right )}{2 x^{2}}-\frac {3 a^{2} \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}-\frac {3 a^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {3 a^{2} \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {3 a^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/(a*x-1)+1/2*(arctanh(a*x)+1)*a^2*(-(a*x-1)*(a*x+1))^(1/2)/(
a*x+1)-1/2*(-(a*x-1)*(a*x+1))^(1/2)*(a*x+arctanh(a*x))/x^2-3/2*a^2*arctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2
))-3/2*a^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+3/2*a^2*arctanh(a*x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3/2*a^
2*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 \[ \int \frac {\operatorname {artanh}\left (a x\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{x^{3} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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