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

Optimal. Leaf size=127 \[ \frac {2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6030, 6020, 4182, 2531, 2282, 6589, 5994, 5958} \[ -2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+2 \text {PolyLog}\left (3,-e^{\tanh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right )+\frac {2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5958

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> -Simp[b/(c*d*Sqrt[d + e*x^2]
), x] + Simp[(x*(a + b*ArcTanh[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0
]

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 6020

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[1/Sqrt[d], Su
bst[Int[(a + b*x)^p*Csch[x], x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGt
Q[p, 0] && GtQ[d, 0]

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]

Rule 6589

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

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )^{3/2}} \, dx &=a^2 \int \frac {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x \sqrt {1-a^2 x^2}} \, dx\\ &=\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-(2 a) \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\operatorname {Subst}\left (\int x^2 \text {csch}(x) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=\frac {2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \operatorname {Subst}\left (\int x \log \left (1-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )+2 \operatorname {Subst}\left (\int x \log \left (1+e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=\frac {2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 \operatorname {Subst}\left (\int \text {Li}_2\left (-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )-2 \operatorname {Subst}\left (\int \text {Li}_2\left (e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=\frac {2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )-2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )\\ &=\frac {2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 159, normalized size = 1.25 \[ \frac {2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-\frac {2 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{-\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x)^2 \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^2 \log \left (e^{-\tanh ^{-1}(a x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.45, size = 232, normalized size = 1.83 \[ -\frac {\left (\arctanh \left (a x \right )^{2}-2 \arctanh \left (a x \right )+2\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )^{2}+2 \arctanh \left (a x \right )+2\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a x +2}-\arctanh \left (a x \right )^{2} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-2 \arctanh \left (a x \right ) \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\arctanh \left (a x \right )^{2} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 \arctanh \left (a x \right ) \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-2 \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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