3.4.92 \(\int \frac {\tanh ^{-1}(a x)}{x (1-a^2 x^2)^{3/2}} \, dx\) [392]

Optimal. Leaf size=112 \[ -\frac {a x}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6177, 6165, 6141, 197} \begin {gather*} -\frac {a x}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

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 6141

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 6165

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

Rule 6177

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

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Mathematica [A]
time = 0.15, size = 97, normalized size = 0.87 \begin {gather*} -\frac {a x}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+\text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.72, size = 157, normalized size = 1.40

method	result	size
default	\(-\frac {\left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a x +2}+\arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/(a*x-1)+1/2*(arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/(a*x+1)+a
rctanh(a*x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-arctanh(a*x)*ln(1+(a*x+1)/(
-a^2*x^2+1)^(1/2))-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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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 )}{x\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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