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

Optimal. Leaf size=130 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+i a \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )-i a \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )+2 a \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6014, 6008, 266, 63, 208, 5950} \[ i a \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )-i a \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+2 a \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 208

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

Rule 266

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 5950

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 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 6014

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

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Mathematica [A]  time = 0.38, size = 121, normalized size = 0.93 \[ a \left (-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a x}+i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+\log \left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(-((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]*Log[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^ArcTan
h[a*x]])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*arctanh(a*x)/x^2, 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)*(-a^2*x^2+1)^(1/2)/x^2,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 = 0.45, size = 188, normalized size = 1.45 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )}{x}-i a \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )-i a \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )+i a \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )+i a \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )+a \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right )-a \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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