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

Optimal. Leaf size=193 \[ -\frac {1}{2} \text {Li}_2\left (1-\frac {2 \sqrt {2}}{x+\sqrt {2}}\right )+\frac {1}{4} \text {Li}_2\left (\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}+1\right )+\frac {1}{4} \text {Li}_2\left (1-\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right )+\log \left (\frac {2 \sqrt {2}}{x+\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \]

[Out]

-1/2*arctanh(1/2*x*2^(1/2))*ln(-4*(1-x)/(2-2^(1/2))/(x+2^(1/2)))+arctanh(1/2*x*2^(1/2))*ln(2*2^(1/2)/(x+2^(1/2
)))-1/2*arctanh(1/2*x*2^(1/2))*ln(4*(1+x)/(2+2^(1/2))/(x+2^(1/2)))+1/4*polylog(2,1+4*(1-x)/(2-2^(1/2))/(x+2^(1
/2)))-1/2*polylog(2,1-2*2^(1/2)/(x+2^(1/2)))+1/4*polylog(2,1-4*(1+x)/(2+2^(1/2))/(x+2^(1/2)))

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Rubi [A]  time = 0.23, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5992, 5920, 2402, 2315, 2447} \[ -\frac {1}{2} \text {PolyLog}\left (2,1-\frac {2 \sqrt {2}}{x+\sqrt {2}}\right )+\frac {1}{4} \text {PolyLog}\left (2,\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}+1\right )+\frac {1}{4} \text {PolyLog}\left (2,1-\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right )+\log \left (\frac {2 \sqrt {2}}{x+\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[x/Sqrt[2]]*Log[(2*Sqrt[2])/(Sqrt[2] + x)] - (ArcTanh[x/Sqrt[2]]*Log[(-4*(1 - x))/((2 - Sqrt[2])*(Sqrt[
2] + x))])/2 - (ArcTanh[x/Sqrt[2]]*Log[(4*(1 + x))/((2 + Sqrt[2])*(Sqrt[2] + x))])/2 - PolyLog[2, 1 - (2*Sqrt[
2])/(Sqrt[2] + x)]/2 + PolyLog[2, 1 + (4*(1 - x))/((2 - Sqrt[2])*(Sqrt[2] + x))]/4 + PolyLog[2, 1 - (4*(1 + x)
)/((2 + Sqrt[2])*(Sqrt[2] + x))]/4

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5920

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

Rule 5992

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTanh[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[
a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 232, normalized size = 1.20 \[ \frac {1}{4} \left (-2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}\right )+\text {Li}_2\left (\left (3-2 \sqrt {2}\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}\right )+\text {Li}_2\left (\left (3+2 \sqrt {2}\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}\right )+4 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}+1\right )-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (\left (2 \sqrt {2}-3\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}+1\right )-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \log \left (1-\left (3+2 \sqrt {2}\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}\right )-4 \sinh ^{-1}(1) \tanh ^{-1}(x)+2 \sinh ^{-1}(1) \log \left (\left (2 \sqrt {2}-3\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}+1\right )-2 \sinh ^{-1}(1) \log \left (1-\left (3+2 \sqrt {2}\right ) e^{-2 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*ArcSinh[1]*ArcTanh[x] + 4*ArcTanh[x/Sqrt[2]]*Log[1 + E^(-2*ArcTanh[x/Sqrt[2]])] + 2*ArcSinh[1]*Log[1 + (-3
+ 2*Sqrt[2])/E^(2*ArcTanh[x/Sqrt[2]])] - 2*ArcTanh[x/Sqrt[2]]*Log[1 + (-3 + 2*Sqrt[2])/E^(2*ArcTanh[x/Sqrt[2]
])] - 2*ArcSinh[1]*Log[1 - (3 + 2*Sqrt[2])/E^(2*ArcTanh[x/Sqrt[2]])] - 2*ArcTanh[x/Sqrt[2]]*Log[1 - (3 + 2*Sqr
t[2])/E^(2*ArcTanh[x/Sqrt[2]])] - 2*PolyLog[2, -E^(-2*ArcTanh[x/Sqrt[2]])] + PolyLog[2, (3 - 2*Sqrt[2])/E^(2*A
rcTanh[x/Sqrt[2]])] + PolyLog[2, (3 + 2*Sqrt[2])/E^(2*ArcTanh[x/Sqrt[2]])])/4

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x \operatorname {artanh}\left (\frac {1}{2} \, \sqrt {2} x\right )}{x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 251, normalized size = 1.30 \[ -\frac {\ln \left (x^{2}-1\right ) \arctanh \left (\frac {x \sqrt {2}}{2}\right )}{2}-\frac {\ln \left (\frac {x \sqrt {2}}{2}-1\right ) \ln \left (x^{2}-1\right )}{4}+\frac {\ln \left (\frac {x \sqrt {2}}{2}-1\right ) \ln \left (\frac {\sqrt {2}-x \sqrt {2}}{-2+\sqrt {2}}\right )}{4}+\frac {\ln \left (\frac {x \sqrt {2}}{2}-1\right ) \ln \left (\frac {\sqrt {2}+x \sqrt {2}}{2+\sqrt {2}}\right )}{4}+\frac {\dilog \left (\frac {\sqrt {2}-x \sqrt {2}}{-2+\sqrt {2}}\right )}{4}+\frac {\dilog \left (\frac {\sqrt {2}+x \sqrt {2}}{2+\sqrt {2}}\right )}{4}+\frac {\ln \left (\frac {x \sqrt {2}}{2}+1\right ) \ln \left (x^{2}-1\right )}{4}-\frac {\ln \left (\frac {x \sqrt {2}}{2}+1\right ) \ln \left (\frac {\sqrt {2}-x \sqrt {2}}{2+\sqrt {2}}\right )}{4}-\frac {\ln \left (\frac {x \sqrt {2}}{2}+1\right ) \ln \left (\frac {\sqrt {2}+x \sqrt {2}}{-2+\sqrt {2}}\right )}{4}-\frac {\dilog \left (\frac {\sqrt {2}-x \sqrt {2}}{2+\sqrt {2}}\right )}{4}-\frac {\dilog \left (\frac {\sqrt {2}+x \sqrt {2}}{-2+\sqrt {2}}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(x^2-1)*arctanh(1/2*x*2^(1/2))-1/4*ln(1/2*x*2^(1/2)-1)*ln(x^2-1)+1/4*ln(1/2*x*2^(1/2)-1)*ln((2^(1/2)-x*
2^(1/2))/(-2+2^(1/2)))+1/4*ln(1/2*x*2^(1/2)-1)*ln((2^(1/2)+x*2^(1/2))/(2+2^(1/2)))+1/4*dilog((2^(1/2)-x*2^(1/2
))/(-2+2^(1/2)))+1/4*dilog((2^(1/2)+x*2^(1/2))/(2+2^(1/2)))+1/4*ln(1/2*x*2^(1/2)+1)*ln(x^2-1)-1/4*ln(1/2*x*2^(
1/2)+1)*ln((2^(1/2)-x*2^(1/2))/(2+2^(1/2)))-1/4*ln(1/2*x*2^(1/2)+1)*ln((2^(1/2)+x*2^(1/2))/(-2+2^(1/2)))-1/4*d
ilog((2^(1/2)-x*2^(1/2))/(2+2^(1/2)))-1/4*dilog((2^(1/2)+x*2^(1/2))/(-2+2^(1/2)))

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maxima [A]  time = 0.43, size = 277, normalized size = 1.44 \[ -\frac {1}{2} \, \operatorname {artanh}\left (\frac {1}{2} \, \sqrt {2} x\right ) \log \left (x^{2} - 1\right ) - \frac {1}{4} \, \log \left (x^{2} - 1\right ) \log \left (\frac {x - \sqrt {2}}{x + \sqrt {2}}\right ) + \frac {1}{8} \, \sqrt {2} {\left (\sqrt {2} \log \left (x^{2} - 1\right ) \log \left (\frac {x - \sqrt {2}}{x + \sqrt {2}}\right ) + \sqrt {2} {\left ({\left (\log \left (2 \, x + 2 \, \sqrt {2}\right ) - \log \left (2 \, x - 2 \, \sqrt {2}\right )\right )} \log \left (x^{2} - 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} + 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} + 1} + 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} - 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} - 1} + 1\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} + 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} + 1}\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} - 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} - 1}\right )\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctanh(1/2*sqrt(2)*x)*log(x^2 - 1) - 1/4*log(x^2 - 1)*log((x - sqrt(2))/(x + sqrt(2))) + 1/8*sqrt(2)*(sq
rt(2)*log(x^2 - 1)*log((x - sqrt(2))/(x + sqrt(2))) + sqrt(2)*((log(2*x + 2*sqrt(2)) - log(2*x - 2*sqrt(2)))*l
og(x^2 - 1) - log(x + sqrt(2))*log(-(x + sqrt(2))/(sqrt(2) + 1) + 1) + log(x - sqrt(2))*log((x - sqrt(2))/(sqr
t(2) + 1) + 1) - log(x + sqrt(2))*log(-(x + sqrt(2))/(sqrt(2) - 1) + 1) + log(x - sqrt(2))*log((x - sqrt(2))/(
sqrt(2) - 1) + 1) - dilog((x + sqrt(2))/(sqrt(2) + 1)) + dilog(-(x - sqrt(2))/(sqrt(2) + 1)) - dilog((x + sqrt
(2))/(sqrt(2) - 1)) + dilog(-(x - sqrt(2))/(sqrt(2) - 1))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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