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

Optimal. Leaf size=88 \[ -\frac{\text{PolyLog}\left (2,-\frac{\sqrt{2}-2 x}{2-\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{\text{PolyLog}\left (2,\frac{\sqrt{2}-2 x}{2+\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right ) \log \left (1-\sqrt{2} x\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0660324, antiderivative size = 108, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {5920, 2402, 2315, 2447} \[ -\frac{\text{PolyLog}\left (2,1-\frac{2}{x+1}\right )}{2 \sqrt{2}}+\frac{\text{PolyLog}\left (2,\frac{2 \left (1+\sqrt{2}\right ) \left (1-\sqrt{2} x\right )}{x+1}+1\right )}{2 \sqrt{2}}+\frac{\log \left (\frac{2}{x+1}\right ) \tanh ^{-1}(x)}{\sqrt{2}}-\frac{\log \left (-\frac{2 \left (1+\sqrt{2}\right ) \left (1-\sqrt{2} x\right )}{x+1}\right ) \tanh ^{-1}(x)}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x]/(1 - Sqrt[2]*x),x]

[Out]

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

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 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 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 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]]

Rubi steps

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

Mathematica [C]  time = 0.0989687, size = 272, normalized size = 3.09 \[ \frac{4 \text{PolyLog}\left (2,e^{2 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-2 \tanh ^{-1}(x)}\right )+4 \text{PolyLog}\left (2,-e^{2 \tanh ^{-1}(x)}\right )-4 i \pi \log \left (\frac{2}{\sqrt{1-x^2}}\right )-8 \log \left (\frac{2}{\sqrt{1-x^2}}\right ) \tanh ^{-1}(x)-4 \log \left (1-x^2\right ) \tanh ^{-1}(x)-8 \tanh ^{-1}(x)^2+8 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right ) \tanh ^{-1}(x)-4 i \pi \tanh ^{-1}(x)-8 \tanh ^{-1}(x) \log \left (1-e^{2 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-2 \tanh ^{-1}(x)}\right )+8 \tanh ^{-1}(x) \log \left (e^{2 \tanh ^{-1}(x)}+1\right )+8 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right ) \log \left (1-e^{2 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-2 \tanh ^{-1}(x)}\right )+4 i \pi \log \left (e^{2 \tanh ^{-1}(x)}+1\right )-8 \tanh ^{-1}(x) \log \left (-i \sinh \left (\tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-\tanh ^{-1}(x)\right )\right )+8 \tanh ^{-1}(x) \log \left (-2 i \sinh \left (\tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-\tanh ^{-1}(x)\right )\right )-8 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right ) \log \left (-2 i \sinh \left (\tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )-\tanh ^{-1}(x)\right )\right )+\pi ^2-4 \tanh ^{-1}\left (\frac{1}{\sqrt{2}}\right )^2}{8 \sqrt{2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Pi^2 - 4*ArcTanh[1/Sqrt[2]]^2 - (4*I)*Pi*ArcTanh[x] + 8*ArcTanh[1/Sqrt[2]]*ArcTanh[x] - 8*ArcTanh[x]^2 + 8*Ar
cTanh[1/Sqrt[2]]*Log[1 - E^(2*ArcTanh[1/Sqrt[2]] - 2*ArcTanh[x])] - 8*ArcTanh[x]*Log[1 - E^(2*ArcTanh[1/Sqrt[2
]] - 2*ArcTanh[x])] + (4*I)*Pi*Log[1 + E^(2*ArcTanh[x])] + 8*ArcTanh[x]*Log[1 + E^(2*ArcTanh[x])] - (4*I)*Pi*L
og[2/Sqrt[1 - x^2]] - 8*ArcTanh[x]*Log[2/Sqrt[1 - x^2]] - 4*ArcTanh[x]*Log[1 - x^2] - 8*ArcTanh[x]*Log[(-I)*Si
nh[ArcTanh[1/Sqrt[2]] - ArcTanh[x]]] - 8*ArcTanh[1/Sqrt[2]]*Log[(-2*I)*Sinh[ArcTanh[1/Sqrt[2]] - ArcTanh[x]]]
+ 8*ArcTanh[x]*Log[(-2*I)*Sinh[ArcTanh[1/Sqrt[2]] - ArcTanh[x]]] + 4*PolyLog[2, E^(2*ArcTanh[1/Sqrt[2]] - 2*Ar
cTanh[x])] + 4*PolyLog[2, -E^(2*ArcTanh[x])])/(8*Sqrt[2])

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Maple [A]  time = 0.036, size = 127, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ( x\sqrt{2}-1 \right ) \sqrt{2}{\it Artanh} \left ( x \right ) }{2}}-{\frac{\ln \left ( x\sqrt{2}-1 \right ) \sqrt{2}}{4}\ln \left ({\frac{\sqrt{2}-x\sqrt{2}}{\sqrt{2}-1}} \right ) }+{\frac{\ln \left ( x\sqrt{2}-1 \right ) \sqrt{2}}{4}\ln \left ({\frac{\sqrt{2}+x\sqrt{2}}{1+\sqrt{2}}} \right ) }-{\frac{\sqrt{2}}{4}{\it dilog} \left ({\frac{\sqrt{2}-x\sqrt{2}}{\sqrt{2}-1}} \right ) }+{\frac{\sqrt{2}}{4}{\it dilog} \left ({\frac{\sqrt{2}+x\sqrt{2}}{1+\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.44954, size = 194, normalized size = 2.2 \begin{align*} \frac{1}{4} \, \sqrt{2}{\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (\sqrt{2} x - 1\right ) - \frac{1}{2} \, \sqrt{2} \operatorname{artanh}\left (x\right ) \log \left (\sqrt{2} x - 1\right ) - \frac{1}{4} \, \sqrt{2}{\left (\log \left (x + 1\right ) \log \left (-\frac{\sqrt{2} x + \sqrt{2}}{\sqrt{2} + 1} + 1\right ) +{\rm Li}_2\left (\frac{\sqrt{2} x + \sqrt{2}}{\sqrt{2} + 1}\right )\right )} + \frac{1}{4} \, \sqrt{2}{\left (\log \left (x - 1\right ) \log \left (\frac{\sqrt{2} x - \sqrt{2}}{\sqrt{2} - 1} + 1\right ) +{\rm Li}_2\left (-\frac{\sqrt{2} x - \sqrt{2}}{\sqrt{2} - 1}\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*(log(x + 1) - log(x - 1))*log(sqrt(2)*x - 1) - 1/2*sqrt(2)*arctanh(x)*log(sqrt(2)*x - 1) - 1/4*sqr
t(2)*(log(x + 1)*log(-(sqrt(2)*x + sqrt(2))/(sqrt(2) + 1) + 1) + dilog((sqrt(2)*x + sqrt(2))/(sqrt(2) + 1))) +
1/4*sqrt(2)*(log(x - 1)*log((sqrt(2)*x - sqrt(2))/(sqrt(2) - 1) + 1) + dilog(-(sqrt(2)*x - sqrt(2))/(sqrt(2)
- 1)))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (\sqrt{2} x + 1\right )} \operatorname{artanh}\left (x\right )}{2 \, x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{atanh}{\left (x \right )}}{\sqrt{2} x - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (x\right )}{\sqrt{2} x - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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