3.1.22 \(\int \frac {\tanh ^{-1}(x)}{1-\sqrt {2} x} \, dx\) [22]

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

[Out]

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

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Rubi [A]
time = 0.05, 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 = {6057, 2449, 2352, 2497} \begin {gather*} -\frac {\text {Li}_2\left (1-\frac {2}{x+1}\right )}{2 \sqrt {2}}+\frac {\text {Li}_2\left (\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}} \end {gather*}

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 2352

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

Rule 2449

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 2497

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 6057

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]

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 {\text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.07, size = 272, normalized size = 3.09 \begin {gather*} \frac {\pi ^2-4 \tanh ^{-1}\left (\frac {1}{\sqrt {2}}\right )^2-4 i \pi \tanh ^{-1}(x)+8 \tanh ^{-1}\left (\frac {1}{\sqrt {2}}\right ) \tanh ^{-1}(x)-8 \tanh ^{-1}(x)^2+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 )-8 \tanh ^{-1}(x) \log \left (1-e^{2 \tanh ^{-1}\left (\frac {1}{\sqrt {2}}\right )-2 \tanh ^{-1}(x)}\right )+4 i \pi \log \left (1+e^{2 \tanh ^{-1}(x)}\right )+8 \tanh ^{-1}(x) \log \left (1+e^{2 \tanh ^{-1}(x)}\right )-4 i \pi \log \left (\frac {2}{\sqrt {1-x^2}}\right )-8 \tanh ^{-1}(x) \log \left (\frac {2}{\sqrt {1-x^2}}\right )-4 \tanh ^{-1}(x) \log \left (1-x^2\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}\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 )+8 \tanh ^{-1}(x) \log \left (-2 i \sinh \left (\tanh ^{-1}\left (\frac {1}{\sqrt {2}}\right )-\tanh ^{-1}(x)\right )\right )+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 )}{8 \sqrt {2}} \end {gather*}

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 = 0.18, size = 127, normalized size = 1.44

method	result	size
default	\(-\frac {\ln \left (x \sqrt {2}-1\right ) \sqrt {2}\, \arctanh \left (x \right )}{2}-\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}\)	\(127\)
risch	\(\frac {\sqrt {2}\, \ln \left (\frac {-2 x +\sqrt {2}}{2+\sqrt {2}}\right ) \ln \left (\frac {2+2 x}{2+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \ln \left (\frac {-2 x +\sqrt {2}}{2+\sqrt {2}}\right ) \ln \left (1+x \right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {2+2 x}{2+\sqrt {2}}\right )}{4}+\frac {\sqrt {2}\, \ln \left (\frac {2 x -\sqrt {2}}{2-\sqrt {2}}\right ) \ln \left (1-x \right )}{4}-\frac {\sqrt {2}\, \ln \left (\frac {2 x -\sqrt {2}}{2-\sqrt {2}}\right ) \ln \left (\frac {2-2 x}{2-\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {2-2 x}{2-\sqrt {2}}\right )}{4}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (69) = 138\).
time = 0.46, size = 144, normalized size = 1.64 \begin {gather*} \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 {gather*}

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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\operatorname {atanh}{\left (x \right )}}{\sqrt {2} x - 1}\, dx \end {gather*}

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 = 0.00, 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(x)/(1-x*2^(1/2)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\mathrm {atanh}\left (x\right )}{\sqrt {2}\,x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(x)/(2^(1/2)*x - 1),x)

[Out]

-int(atanh(x)/(2^(1/2)*x - 1), x)

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