3.358 \(\int \frac {\log (1-x^2)}{2-x^2} \, dx\)

Optimal. Leaf size=239 \[ -\frac {\text {Li}_2\left (1-\frac {2 \sqrt {2}}{x+\sqrt {2}}\right )}{\sqrt {2}}+\frac {\text {Li}_2\left (\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}+1\right )}{2 \sqrt {2}}+\frac {\text {Li}_2\left (1-\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right )}{2 \sqrt {2}}+\frac {\log \left (1-x^2\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}+\sqrt {2} \log \left (\frac {2 \sqrt {2}}{x+\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {\log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\log \left (\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {206, 2470, 12, 5992, 5920, 2402, 2315, 2447} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2 \sqrt {2}}{x+\sqrt {2}}\right )}{\sqrt {2}}+\frac {\text {PolyLog}\left (2,\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}+1\right )}{2 \sqrt {2}}+\frac {\text {PolyLog}\left (2,1-\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right )}{2 \sqrt {2}}+\frac {\log \left (1-x^2\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}+\sqrt {2} \log \left (\frac {2 \sqrt {2}}{x+\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {\log \left (-\frac {4 (1-x)}{\left (2-\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\log \left (\frac {4 (x+1)}{\left (2+\sqrt {2}\right ) \left (x+\sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[Log[1 - x^2]/(2 - x^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

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

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Mathematica [A]  time = 0.13, size = 248, normalized size = 1.04 \[ \frac {\text {Li}_2\left (\frac {x-1}{-1-\sqrt {2}}\right )+\log \left (1-\frac {x-1}{-1-\sqrt {2}}\right ) \log (x-1)}{2 \sqrt {2}}-\frac {\text {Li}_2\left (\frac {x-1}{-1+\sqrt {2}}\right )+\log \left (1-\frac {x-1}{\sqrt {2}-1}\right ) \log (x-1)}{2 \sqrt {2}}+\frac {\text {Li}_2\left (\frac {x+1}{1-\sqrt {2}}\right )+\log (x+1) \log \left (1-\frac {x+1}{1-\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\text {Li}_2\left (\frac {x+1}{1+\sqrt {2}}\right )+\log (x+1) \log \left (1-\frac {x+1}{1+\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\left (\log \left (\sqrt {2}-x\right )-\log \left (x+\sqrt {2}\right )\right ) \left (\log \left (1-x^2\right )-\log (x-1)-\log (x+1)\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 - x^2]/(2 - x^2),x]

[Out]

-1/2*((Log[Sqrt[2] - x] - Log[Sqrt[2] + x])*(-Log[-1 + x] - Log[1 + x] + Log[1 - x^2]))/Sqrt[2] + (Log[1 - (-1
+ x)/(-1 - Sqrt[2])]*Log[-1 + x] + PolyLog[2, (-1 + x)/(-1 - Sqrt[2])])/(2*Sqrt[2]) - (Log[1 - (-1 + x)/(-1 +
Sqrt[2])]*Log[-1 + x] + PolyLog[2, (-1 + x)/(-1 + Sqrt[2])])/(2*Sqrt[2]) + (Log[1 + x]*Log[1 - (1 + x)/(1 - S
qrt[2])] + PolyLog[2, (1 + x)/(1 - Sqrt[2])])/(2*Sqrt[2]) - (Log[1 + x]*Log[1 - (1 + x)/(1 + Sqrt[2])] + PolyL
og[2, (1 + x)/(1 + Sqrt[2])])/(2*Sqrt[2])

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\log \left (-x^{2} + 1\right )}{x^{2} - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-log(-x^2 + 1)/(x^2 - 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-log(-x^2 + 1)/(x^2 - 2), x)

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maple [A]  time = 0.34, size = 214, normalized size = 0.90 \[ -\frac {\sqrt {2}\, \ln \left (\frac {x -1}{-\sqrt {2}-1}\right ) \ln \left (x +\sqrt {2}\right )}{4}+\frac {\sqrt {2}\, \ln \left (\frac {x -1}{\sqrt {2}-1}\right ) \ln \left (x -\sqrt {2}\right )}{4}+\frac {\sqrt {2}\, \ln \left (\frac {x +1}{1+\sqrt {2}}\right ) \ln \left (x -\sqrt {2}\right )}{4}-\frac {\sqrt {2}\, \ln \left (\frac {x +1}{-\sqrt {2}+1}\right ) \ln \left (x +\sqrt {2}\right )}{4}-\frac {\sqrt {2}\, \ln \left (x -\sqrt {2}\right ) \ln \left (-x^{2}+1\right )}{4}+\frac {\sqrt {2}\, \ln \left (x +\sqrt {2}\right ) \ln \left (-x^{2}+1\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {x -1}{-\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {x -1}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \dilog \left (\frac {x +1}{1+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \dilog \left (\frac {x +1}{-\sqrt {2}+1}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-x^2+1)/(-x^2+2),x)

[Out]

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

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maxima [A]  time = 1.06, size = 208, normalized size = 0.87 \[ \frac {1}{4} \, \sqrt {2} {\left ({\left (\log \left (2 \, x + 2 \, \sqrt {2}\right ) - \log \left (2 \, x - 2 \, \sqrt {2}\right )\right )} \log \left (-x^{2} + 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} + 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} + 1} + 1\right ) - \log \left (x + \sqrt {2}\right ) \log \left (-\frac {x + \sqrt {2}}{\sqrt {2} - 1} + 1\right ) + \log \left (x - \sqrt {2}\right ) \log \left (\frac {x - \sqrt {2}}{\sqrt {2} - 1} + 1\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} + 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} + 1}\right ) - {\rm Li}_2\left (\frac {x + \sqrt {2}}{\sqrt {2} - 1}\right ) + {\rm Li}_2\left (-\frac {x - \sqrt {2}}{\sqrt {2} - 1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*((log(2*x + 2*sqrt(2)) - log(2*x - 2*sqrt(2)))*log(-x^2 + 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) - 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 = -0.00 \[ -\int \frac {\ln \left (1-x^2\right )}{x^2-2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(1 - x^2)/(x^2 - 2),x)

[Out]

-int(log(1 - x^2)/(x^2 - 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(log(1 - x**2)/(x**2 - 2), x)

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