3.359 \(\int \frac {\log (d+e x^2)}{1-x^2} \, dx\)

Optimal. Leaf size=217 \[ \frac {1}{2} \text {Li}_2\left (1-\frac {2 \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (\sqrt {-d}-\sqrt {e}\right ) (x+1)}\right )+\frac {1}{2} \text {Li}_2\left (1-\frac {2 \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d}+\sqrt {e}\right ) (x+1)}\right )+\tanh ^{-1}(x) \log \left (d+e x^2\right )-\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {-d}-\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}-\sqrt {e}\right )}\right )-\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {-d}+\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}+\sqrt {e}\right )}\right )-\text {Li}_2\left (1-\frac {2}{x+1}\right )+2 \log \left (\frac {2}{x+1}\right ) \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {206, 2470, 5992, 5920, 2402, 2315, 2447} \[ \frac {1}{2} \text {PolyLog}\left (2,1-\frac {2 \left (\sqrt {-d}-\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}-\sqrt {e}\right )}\right )+\frac {1}{2} \text {PolyLog}\left (2,1-\frac {2 \left (\sqrt {-d}+\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}+\sqrt {e}\right )}\right )-\text {PolyLog}\left (2,1-\frac {2}{x+1}\right )+\tanh ^{-1}(x) \log \left (d+e x^2\right )-\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {-d}-\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}-\sqrt {e}\right )}\right )-\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {-d}+\sqrt {e} x\right )}{(x+1) \left (\sqrt {-d}+\sqrt {e}\right )}\right )+2 \log \left (\frac {2}{x+1}\right ) \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Log[d + e*x^2]/(1 - x^2),x]

[Out]

2*ArcTanh[x]*Log[2/(1 + x)] - ArcTanh[x]*Log[(2*(Sqrt[-d] - Sqrt[e]*x))/((Sqrt[-d] - Sqrt[e])*(1 + x))] - ArcT
anh[x]*Log[(2*(Sqrt[-d] + Sqrt[e]*x))/((Sqrt[-d] + Sqrt[e])*(1 + x))] + ArcTanh[x]*Log[d + e*x^2] - PolyLog[2,
1 - 2/(1 + x)] + PolyLog[2, 1 - (2*(Sqrt[-d] - Sqrt[e]*x))/((Sqrt[-d] - Sqrt[e])*(1 + x))]/2 + PolyLog[2, 1 -
(2*(Sqrt[-d] + Sqrt[e]*x))/((Sqrt[-d] + Sqrt[e])*(1 + x))]/2

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

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Mathematica [C]  time = 0.13, size = 468, normalized size = 2.16 \[ \frac {1}{2} \left (-\text {Li}_2\left (\frac {\sqrt {d}-i \sqrt {e} x}{\sqrt {d}-i \sqrt {e}}\right )+\text {Li}_2\left (\frac {\sqrt {d}-i \sqrt {e} x}{\sqrt {d}+i \sqrt {e}}\right )+\text {Li}_2\left (\frac {i \sqrt {e} x+\sqrt {d}}{\sqrt {d}-i \sqrt {e}}\right )-\text {Li}_2\left (\frac {i \sqrt {e} x+\sqrt {d}}{\sqrt {d}+i \sqrt {e}}\right )-\log (1-x) \log \left (d+e x^2\right )+\log (x+1) \log \left (d+e x^2\right )+\log (1-x) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )-\log \left (\frac {\sqrt {e} (x-1)}{-\sqrt {e}+i \sqrt {d}}\right ) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )-\log (x+1) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )+\log \left (-\frac {i \sqrt {e} (x+1)}{\sqrt {d}-i \sqrt {e}}\right ) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )+\log (1-x) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )-\log \left (\frac {\sqrt {e} (x-1)}{-\sqrt {e}-i \sqrt {d}}\right ) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )-\log (x+1) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )+\log \left (\frac {i \sqrt {e} (x+1)}{\sqrt {d}+i \sqrt {e}}\right ) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[d + e*x^2]/(1 - x^2),x]

[Out]

(Log[1 - x]*Log[((-I)*Sqrt[d])/Sqrt[e] + x] - Log[(Sqrt[e]*(-1 + x))/(I*Sqrt[d] - Sqrt[e])]*Log[((-I)*Sqrt[d])
/Sqrt[e] + x] - Log[1 + x]*Log[((-I)*Sqrt[d])/Sqrt[e] + x] + Log[((-I)*Sqrt[e]*(1 + x))/(Sqrt[d] - I*Sqrt[e])]
*Log[((-I)*Sqrt[d])/Sqrt[e] + x] + Log[1 - x]*Log[(I*Sqrt[d])/Sqrt[e] + x] - Log[(Sqrt[e]*(-1 + x))/((-I)*Sqrt
[d] - Sqrt[e])]*Log[(I*Sqrt[d])/Sqrt[e] + x] - Log[1 + x]*Log[(I*Sqrt[d])/Sqrt[e] + x] + Log[(I*Sqrt[e]*(1 + x
))/(Sqrt[d] + I*Sqrt[e])]*Log[(I*Sqrt[d])/Sqrt[e] + x] - Log[1 - x]*Log[d + e*x^2] + Log[1 + x]*Log[d + e*x^2]
- PolyLog[2, (Sqrt[d] - I*Sqrt[e]*x)/(Sqrt[d] - I*Sqrt[e])] + PolyLog[2, (Sqrt[d] - I*Sqrt[e]*x)/(Sqrt[d] + I
*Sqrt[e])] + PolyLog[2, (Sqrt[d] + I*Sqrt[e]*x)/(Sqrt[d] - I*Sqrt[e])] - PolyLog[2, (Sqrt[d] + I*Sqrt[e]*x)/(S
qrt[d] + I*Sqrt[e])])/2

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\log \left (e x^{2} + d\right )}{x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-log(e*x^2 + d)/(x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 282, normalized size = 1.30 \[ -\frac {\ln \left (\frac {-\left (x +1\right ) e +e +\sqrt {-d e}}{e +\sqrt {-d e}}\right ) \ln \left (x +1\right )}{2}-\frac {\ln \left (\frac {\left (x +1\right ) e -e +\sqrt {-d e}}{-e +\sqrt {-d e}}\right ) \ln \left (x +1\right )}{2}+\frac {\ln \left (\frac {-\left (x -1\right ) e -e +\sqrt {-d e}}{-e +\sqrt {-d e}}\right ) \ln \left (x -1\right )}{2}+\frac {\ln \left (\frac {\left (x -1\right ) e +e +\sqrt {-d e}}{e +\sqrt {-d e}}\right ) \ln \left (x -1\right )}{2}-\frac {\ln \left (x -1\right ) \ln \left (e \,x^{2}+d \right )}{2}+\frac {\ln \left (x +1\right ) \ln \left (e \,x^{2}+d \right )}{2}-\frac {\dilog \left (\frac {-\left (x +1\right ) e +e +\sqrt {-d e}}{e +\sqrt {-d e}}\right )}{2}-\frac {\dilog \left (\frac {\left (x +1\right ) e -e +\sqrt {-d e}}{-e +\sqrt {-d e}}\right )}{2}+\frac {\dilog \left (\frac {-\left (x -1\right ) e -e +\sqrt {-d e}}{-e +\sqrt {-d e}}\right )}{2}+\frac {\dilog \left (\frac {\left (x -1\right ) e +e +\sqrt {-d e}}{e +\sqrt {-d e}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(d + e*x^2)/(x^2 - 1),x)

[Out]

-int(log(d + e*x^2)/(x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(log(d + e*x**2)/(x**2 - 1), x)

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