\(\int \frac {1-x^2}{1-4 x^2+x^4} \, dx\) [90]

Optimal result

Integrand size = 20, antiderivative size = 47 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=-\frac {\text {arctanh}\left (\frac {1-\sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}}+\frac {\text {arctanh}\left (\frac {1+\sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \]

[Out]

-1/6*arctanh(1/3*(1-x*2^(1/2))*3^(1/2))*6^(1/2)+1/6*arctanh(1/3*(1+x*2^(1/2))*3^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1175, 632, 212} \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {2} x+1}{\sqrt {3}}\right )}{\sqrt {6}}-\frac {\text {arctanh}\left (\frac {1-\sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \]

[In]

Int[(1 - x^2)/(1 - 4*x^2 + x^4),x]

[Out]

-(ArcTanh[(1 - Sqrt[2]*x)/Sqrt[3]]/Sqrt[6]) + ArcTanh[(1 + Sqrt[2]*x)/Sqrt[3]]/Sqrt[6]

Rule 212

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int \frac {1}{-1-\sqrt {2} x+x^2} \, dx\right )-\frac {1}{2} \int \frac {1}{-1+\sqrt {2} x+x^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{6-x^2} \, dx,x,-\sqrt {2}+2 x\right )+\text {Subst}\left (\int \frac {1}{6-x^2} \, dx,x,\sqrt {2}+2 x\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {-1+\sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}}+\frac {\tanh ^{-1}\left (\frac {1+\sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=\frac {-\log \left (-1+\sqrt {6} x-x^2\right )+\log \left (1+\sqrt {6} x+x^2\right )}{2 \sqrt {6}} \]

[In]

Integrate[(1 - x^2)/(1 - 4*x^2 + x^4),x]

[Out]

(-Log[-1 + Sqrt[6]*x - x^2] + Log[1 + Sqrt[6]*x + x^2])/(2*Sqrt[6])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\sqrt {6}\, \ln \left (x^{2}+x \sqrt {6}+1\right )}{12}-\frac {\sqrt {6}\, \ln \left (x^{2}-x \sqrt {6}+1\right )}{12}\)	\(35\)
default	\(\frac {\left (1+\sqrt {3}\right ) \sqrt {3}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {6}+\sqrt {2}}\right )}{3 \sqrt {6}+3 \sqrt {2}}+\frac {\left (\sqrt {3}-1\right ) \sqrt {3}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {6}-\sqrt {2}}\right )}{3 \sqrt {6}-3 \sqrt {2}}\)	\(70\)

[In]

int((-x^2+1)/(x^4-4*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/12*6^(1/2)*ln(x^2+x*6^(1/2)+1)-1/12*6^(1/2)*ln(x^2-x*6^(1/2)+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=\frac {1}{12} \, \sqrt {6} \log \left (\frac {x^{4} + 8 \, x^{2} + 2 \, \sqrt {6} {\left (x^{3} + x\right )} + 1}{x^{4} - 4 \, x^{2} + 1}\right ) \]

[In]

integrate((-x^2+1)/(x^4-4*x^2+1),x, algorithm="fricas")

[Out]

1/12*sqrt(6)*log((x^4 + 8*x^2 + 2*sqrt(6)*(x^3 + x) + 1)/(x^4 - 4*x^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=- \frac {\sqrt {6} \log {\left (x^{2} - \sqrt {6} x + 1 \right )}}{12} + \frac {\sqrt {6} \log {\left (x^{2} + \sqrt {6} x + 1 \right )}}{12} \]

[In]

integrate((-x**2+1)/(x**4-4*x**2+1),x)

[Out]

-sqrt(6)*log(x**2 - sqrt(6)*x + 1)/12 + sqrt(6)*log(x**2 + sqrt(6)*x + 1)/12

Maxima [F]

\[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=\int { -\frac {x^{2} - 1}{x^{4} - 4 \, x^{2} + 1} \,d x } \]

[In]

integrate((-x^2+1)/(x^4-4*x^2+1),x, algorithm="maxima")

[Out]

-integrate((x^2 - 1)/(x^4 - 4*x^2 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=-\frac {1}{12} \, \sqrt {6} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {6} + \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {6} + \frac {2}{x} \right |}}\right ) \]

[In]

integrate((-x^2+1)/(x^4-4*x^2+1),x, algorithm="giac")

[Out]

-1/12*sqrt(6)*log(abs(2*x - 2*sqrt(6) + 2/x)/abs(2*x + 2*sqrt(6) + 2/x))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.38 \[ \int \frac {1-x^2}{1-4 x^2+x^4} \, dx=\frac {\sqrt {6}\,\mathrm {atanh}\left (\frac {\sqrt {6}\,x}{x^2+1}\right )}{6} \]

[In]

int(-(x^2 - 1)/(x^4 - 4*x^2 + 1),x)

[Out]

(6^(1/2)*atanh((6^(1/2)*x)/(x^2 + 1)))/6