\(\int \frac {1-x^2}{1-3 x^2+x^4} \, dx\) [89]

Optimal result

Integrand size = 20, antiderivative size = 38 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=-\frac {\text {arctanh}\left (\frac {1-2 x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {\text {arctanh}\left (\frac {1+2 x}{\sqrt {5}}\right )}{\sqrt {5}} \]

[Out]

-1/5*arctanh(1/5*(1-2*x)*5^(1/2))*5^(1/2)+1/5*arctanh(1/5*(1+2*x)*5^(1/2))*5^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, 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} \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\frac {2 x+1}{\sqrt {5}}\right )}{\sqrt {5}}-\frac {\text {arctanh}\left (\frac {1-2 x}{\sqrt {5}}\right )}{\sqrt {5}} \]

[In]

Int[(1 - x^2)/(1 - 3*x^2 + x^4),x]

[Out]

-(ArcTanh[(1 - 2*x)/Sqrt[5]]/Sqrt[5]) + ArcTanh[(1 + 2*x)/Sqrt[5]]/Sqrt[5]

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-x+x^2} \, dx\right )-\frac {1}{2} \int \frac {1}{-1+x+x^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{5-x^2} \, dx,x,-1+2 x\right )+\text {Subst}\left (\int \frac {1}{5-x^2} \, dx,x,1+2 x\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {-1+2 x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {\tanh ^{-1}\left (\frac {1+2 x}{\sqrt {5}}\right )}{\sqrt {5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=\frac {-\log \left (-1+\sqrt {5} x-x^2\right )+\log \left (1+\sqrt {5} x+x^2\right )}{2 \sqrt {5}} \]

[In]

Integrate[(1 - x^2)/(1 - 3*x^2 + x^4),x]

[Out]

(-Log[-1 + Sqrt[5]*x - x^2] + Log[1 + Sqrt[5]*x + x^2])/(2*Sqrt[5])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {\left (1+2 x \right ) \sqrt {5}}{5}\right ) \sqrt {5}}{5}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x -1\right ) \sqrt {5}}{5}\right )}{5}\)	\(34\)
risch	\(\frac {\sqrt {5}\, \ln \left (x^{2}+x \sqrt {5}+1\right )}{10}-\frac {\sqrt {5}\, \ln \left (x^{2}-x \sqrt {5}+1\right )}{10}\)	\(35\)

[In]

int((-x^2+1)/(x^4-3*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/5*arctanh(1/5*(1+2*x)*5^(1/2))*5^(1/2)+1/5*5^(1/2)*arctanh(1/5*(2*x-1)*5^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=\frac {1}{10} \, \sqrt {5} \log \left (\frac {x^{4} + 7 \, x^{2} + 2 \, \sqrt {5} {\left (x^{3} + x\right )} + 1}{x^{4} - 3 \, x^{2} + 1}\right ) \]

[In]

integrate((-x^2+1)/(x^4-3*x^2+1),x, algorithm="fricas")

[Out]

1/10*sqrt(5)*log((x^4 + 7*x^2 + 2*sqrt(5)*(x^3 + x) + 1)/(x^4 - 3*x^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=- \frac {\sqrt {5} \log {\left (x^{2} - \sqrt {5} x + 1 \right )}}{10} + \frac {\sqrt {5} \log {\left (x^{2} + \sqrt {5} x + 1 \right )}}{10} \]

[In]

integrate((-x**2+1)/(x**4-3*x**2+1),x)

[Out]

-sqrt(5)*log(x**2 - sqrt(5)*x + 1)/10 + sqrt(5)*log(x**2 + sqrt(5)*x + 1)/10

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.45 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=-\frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x - \sqrt {5} + 1}{2 \, x + \sqrt {5} + 1}\right ) - \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x - \sqrt {5} - 1}{2 \, x + \sqrt {5} - 1}\right ) \]

[In]

integrate((-x^2+1)/(x^4-3*x^2+1),x, algorithm="maxima")

[Out]

-1/10*sqrt(5)*log((2*x - sqrt(5) + 1)/(2*x + sqrt(5) + 1)) - 1/10*sqrt(5)*log((2*x - sqrt(5) - 1)/(2*x + sqrt(
5) - 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=-\frac {1}{10} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {5} + \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {5} + \frac {2}{x} \right |}}\right ) \]

[In]

integrate((-x^2+1)/(x^4-3*x^2+1),x, algorithm="giac")

[Out]

-1/10*sqrt(5)*log(abs(2*x - 2*sqrt(5) + 2/x)/abs(2*x + 2*sqrt(5) + 2/x))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.47 \[ \int \frac {1-x^2}{1-3 x^2+x^4} \, dx=\frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,x}{x^2+1}\right )}{5} \]

[In]

int(-(x^2 - 1)/(x^4 - 3*x^2 + 1),x)

[Out]

(5^(1/2)*atanh((5^(1/2)*x)/(x^2 + 1)))/5