\(\int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx\) [67]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=-\frac {\text {arctanh}\left (\frac {1-2 \sqrt {2} x}{\sqrt {5}}\right )}{\sqrt {10}}+\frac {\text {arctanh}\left (\frac {1+2 \sqrt {2} x}{\sqrt {5}}\right )}{\sqrt {10}} \]

[Out]

-1/10*arctanh(1/5*(1-2*x*2^(1/2))*5^(1/2))*10^(1/2)+1/10*arctanh(1/5*(1+2*x*2^(1/2))*5^(1/2))*10^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1175, 632, 212} \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\text {arctanh}\left (\frac {2 \sqrt {2} x+1}{\sqrt {5}}\right )}{\sqrt {10}}-\frac {\text {arctanh}\left (\frac {1-2 \sqrt {2} x}{\sqrt {5}}\right )}{\sqrt {10}} \]

[In]

Int[(1 - 2*x^2)/(1 - 6*x^2 + 4*x^4),x]

[Out]

-(ArcTanh[(1 - 2*Sqrt[2]*x)/Sqrt[5]]/Sqrt[10]) + ArcTanh[(1 + 2*Sqrt[2]*x)/Sqrt[5]]/Sqrt[10]

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}{4} \int \frac {1}{-\frac {1}{2}-\frac {x}{\sqrt {2}}+x^2} \, dx\right )-\frac {1}{4} \int \frac {1}{-\frac {1}{2}+\frac {x}{\sqrt {2}}+x^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\frac {5}{2}-x^2} \, dx,x,-\frac {1}{\sqrt {2}}+2 x\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\frac {5}{2}-x^2} \, dx,x,\frac {1}{\sqrt {2}}+2 x\right ) \\ & = -\frac {\tanh ^{-1}\left (\frac {1-2 \sqrt {2} x}{\sqrt {5}}\right )}{\sqrt {10}}+\frac {\tanh ^{-1}\left (\frac {1+2 \sqrt {2} x}{\sqrt {5}}\right )}{\sqrt {10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {-\log \left (-1+\sqrt {10} x-2 x^2\right )+\log \left (1+\sqrt {10} x+2 x^2\right )}{2 \sqrt {10}} \]

[In]

Integrate[(1 - 2*x^2)/(1 - 6*x^2 + 4*x^4),x]

[Out]

(-Log[-1 + Sqrt[10]*x - 2*x^2] + Log[1 + Sqrt[10]*x + 2*x^2])/(2*Sqrt[10])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\sqrt {10}\, \ln \left (\sqrt {10}\, x +2 x^{2}+1\right )}{20}-\frac {\sqrt {10}\, \ln \left (-\sqrt {10}\, x +2 x^{2}+1\right )}{20}\)	\(39\)
default	\(\frac {2 \left (\sqrt {5}-1\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}+\frac {2 \left (\sqrt {5}+1\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )}\)	\(82\)

[In]

int((-2*x^2+1)/(4*x^4-6*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/20*10^(1/2)*ln(10^(1/2)*x+2*x^2+1)-1/20*10^(1/2)*ln(-10^(1/2)*x+2*x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {1}{20} \, \sqrt {10} \log \left (\frac {4 \, x^{4} + 14 \, x^{2} + 2 \, \sqrt {10} {\left (2 \, x^{3} + x\right )} + 1}{4 \, x^{4} - 6 \, x^{2} + 1}\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="fricas")

[Out]

1/20*sqrt(10)*log((4*x^4 + 14*x^2 + 2*sqrt(10)*(2*x^3 + x) + 1)/(4*x^4 - 6*x^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=- \frac {\sqrt {10} \log {\left (x^{2} - \frac {\sqrt {10} x}{2} + \frac {1}{2} \right )}}{20} + \frac {\sqrt {10} \log {\left (x^{2} + \frac {\sqrt {10} x}{2} + \frac {1}{2} \right )}}{20} \]

[In]

integrate((-2*x**2+1)/(4*x**4-6*x**2+1),x)

[Out]

-sqrt(10)*log(x**2 - sqrt(10)*x/2 + 1/2)/20 + sqrt(10)*log(x**2 + sqrt(10)*x/2 + 1/2)/20

Maxima [F]

\[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\int { -\frac {2 \, x^{2} - 1}{4 \, x^{4} - 6 \, x^{2} + 1} \,d x } \]

[In]

integrate((-2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="maxima")

[Out]

-integrate((2*x^2 - 1)/(4*x^4 - 6*x^2 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.60 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {1}{20} \, \sqrt {10} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) + \frac {1}{20} \, \sqrt {10} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) - \frac {1}{20} \, \sqrt {10} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) - \frac {1}{20} \, \sqrt {10} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="giac")

[Out]

1/20*sqrt(10)*log(abs(x + 1/4*sqrt(10) + 1/4*sqrt(2))) + 1/20*sqrt(10)*log(abs(x + 1/4*sqrt(10) - 1/4*sqrt(2))
) - 1/20*sqrt(10)*log(abs(x - 1/4*sqrt(10) + 1/4*sqrt(2))) - 1/20*sqrt(10)*log(abs(x - 1/4*sqrt(10) - 1/4*sqrt
(2)))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.42 \[ \int \frac {1-2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\sqrt {10}\,\mathrm {atanh}\left (\frac {\sqrt {10}\,x}{2\,x^2+1}\right )}{10} \]

[In]

int(-(2*x^2 - 1)/(4*x^4 - 6*x^2 + 1),x)

[Out]

(10^(1/2)*atanh((10^(1/2)*x)/(2*x^2 + 1)))/10