\(\int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx\) [55]

Optimal result

Integrand size = 22, antiderivative size = 46 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {2}}-\frac {\arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {2}} \]

[Out]

1/2*arctan(2*x/(1/2*10^(1/2)-1/2*2^(1/2)))*2^(1/2)-1/2*arctan(2*x/(1/2*10^(1/2)+1/2*2^(1/2)))*2^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1177, 209} \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {2}}-\frac {\arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {2}} \]

[In]

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

[Out]

ArcTan[(2*x)/Sqrt[3 - Sqrt[5]]]/Sqrt[2] - ArcTan[(2*x)/Sqrt[3 + Sqrt[5]]]/Sqrt[2]

Rule 209

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

Rule 1177

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), 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[b^2
- 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \left (-1-\sqrt {5}\right ) \int \frac {1}{3+\sqrt {5}+4 x^2} \, dx+\left (-1+\sqrt {5}\right ) \int \frac {1}{3-\sqrt {5}+4 x^2} \, dx \\ & = \frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.83 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {-\left (\left (-5+\sqrt {5}\right ) \sqrt {3+\sqrt {5}} \arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )\right )-\sqrt {3-\sqrt {5}} \left (5+\sqrt {5}\right ) \arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{4 \sqrt {5}} \]

[In]

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

[Out]

(-((-5 + Sqrt[5])*Sqrt[3 + Sqrt[5]]*ArcTan[(2*x)/Sqrt[3 - Sqrt[5]]]) - Sqrt[3 - Sqrt[5]]*(5 + Sqrt[5])*ArcTan[
(2*x)/Sqrt[3 + Sqrt[5]]])/(4*Sqrt[5])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\arctan \left (x \sqrt {2}\right ) \sqrt {2}}{2}+\frac {\sqrt {2}\, \arctan \left (2 x^{3} \sqrt {2}+2 x \sqrt {2}\right )}{2}\)	\(34\)
default	\(-\frac {2 \sqrt {5}\, \left (5+\sqrt {5}\right ) \arctan \left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )}-\frac {2 \left (-5+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}\)	\(82\)

[In]

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

[Out]

-1/2*arctan(x*2^(1/2))*2^(1/2)+1/2*2^(1/2)*arctan(2*x^3*2^(1/2)+2*x*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (2 \, \sqrt {2} {\left (x^{3} + x\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) \]

[In]

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

[Out]

1/2*sqrt(2)*arctan(2*sqrt(2)*(x^3 + x)) - 1/2*sqrt(2)*arctan(sqrt(2)*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=- \frac {\sqrt {2} \cdot \left (2 \operatorname {atan}{\left (\sqrt {2} x \right )} - 2 \operatorname {atan}{\left (2 \sqrt {2} x^{3} + 2 \sqrt {2} x \right )}\right )}{4} \]

[In]

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

[Out]

-sqrt(2)*(2*atan(sqrt(2)*x) - 2*atan(2*sqrt(2)*x**3 + 2*sqrt(2)*x))/4

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 = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {4 \, x}{\sqrt {10} + \sqrt {2}}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {4 \, x}{\sqrt {10} - \sqrt {2}}\right ) \]

[In]

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

[Out]

-1/2*sqrt(2)*arctan(4*x/(sqrt(10) + sqrt(2))) + 1/2*sqrt(2)*arctan(4*x/(sqrt(10) - sqrt(2)))

Mupad [B] (verification not implemented)

Time = 13.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \frac {1-2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\sqrt {2}\,\left (\mathrm {atan}\left (2\,\sqrt {2}\,x^3+2\,\sqrt {2}\,x\right )-\mathrm {atan}\left (\sqrt {2}\,x\right )\right )}{2} \]

[In]

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

[Out]

(2^(1/2)*(atan(2*2^(1/2)*x + 2*2^(1/2)*x^3) - atan(2^(1/2)*x)))/2