\(\int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx\) [45]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=-\frac {\arctan \left (\frac {1-2 \sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}}+\frac {\arctan \left (\frac {1+2 \sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \]

[Out]

-1/6*arctan(1/3*(1-2*x*2^(1/2))*3^(1/2))*6^(1/2)+1/6*arctan(1/3*(1+2*x*2^(1/2))*3^(1/2))*6^(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, 210} \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {2 \sqrt {2} x+1}{\sqrt {3}}\right )}{\sqrt {6}}-\frac {\arctan \left (\frac {1-2 \sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \]

[In]

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

[Out]

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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}& = \frac {1}{4} \int \frac {1}{\frac {1}{2}-\frac {x}{\sqrt {2}}+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\frac {x}{\sqrt {2}}+x^2} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {3}{2}-x^2} \, dx,x,-\frac {1}{\sqrt {2}}+2 x\right )\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {3}{2}-x^2} \, dx,x,\frac {1}{\sqrt {2}}+2 x\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1-2 \sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}}+\frac {\tan ^{-1}\left (\frac {1+2 \sqrt {2} x}{\sqrt {3}}\right )}{\sqrt {6}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.06 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {2 x}{\sqrt {1-i \sqrt {3}}}\right )}{2 \sqrt {3 \left (1-i \sqrt {3}\right )}}+\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {2 x}{\sqrt {1+i \sqrt {3}}}\right )}{2 \sqrt {3 \left (1+i \sqrt {3}\right )}} \]

[In]

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

[Out]

((-I + Sqrt[3])*ArcTan[(2*x)/Sqrt[1 - I*Sqrt[3]]])/(2*Sqrt[3*(1 - I*Sqrt[3])]) + ((I + Sqrt[3])*ArcTan[(2*x)/S
qrt[1 + I*Sqrt[3]]])/(2*Sqrt[3*(1 + I*Sqrt[3])])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\sqrt {6}\, \arctan \left (\frac {x \sqrt {6}}{3}\right )}{6}+\frac {\sqrt {6}\, \arctan \left (\frac {2 x^{3} \sqrt {6}}{3}+\frac {2 x \sqrt {6}}{3}\right )}{6}\)	\(35\)
default	\(\frac {\sqrt {6}\, \arctan \left (\frac {\left (4 x -\sqrt {2}\right ) \sqrt {6}}{6}\right )}{6}+\frac {\sqrt {6}\, \arctan \left (\frac {\left (4 x +\sqrt {2}\right ) \sqrt {6}}{6}\right )}{6}\)	\(40\)

[In]

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

[Out]

1/6*6^(1/2)*arctan(1/3*x*6^(1/2))+1/6*6^(1/2)*arctan(2/3*x^3*6^(1/2)+2/3*x*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {1}{6} \, \sqrt {6} \arctan \left (\frac {2}{3} \, \sqrt {6} {\left (x^{3} + x\right )}\right ) + \frac {1}{6} \, \sqrt {6} \arctan \left (\frac {1}{3} \, \sqrt {6} x\right ) \]

[In]

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

[Out]

1/6*sqrt(6)*arctan(2/3*sqrt(6)*(x^3 + x)) + 1/6*sqrt(6)*arctan(1/3*sqrt(6)*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {\sqrt {6} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {6} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {2 \sqrt {6} x^{3}}{3} + \frac {2 \sqrt {6} x}{3} \right )}\right )}{12} \]

[In]

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

[Out]

sqrt(6)*(2*atan(sqrt(6)*x/3) + 2*atan(2*sqrt(6)*x**3/3 + 2*sqrt(6)*x/3))/12

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {1}{6} \, \sqrt {6} \arctan \left (\frac {4}{3} \, \sqrt {3} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (2 \, x + \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{6} \, \sqrt {6} \arctan \left (\frac {4}{3} \, \sqrt {3} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (2 \, x - \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) \]

[In]

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

[Out]

1/6*sqrt(6)*arctan(4/3*sqrt(3)*(1/4)^(3/4)*(2*x + (1/4)^(1/4))) + 1/6*sqrt(6)*arctan(4/3*sqrt(3)*(1/4)^(3/4)*(
2*x - (1/4)^(1/4)))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60 \[ \int \frac {1+2 x^2}{1+2 x^2+4 x^4} \, dx=\frac {\sqrt {6}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {6}\,x^3}{3}+\frac {2\,\sqrt {6}\,x}{3}\right )+\mathrm {atan}\left (\frac {\sqrt {6}\,x}{3}\right )\right )}{6} \]

[In]

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

[Out]

(6^(1/2)*(atan((2*6^(1/2)*x)/3 + (2*6^(1/2)*x^3)/3) + atan((6^(1/2)*x)/3)))/6