\(\int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx\) [49]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=-\frac {\arctan \left (\sqrt {3}-2 \sqrt {2} x\right )}{\sqrt {2}}+\frac {\arctan \left (\sqrt {3}+2 \sqrt {2} x\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, 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} \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {\arctan \left (2 \sqrt {2} x+\sqrt {3}\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {3}-2 \sqrt {2} x\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.25 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {\left (-3 i+\sqrt {3}\right ) \arctan \left (\frac {2 x}{\sqrt {-1-i \sqrt {3}}}\right )}{2 \sqrt {3 \left (-1-i \sqrt {3}\right )}}+\frac {\left (3 i+\sqrt {3}\right ) \arctan \left (\frac {2 x}{\sqrt {-1+i \sqrt {3}}}\right )}{2 \sqrt {3 \left (-1+i \sqrt {3}\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {\arctan \left (x \sqrt {2}\right ) \sqrt {2}}{2}+\frac {\sqrt {2}\, \arctan \left (2 x^{3} \sqrt {2}\right )}{2}\)	\(27\)
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\left (4 x -\sqrt {6}\right ) \sqrt {2}}{2}\right )}{2}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (4 x +\sqrt {6}\right ) \sqrt {2}}{2}\right )}{2}\)	\(40\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.59 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (2 \, \sqrt {2} x^{3}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {\sqrt {2} \cdot \left (2 \operatorname {atan}{\left (\sqrt {2} x \right )} + 2 \operatorname {atan}{\left (2 \sqrt {2} x^{3} \right )}\right )}{4} \]

[In]

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

[Out]

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

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 = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {3} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {3} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.48 \[ \int \frac {1+2 x^2}{1-2 x^2+4 x^4} \, dx=\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\sqrt {2}\,x\right )+\mathrm {atan}\left (2\,\sqrt {2}\,x^3\right )\right )}{2} \]

[In]

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

[Out]

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