\(\int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx\) [50]

Optimal result

Integrand size = 22, antiderivative size = 23 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=-\arctan \left (\sqrt {7}-4 x\right )+\arctan \left (\sqrt {7}+4 x\right ) \]

[Out]

arctan(4*x-7^(1/2))+arctan(4*x+7^(1/2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, 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} \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\arctan \left (4 x+\sqrt {7}\right )-\arctan \left (\sqrt {7}-4 x\right ) \]

[In]

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

[Out]

-ArcTan[Sqrt[7] - 4*x] + ArcTan[Sqrt[7] + 4*x]

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 {\sqrt {7} x}{2}+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\frac {\sqrt {7} x}{2}+x^2} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {1}{4}-x^2} \, dx,x,-\frac {\sqrt {7}}{2}+2 x\right )\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {1}{4}-x^2} \, dx,x,\frac {\sqrt {7}}{2}+2 x\right ) \\ & = -\tan ^{-1}\left (\sqrt {7}-4 x\right )+\tan ^{-1}\left (\sqrt {7}+4 x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=-\arctan \left (\frac {x}{-1+2 x^2}\right ) \]

[In]

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

[Out]

-ArcTan[x/(-1 + 2*x^2)]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
risch	\(\arctan \left (4 x^{3}-x \right )+\arctan \left (2 x \right )\)	\(16\)
default	\(\arctan \left (4 x -\sqrt {7}\right )+\arctan \left (4 x +\sqrt {7}\right )\)	\(20\)
parallelrisch	\(-\frac {i \ln \left (x^{2}-\frac {1}{2} i x -\frac {1}{2}\right )}{2}+\frac {i \ln \left (x^{2}+\frac {1}{2} i x -\frac {1}{2}\right )}{2}\)	\(28\)

[In]

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

[Out]

arctan(4*x^3-x)+arctan(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\arctan \left (4 \, x^{3} - x\right ) + \arctan \left (2 \, x\right ) \]

[In]

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

[Out]

arctan(4*x^3 - x) + arctan(2*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\operatorname {atan}{\left (2 x \right )} + \operatorname {atan}{\left (4 x^{3} - x \right )} \]

[In]

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

[Out]

atan(2*x) + atan(4*x**3 - x)

Maxima [F]

\[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\int { \frac {2 \, x^{2} + 1}{4 \, x^{4} - 3 \, x^{2} + 1} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\arctan \left (2 \, \sqrt {2} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (4 \, x + \sqrt {14} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) + \arctan \left (2 \, \sqrt {2} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (4 \, x - \sqrt {14} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) \]

[In]

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

[Out]

arctan(2*sqrt(2)*(1/4)^(3/4)*(4*x + sqrt(14)*(1/4)^(1/4))) + arctan(2*sqrt(2)*(1/4)^(3/4)*(4*x - sqrt(14)*(1/4
)^(1/4)))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {1+2 x^2}{1-3 x^2+4 x^4} \, dx=\mathrm {atan}\left (2\,x\right )-\mathrm {atan}\left (x-4\,x^3\right ) \]

[In]

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

[Out]

atan(2*x) - atan(x - 4*x^3)