3.535 \(\int \frac{1}{1+2 x^2+x^4} \, dx\)

Optimal. Leaf size=19 \[ \frac{x}{2 \left (x^2+1\right )}+\frac{1}{2} \tan ^{-1}(x) \]

[Out]

x/(2*(1 + x^2)) + ArcTan[x]/2

________________________________________________________________________________________

Rubi [A]  time = 0.0031113, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {28, 199, 203} \[ \frac{x}{2 \left (x^2+1\right )}+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2 + x^4)^(-1),x]

[Out]

x/(2*(1 + x^2)) + ArcTan[x]/2

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.005355, size = 16, normalized size = 0.84 \[ \frac{1}{2} \left (\frac{x}{x^2+1}+\tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2 + x^4)^(-1),x]

[Out]

(x/(1 + x^2) + ArcTan[x])/2

________________________________________________________________________________________

Maple [A]  time = 0.044, size = 16, normalized size = 0.8 \begin{align*}{\frac{x}{2\,{x}^{2}+2}}+{\frac{\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.48803, size = 20, normalized size = 1.05 \begin{align*} \frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x/(x^2 + 1) + 1/2*arctan(x)

________________________________________________________________________________________

Fricas [A]  time = 1.48237, size = 55, normalized size = 2.89 \begin{align*} \frac{{\left (x^{2} + 1\right )} \arctan \left (x\right ) + x}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.098102, size = 12, normalized size = 0.63 \begin{align*} \frac{x}{2 x^{2} + 2} + \frac{\operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*x**2 + 2) + atan(x)/2

________________________________________________________________________________________

Giac [A]  time = 1.12473, size = 20, normalized size = 1.05 \begin{align*} \frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x/(x^2 + 1) + 1/2*arctan(x)