\(\int \frac {x}{10+2 x^2+x^4} \, dx\) [292]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \arctan \left (\frac {1}{3} \left (1+x^2\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1121, 632, 210} \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \arctan \left (\frac {1}{3} \left (x^2+1\right )\right ) \]

[In]

Int[x/(10 + 2*x^2 + x^4),x]

[Out]

ArcTan[(1 + x^2)/3]/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 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{10+2 x+x^2} \, dx,x,x^2\right ) \\ & = -\text {Subst}\left (\int \frac {1}{-36-x^2} \, dx,x,2 \left (1+x^2\right )\right ) \\ & = \frac {1}{6} \arctan \left (\frac {1}{3} \left (1+x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \arctan \left (\frac {1}{3} \left (1+x^2\right )\right ) \]

[In]

Integrate[x/(10 + 2*x^2 + x^4),x]

[Out]

ArcTan[(1 + x^2)/3]/6

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\arctan \left (\frac {x^{2}}{3}+\frac {1}{3}\right )}{6}\)	\(11\)
risch	\(\frac {\arctan \left (\frac {x^{2}}{3}+\frac {1}{3}\right )}{6}\)	\(11\)
parallelrisch	\(\frac {i \ln \left (x^{2}+3 i+1\right )}{12}-\frac {i \ln \left (x^{2}-3 i+1\right )}{12}\)	\(24\)

[In]

int(x/(x^4+2*x^2+10),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="fricas")

[Out]

1/6*arctan(1/3*x^2 + 1/3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {\operatorname {atan}{\left (\frac {x^{2}}{3} + \frac {1}{3} \right )}}{6} \]

[In]

integrate(x/(x**4+2*x**2+10),x)

[Out]

atan(x**2/3 + 1/3)/6

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="maxima")

[Out]

1/6*arctan(1/3*x^2 + 1/3)

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="giac")

[Out]

1/6*arctan(1/3*x^2 + 1/3)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x}{10+2 x^2+x^4} \, dx=\frac {\mathrm {atan}\left (\frac {x^2}{3}+\frac {1}{3}\right )}{6} \]

[In]

int(x/(2*x^2 + x^4 + 10),x)

[Out]

atan(x^2/3 + 1/3)/6