\(\int \frac {x^2}{1+x^6} \, dx\) [1367]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 8 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {\arctan \left (x^3\right )}{3} \]

[Out]

1/3*arctan(x^3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 209} \[ \int \frac {x^2}{1+x^6} \, dx=\frac {\arctan \left (x^3\right )}{3} \]

[In]

Int[x^2/(1 + x^6),x]

[Out]

ArcTan[x^3]/3

Rule 209

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {\arctan \left (x^3\right )}{3} \]

[In]

Integrate[x^2/(1 + x^6),x]

[Out]

ArcTan[x^3]/3

Maple [A] (verified)

Time = 4.56 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method result size
default \(\frac {\arctan \left (x^{3}\right )}{3}\) \(7\)
meijerg \(\frac {\arctan \left (x^{3}\right )}{3}\) \(7\)
risch \(\frac {\arctan \left (x^{3}\right )}{3}\) \(7\)
parallelrisch \(\frac {i \ln \left (x -i\right )}{6}-\frac {i \ln \left (x +i\right )}{6}-\frac {i \ln \left (x^{2}-i x -1\right )}{6}+\frac {i \ln \left (x^{2}+i x -1\right )}{6}\) \(44\)

[In]

int(x^2/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

1/3*arctan(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (x^{3}\right ) \]

[In]

integrate(x^2/(x^6+1),x, algorithm="fricas")

[Out]

1/3*arctan(x^3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {\operatorname {atan}{\left (x^{3} \right )}}{3} \]

[In]

integrate(x**2/(x**6+1),x)

[Out]

atan(x**3)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (x^{3}\right ) \]

[In]

integrate(x^2/(x^6+1),x, algorithm="maxima")

[Out]

1/3*arctan(x^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (x^{3}\right ) \]

[In]

integrate(x^2/(x^6+1),x, algorithm="giac")

[Out]

1/3*arctan(x^3)

Mupad [B] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{1+x^6} \, dx=\frac {\mathrm {atan}\left (x^3\right )}{3} \]

[In]

int(x^2/(x^6 + 1),x)

[Out]

atan(x^3)/3