\(\int \frac {x^2}{2+x^3+x^6} \, dx\) [183]

Optimal result

Integrand size = 14, antiderivative size = 23 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2 \arctan \left (\frac {1+2 x^3}{\sqrt {7}}\right )}{3 \sqrt {7}} \]

[Out]

2/21*arctan(1/7*(2*x^3+1)*7^(1/2))*7^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, 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 = {1366, 632, 210} \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2 \arctan \left (\frac {2 x^3+1}{\sqrt {7}}\right )}{3 \sqrt {7}} \]

[In]

Int[x^2/(2 + x^3 + x^6),x]

[Out]

(2*ArcTan[(1 + 2*x^3)/Sqrt[7]])/(3*Sqrt[7])

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 1366

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,x^3\right ) \\ & = -\left (\frac {2}{3} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 x^3\right )\right ) \\ & = \frac {2 \tan ^{-1}\left (\frac {1+2 x^3}{\sqrt {7}}\right )}{3 \sqrt {7}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2 \arctan \left (\frac {1+2 x^3}{\sqrt {7}}\right )}{3 \sqrt {7}} \]

[In]

Integrate[x^2/(2 + x^3 + x^6),x]

[Out]

(2*ArcTan[(1 + 2*x^3)/Sqrt[7]])/(3*Sqrt[7])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {2 \arctan \left (\frac {\left (2 x^{3}+1\right ) \sqrt {7}}{7}\right ) \sqrt {7}}{21}\)	\(19\)
risch	\(\frac {2 \arctan \left (\frac {\left (2 x^{3}+1\right ) \sqrt {7}}{7}\right ) \sqrt {7}}{21}\)	\(19\)

[In]

int(x^2/(x^6+x^3+2),x,method=_RETURNVERBOSE)

[Out]

2/21*arctan(1/7*(2*x^3+1)*7^(1/2))*7^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2}{21} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{3} + 1\right )}\right ) \]

[In]

integrate(x^2/(x^6+x^3+2),x, algorithm="fricas")

[Out]

2/21*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^3 + 1))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2 \sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x^{3}}{7} + \frac {\sqrt {7}}{7} \right )}}{21} \]

[In]

integrate(x**2/(x**6+x**3+2),x)

[Out]

2*sqrt(7)*atan(2*sqrt(7)*x**3/7 + sqrt(7)/7)/21

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2}{21} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{3} + 1\right )}\right ) \]

[In]

integrate(x^2/(x^6+x^3+2),x, algorithm="maxima")

[Out]

2/21*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^3 + 1))

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2}{21} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{3} + 1\right )}\right ) \]

[In]

integrate(x^2/(x^6+x^3+2),x, algorithm="giac")

[Out]

2/21*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^3 + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{2+x^3+x^6} \, dx=\frac {2\,\sqrt {7}\,\mathrm {atan}\left (\frac {2\,\sqrt {7}\,x^3}{7}+\frac {\sqrt {7}}{7}\right )}{21} \]

[In]

int(x^2/(x^3 + x^6 + 2),x)

[Out]

(2*7^(1/2)*atan(7^(1/2)/7 + (2*7^(1/2)*x^3)/7))/21