\(\int \frac {x^5}{2+3 x^4} \, dx\) [689]

Optimal result

Integrand size = 13, antiderivative size = 29 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {x^2}{6}-\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{3 \sqrt {6}} \]

[Out]

1/6*x^2-1/18*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 327, 209} \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {x^2}{6}-\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{3 \sqrt {6}} \]

[In]

Int[x^5/(2 + 3*x^4),x]

[Out]

x^2/6 - ArcTan[Sqrt[3/2]*x^2]/(3*Sqrt[6])

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]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {x^2}{6}-\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{3 \sqrt {6}} \]

[In]

Integrate[x^5/(2 + 3*x^4),x]

[Out]

x^2/6 - ArcTan[Sqrt[3/2]*x^2]/(3*Sqrt[6])

Maple [A] (verified)

Time = 3.99 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {x^{2}}{6}-\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{18}\)	\(21\)
risch	\(\frac {x^{2}}{6}-\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{18}\)	\(21\)
meijerg	\(\frac {\sqrt {6}\, \left (x^{2} \sqrt {6}-2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )\right )}{36}\)	\(28\)

[In]

int(x^5/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/6*x^2-1/18*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {1}{6} \, x^{2} - \frac {1}{18} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

integrate(x^5/(3*x^4+2),x, algorithm="fricas")

[Out]

1/6*x^2 - 1/18*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {x^{2}}{6} - \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{18} \]

[In]

integrate(x**5/(3*x**4+2),x)

[Out]

x**2/6 - sqrt(6)*atan(sqrt(6)*x**2/2)/18

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {1}{6} \, x^{2} - \frac {1}{18} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

integrate(x^5/(3*x^4+2),x, algorithm="maxima")

[Out]

1/6*x^2 - 1/18*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {1}{6} \, x^{2} - \frac {1}{18} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

integrate(x^5/(3*x^4+2),x, algorithm="giac")

[Out]

1/6*x^2 - 1/18*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x^5}{2+3 x^4} \, dx=\frac {x^2}{6}-\frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{18} \]

[In]

int(x^5/(3*x^4 + 2),x)

[Out]

x^2/6 - (6^(1/2)*atan((6^(1/2)*x^2)/2))/18