\(\int \frac {x^3}{(1+x^6)^{2/3}} \, dx\) [1205]

Optimal result

Integrand size = 13, antiderivative size = 88 \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{1+x^6}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (-x^2+\sqrt [3]{1+x^6}\right )+\frac {1}{12} \log \left (x^4+x^2 \sqrt [3]{1+x^6}+\left (1+x^6\right )^{2/3}\right ) \]

[Out]

-1/6*arctan(3^(1/2)*x^2/(x^2+2*(x^6+1)^(1/3)))*3^(1/2)-1/6*ln(-x^2+(x^6+1)^(1/3))+1/12*ln(x^4+x^2*(x^6+1)^(1/3
)+(x^6+1)^(2/3))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {281, 337} \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=-\frac {\arctan \left (\frac {\frac {2 x^2}{\sqrt [3]{x^6+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{4} \log \left (x^2-\sqrt [3]{x^6+1}\right ) \]

[In]

Int[x^3/(1 + x^6)^(2/3),x]

[Out]

-1/2*ArcTan[(1 + (2*x^2)/(1 + x^6)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[x^2 - (1 + x^6)^(1/3)]/4

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 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx,x,x^2\right ) \\ & = -\frac {\arctan \left (\frac {1+\frac {2 x^2}{\sqrt [3]{1+x^6}}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{4} \log \left (x^2-\sqrt [3]{1+x^6}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\frac {1}{12} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{1+x^6}}\right )-2 \log \left (-x^2+\sqrt [3]{1+x^6}\right )+\log \left (x^4+x^2 \sqrt [3]{1+x^6}+\left (1+x^6\right )^{2/3}\right )\right ) \]

[In]

Integrate[x^3/(1 + x^6)^(2/3),x]

[Out]

(-2*Sqrt[3]*ArcTan[(Sqrt[3]*x^2)/(x^2 + 2*(1 + x^6)^(1/3))] - 2*Log[-x^2 + (1 + x^6)^(1/3)] + Log[x^4 + x^2*(1
+ x^6)^(1/3) + (1 + x^6)^(2/3)])/12

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 5.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.19

method	result	size
meijerg	\(\frac {x^{4} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{6}\right )}{4}\)	\(17\)
pseudoelliptic	\(-\frac {\ln \left (\frac {-x^{2}+\left (x^{6}+1\right )^{\frac {1}{3}}}{x^{2}}\right )}{6}+\frac {\ln \left (\frac {x^{4}+x^{2} \left (x^{6}+1\right )^{\frac {1}{3}}+\left (x^{6}+1\right )^{\frac {2}{3}}}{x^{4}}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{2}+2 \left (x^{6}+1\right )^{\frac {1}{3}}\right )}{3 x^{2}}\right )}{6}\)	\(78\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{6}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{6}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+x^{6}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1\right )}{6}-\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{6}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{6}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+4 x^{6}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}+3 x^{4} \left (x^{6}+1\right )^{\frac {1}{3}}+3 x^{2} \left (x^{6}+1\right )^{\frac {2}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )}{6}+\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{6}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{6}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+4 x^{6}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}+3 x^{4} \left (x^{6}+1\right )^{\frac {1}{3}}+3 x^{2} \left (x^{6}+1\right )^{\frac {2}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+2\right )}{6}\)	\(340\)

[In]

int(x^3/(x^6+1)^(2/3),x,method=_RETURNVERBOSE)

[Out]

1/4*x^4*hypergeom([2/3,2/3],[5/3],-x^6)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2} + 2 \, \sqrt {3} {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{3 \, x^{2}}\right ) - \frac {1}{6} \, \log \left (-\frac {x^{2} - {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}}\right ) + \frac {1}{12} \, \log \left (\frac {x^{4} + {\left (x^{6} + 1\right )}^{\frac {1}{3}} x^{2} + {\left (x^{6} + 1\right )}^{\frac {2}{3}}}{x^{4}}\right ) \]

[In]

integrate(x^3/(x^6+1)^(2/3),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*arctan(1/3*(sqrt(3)*x^2 + 2*sqrt(3)*(x^6 + 1)^(1/3))/x^2) - 1/6*log(-(x^2 - (x^6 + 1)^(1/3))/x^2)
+ 1/12*log((x^4 + (x^6 + 1)^(1/3)*x^2 + (x^6 + 1)^(2/3))/x^4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.33 \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\frac {x^{4} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{6} e^{i \pi }} \right )}}{6 \Gamma \left (\frac {5}{3}\right )} \]

[In]

integrate(x**3/(x**6+1)**(2/3),x)

[Out]

x**4*gamma(2/3)*hyper((2/3, 2/3), (5/3,), x**6*exp_polar(I*pi))/(6*gamma(5/3))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} + 1\right )}\right ) + \frac {1}{12} \, \log \left (\frac {{\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} + \frac {{\left (x^{6} + 1\right )}^{\frac {2}{3}}}{x^{4}} + 1\right ) - \frac {1}{6} \, \log \left (\frac {{\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} - 1\right ) \]

[In]

integrate(x^3/(x^6+1)^(2/3),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^6 + 1)^(1/3)/x^2 + 1)) + 1/12*log((x^6 + 1)^(1/3)/x^2 + (x^6 + 1)^(2/3)/x
^4 + 1) - 1/6*log((x^6 + 1)^(1/3)/x^2 - 1)

Giac [F]

\[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\int { \frac {x^{3}}{{\left (x^{6} + 1\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(x^3/(x^6+1)^(2/3),x, algorithm="giac")

[Out]

integrate(x^3/(x^6 + 1)^(2/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx=\int \frac {x^3}{{\left (x^6+1\right )}^{2/3}} \,d x \]

[In]

int(x^3/(x^6 + 1)^(2/3),x)

[Out]

int(x^3/(x^6 + 1)^(2/3), x)