\(\int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx\) [1492]

Optimal result

Integrand size = 15, antiderivative size = 104 \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{-x^2+x^6}}\right )-\frac {1}{4} \log \left (-x^2+\sqrt [3]{-x^2+x^6}\right )+\frac {1}{8} \log \left (x^4+x^2 \sqrt [3]{-x^2+x^6}+\left (-x^2+x^6\right )^{2/3}\right ) \]

[Out]

1/4*3^(1/2)*arctan(3^(1/2)*x^2/(x^2+2*(x^6-x^2)^(1/3)))-1/4*ln(-x^2+(x^6-x^2)^(1/3))+1/8*ln(x^4+x^2*(x^6-x^2)^
(1/3)+(x^6-x^2)^(2/3))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2038, 2036, 335, 281, 245} \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=\frac {\sqrt {3} \sqrt [3]{x^2} \sqrt [3]{x^4-1} \arctan \left (\frac {\frac {2 \left (x^2\right )^{2/3}}{\sqrt [3]{x^4-1}}+1}{\sqrt {3}}\right )}{4 \sqrt [3]{x^6-x^2}}-\frac {3 \sqrt [3]{x^2} \sqrt [3]{x^4-1} \log \left (\left (x^2\right )^{2/3}-\sqrt [3]{x^4-1}\right )}{8 \sqrt [3]{x^6-x^2}} \]

[In]

Int[x/(-x^2 + x^6)^(1/3),x]

[Out]

(Sqrt[3]*(x^2)^(1/3)*(-1 + x^4)^(1/3)*ArcTan[(1 + (2*(x^2)^(2/3))/(-1 + x^4)^(1/3))/Sqrt[3]])/(4*(-x^2 + x^6)^
(1/3)) - (3*(x^2)^(1/3)*(-1 + x^4)^(1/3)*Log[(x^2)^(2/3) - (-1 + x^4)^(1/3)])/(8*(-x^2 + x^6)^(1/3))

Rule 245

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

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 335

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

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2038

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-x+x^3}} \, dx,x,x^2\right ) \\ & = \frac {\left (\sqrt [3]{x^2} \sqrt [3]{-1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \sqrt [3]{-1+x^2}} \, dx,x,x^2\right )}{2 \sqrt [3]{-x^2+x^6}} \\ & = \frac {\left (3 \sqrt [3]{x^2} \sqrt [3]{-1+x^4}\right ) \text {Subst}\left (\int \frac {x}{\sqrt [3]{-1+x^6}} \, dx,x,\sqrt [3]{x^2}\right )}{2 \sqrt [3]{-x^2+x^6}} \\ & = \frac {\left (3 \sqrt [3]{x^2} \sqrt [3]{-1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x^3}} \, dx,x,\left (x^2\right )^{2/3}\right )}{4 \sqrt [3]{-x^2+x^6}} \\ & = \frac {\sqrt {3} \sqrt [3]{x^2} \sqrt [3]{-1+x^4} \arctan \left (\frac {1+\frac {2 \left (x^2\right )^{2/3}}{\sqrt [3]{-1+x^4}}}{\sqrt {3}}\right )}{4 \sqrt [3]{-x^2+x^6}}-\frac {3 \sqrt [3]{x^2} \sqrt [3]{-1+x^4} \log \left (\left (x^2\right )^{2/3}-\sqrt [3]{-1+x^4}\right )}{8 \sqrt [3]{-x^2+x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.61 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=\frac {x^{2/3} \sqrt [3]{-1+x^4} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{4/3}}{x^{4/3}+2 \sqrt [3]{-1+x^4}}\right )-2 \log \left (-x^{4/3}+\sqrt [3]{-1+x^4}\right )+\log \left (x^{8/3}+x^{4/3} \sqrt [3]{-1+x^4}+\left (-1+x^4\right )^{2/3}\right )\right )}{8 \sqrt [3]{x^2 \left (-1+x^4\right )}} \]

[In]

Integrate[x/(-x^2 + x^6)^(1/3),x]

[Out]

(x^(2/3)*(-1 + x^4)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(4/3))/(x^(4/3) + 2*(-1 + x^4)^(1/3))] - 2*Log[-x^(4/3)
+ (-1 + x^4)^(1/3)] + Log[x^(8/3) + x^(4/3)*(-1 + x^4)^(1/3) + (-1 + x^4)^(2/3)]))/(8*(x^2*(-1 + x^4))^(1/3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.32

method	result	size
meijerg	\(\frac {3 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{3}} x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{4}\right )}{4 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{3}}}\)	\(33\)
pseudoelliptic	\(\frac {\ln \left (\frac {x^{4}+x^{2} \left (x^{6}-x^{2}\right )^{\frac {1}{3}}+\left (x^{6}-x^{2}\right )^{\frac {2}{3}}}{x^{4}}\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{2}+2 \left (x^{6}-x^{2}\right )^{\frac {1}{3}}\right )}{3 x^{2}}\right )}{4}-\frac {\ln \left (\frac {-x^{2}+\left (x^{6}-x^{2}\right )^{\frac {1}{3}}}{x^{2}}\right )}{4}\)	\(94\)
trager	\(-\frac {\ln \left (-1534273 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}-133699480 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{3}} x^{2}[phone] x^{4}-209167776 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}-x^{2}\right )^{\frac {2}{3}}-418335552 x^{2} \left (x^{6}-x^{2}\right )^{\frac {1}{3}}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-273536052 \left (x^{6}-x^{2}\right )^{\frac {2}{3}}-73229986 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-223609080\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (1863409 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}-176694719 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{3}} x^{2}[phone] x^{4}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}-x^{2}\right )^{\frac {2}{3}}[phone] x^{2} \left (x^{6}-x^{2}\right )^{\frac {1}{3}}-29814544 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-345935802 \left (x^{6}-x^{2}\right )^{\frac {2}{3}}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-95124926\right )}{8}\)	\(291\)

[In]

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

[Out]

3/4/signum(x^4-1)^(1/3)*(-signum(x^4-1))^(1/3)*x^(4/3)*hypergeom([1/3,1/3],[4/3],x^4)

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (-\frac {44032959556 \, \sqrt {3} {\left (x^{6} - x^{2}\right )}^{\frac {1}{3}} x^{2} + \sqrt {3} {\left (16754327161 \, x^{4} - 2707204793\right )} - 10524305234 \, \sqrt {3} {\left (x^{6} - x^{2}\right )}^{\frac {2}{3}}}{81835897185 \, x^{4} - 1102302937}\right ) - \frac {1}{8} \, \log \left (-3 \, {\left (x^{6} - x^{2}\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{6} - x^{2}\right )}^{\frac {2}{3}} + 1\right ) \]

[In]

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

[Out]

1/4*sqrt(3)*arctan(-(44032959556*sqrt(3)*(x^6 - x^2)^(1/3)*x^2 + sqrt(3)*(16754327161*x^4 - 2707204793) - 1052
4305234*sqrt(3)*(x^6 - x^2)^(2/3))/(81835897185*x^4 - 1102302937)) - 1/8*log(-3*(x^6 - x^2)^(1/3)*x^2 + 3*(x^6
- x^2)^(2/3) + 1)

Sympy [F]

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

[In]

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

[Out]

Integral(x/(x**2*(x - 1)*(x + 1)*(x**2 + 1))**(1/3), x)

Maxima [F]

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

[In]

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

[Out]

integrate(x/(x^6 - x^2)^(1/3), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=-\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{8} \, \log \left ({\left (-\frac {1}{x^{4}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{4} \, \log \left ({\left | {\left (-\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-1/x^4 + 1)^(1/3) + 1)) + 1/8*log((-1/x^4 + 1)^(2/3) + (-1/x^4 + 1)^(1/3)
+ 1) - 1/4*log(abs((-1/x^4 + 1)^(1/3) - 1))

Mupad [B] (verification not implemented)

Time = 6.40 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.32 \[ \int \frac {x}{\sqrt [3]{-x^2+x^6}} \, dx=\frac {3\,x^2\,{\left (1-x^4\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^4\right )}{4\,{\left (x^6-x^2\right )}^{1/3}} \]

[In]

int(x/(x^6 - x^2)^(1/3),x)

[Out]

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