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

Optimal. Leaf size=96 \[ -\frac {1}{4} \log \left (\sqrt [3]{x^6+x^2}-x^2\right )+\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{x^6+x^2}}\right )+\frac {1}{8} \log \left (x^4+\sqrt [3]{x^6+x^2} x^2+\left (x^6+x^2\right )^{2/3}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2013, 2011, 329, 275, 239} \begin {gather*} \frac {\sqrt {3} \sqrt [3]{x^2} \sqrt [3]{x^4+1} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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 239

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 275

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 329

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 2011

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 2013

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*} \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx &=\frac {1}{2} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.41 \begin {gather*} \frac {3 \left (x^6+x^2\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};-x^4\right )}{4 \left (x^4+1\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(x^2 + x^6)^(1/3),x]

[Out]

(3*(x^2 + x^6)^(2/3)*Hypergeometric2F1[1/3, 1/3, 4/3, -x^4])/(4*(1 + x^4)^(2/3))

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IntegrateAlgebraic [A]  time = 0.32, size = 96, normalized size = 1.00 \begin {gather*} \frac {1}{4} \sqrt {3} \tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(x^2 + x^6)^(1/3),x]

[Out]

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

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fricas [A]  time = 1.33, size = 92, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} x^{2} - \sqrt {3} {\left (539 \, x^{4} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {2}{3}}}{2205 \, x^{4} + 2197}\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*arctan(-(196*sqrt(3)*(x^6 + x^2)^(1/3)*x^2 - sqrt(3)*(539*x^4 + 507) - 1274*sqrt(3)*(x^6 + x^2)^(2
/3))/(2205*x^4 + 2197)) - 1/8*log(3*(x^6 + x^2)^(1/3)*x^2 - 3*(x^6 + x^2)^(2/3) + 1)

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giac [A]  time = 0.16, size = 55, normalized size = 0.57 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 4.64, size = 17, normalized size = 0.18

method	result	size
meijerg	\(\frac {3 x^{\frac {4}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{4}\right )}{4}\)	\(17\)
trager	\(-\frac {\ln \left (11 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{4}-215 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{4}+789 \left (x^{6}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+800 x^{4}-585 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-585 x^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-204 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-11 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}+138 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+320\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-160 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{4}-629 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{4}+204 \left (x^{6}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+44 x^{4}+585 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}+585 x^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-789 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}+160 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-491 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+33\right )}{4}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(x^6 + x^2)^(1/3), x)

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mupad [B]  time = 0.94, size = 31, normalized size = 0.32 \begin {gather*} \frac {3\,x^2\,{\left (x^4+1\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2 + x^6)^(1/3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt [3]{x^{2} \left (x^{4} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x**2*(x**4 + 1))**(1/3), x)

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