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

Optimal. Leaf size=88 \[ -\frac {1}{6} \log \left (\sqrt [3]{x^6+1}-x^2\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{x^6+1}+x^2}\right )}{2 \sqrt {3}}+\frac {1}{12} \log \left (\left (x^6+1\right )^{2/3}+x^4+\sqrt [3]{x^6+1} x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {275, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} -\frac {1}{6} \log \left (1-\frac {x^2}{\sqrt [3]{x^6+1}}\right )-\frac {\tan ^{-1}\left (\frac {\frac {2 x^2}{\sqrt [3]{x^6+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{12} \log \left (\frac {x^4}{\left (x^6+1\right )^{2/3}}+\frac {x^2}{\sqrt [3]{x^6+1}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 292

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 618

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=-\frac {1}{6} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=-\frac {1}{6} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{12} \log \left (1+\frac {x^4}{\left (1+x^6\right )^{2/3}}+\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^2}{\sqrt [3]{1+x^6}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {1+\frac {2 x^2}{\sqrt [3]{1+x^6}}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{12} \log \left (1+\frac {x^4}{\left (1+x^6\right )^{2/3}}+\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.40 \begin {gather*} \frac {x^4 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x^6}{x^6+1}\right )}{4 \left (x^6+1\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x^6)^(2/3),x]

[Out]

(x^4*Hypergeometric2F1[2/3, 1, 5/3, x^6/(1 + x^6)])/(4*(1 + x^6)^(2/3))

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IntegrateAlgebraic [A]  time = 0.74, size = 88, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 82, normalized size = 0.93 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/(x^6 + 1)^(2/3), x)

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maple [C]  time = 3.12, size = 17, normalized size = 0.19

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.45, size = 69, normalized size = 0.78 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\left (x^6+1\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^6 + 1)^(2/3),x)

[Out]

int(x^3/(x^6 + 1)^(2/3), x)

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sympy [C]  time = 0.78, size = 29, normalized size = 0.33 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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