3.1403 \(\int \frac {x^3}{\sqrt {2+x^6}} \, dx\)

Optimal. Leaf size=354 \[ \frac {\sqrt {x^6+2}}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}+\frac {2^{2/3} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} E\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{2^{5/6} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}} \]

[Out]

(x^6+2)^(1/2)/(x^2+2^(1/3)*(1+3^(1/2)))+1/3*2^(2/3)*(2^(1/3)+x^2)*EllipticF((x^2+2^(1/3)*(1-3^(1/2)))/(x^2+2^(
1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*((2^(2/3)-2^(1/3)*x^2+x^4)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/(x^6+2)
^(1/2)/((2^(1/3)+x^2)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)-1/2*2^(1/6)*3^(1/4)*(2^(1/3)+x^2)*EllipticE((x^2+2^(1
/3)*(1-3^(1/2)))/(x^2+2^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)-1/2*2^(1/2))*((2^(2/3)-2^(1/3)*x^2+x^4)
/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)/(x^6+2)^(1/2)/((2^(1/3)+x^2)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 354, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {275, 303, 218, 1877} \[ \frac {\sqrt {x^6+2}}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}+\frac {2^{2/3} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} E\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{2^{5/6} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[2 + x^6],x]

[Out]

Sqrt[2 + x^6]/(2^(1/3)*(1 + Sqrt[3]) + x^2) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*(2^(1/3) + x^2)*Sqrt[(2^(2/3) - 2^(1/
3)*x^2 + x^4)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*EllipticE[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sq
rt[3]) + x^2)], -7 - 4*Sqrt[3]])/(2^(5/6)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])
+ (2^(2/3)*(2^(1/3) + x^2)*Sqrt[(2^(2/3) - 2^(1/3)*x^2 + x^4)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*EllipticF[ArcS
in[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(3^(1/4)*Sqrt[(2^(1/3) + x^2
)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

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 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {2+x^6}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x}{\sqrt {2+x^3}} \, dx,x,x^2\right )+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {2+x^3}} \, dx,x,x^2\right )}{\sqrt [6]{2} \sqrt {2+\sqrt {3}}}\\ &=\frac {\sqrt {2+x^6}}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} E\left (\sin ^{-1}\left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{2^{5/6} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}+\frac {2^{2/3} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.08 \[ \frac {x^4 \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};-\frac {x^6}{2}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[2 + x^6],x]

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3}}{\sqrt {x^{6} + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3/sqrt(x^6 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x^{6} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/sqrt(x^6 + 2), x)

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maple [C]  time = 0.15, size = 20, normalized size = 0.06 \[ \frac {\sqrt {2}\, x^{4} \hypergeom \left (\left [\frac {1}{2}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -\frac {x^{6}}{2}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/2)*x^4*hypergeom([1/2,2/3],[5/3],-1/2*x^6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x^{6} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/sqrt(x^6 + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{\sqrt {x^6+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^6 + 2)^(1/2),x)

[Out]

int(x^3/(x^6 + 2)^(1/2), x)

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sympy [A]  time = 0.72, size = 36, normalized size = 0.10 \[ \frac {\sqrt {2} x^{4} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{12 \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*x**4*gamma(2/3)*hyper((1/2, 2/3), (5/3,), x**6*exp_polar(I*pi)/2)/(12*gamma(5/3))

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