3.1430 \(\int \frac{x^3}{(2+x^6)^{3/2}} \, dx\)

Optimal. Leaf size=378 \[ -\frac{\left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right ),-7-4 \sqrt{3}\right )}{3 \sqrt [3]{2} \sqrt [4]{3} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}}+\frac{x^4}{6 \sqrt{x^6+2}}-\frac{\sqrt{x^6+2}}{6 \left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )}+\frac{\sqrt{2-\sqrt{3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} E\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right )|-7-4 \sqrt{3}\right )}{2\ 2^{5/6} 3^{3/4} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}} \]

[Out]

x^4/(6*Sqrt[2 + x^6]) - Sqrt[2 + x^6]/(6*(2^(1/3)*(1 + Sqrt[3]) + x^2)) + (Sqrt[2 - Sqrt[3]]*(2^(1/3) + x^2)*S
qrt[(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 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(2*2^(5/6)*3^(3/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqr
t[3]) + x^2)^2]*Sqrt[2 + x^6]) - ((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[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(3*2^(
1/3)*3^(1/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

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Rubi [A]  time = 0.187057, antiderivative size = 378, normalized size of antiderivative = 1., 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 = {275, 290, 303, 218, 1877} \[ \frac{x^4}{6 \sqrt{x^6+2}}-\frac{\sqrt{x^6+2}}{6 \left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )}-\frac{\left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [3]{2} \sqrt [4]{3} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}}+\frac{\sqrt{2-\sqrt{3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} E\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right )|-7-4 \sqrt{3}\right )}{2\ 2^{5/6} 3^{3/4} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(2 + x^6)^(3/2),x]

[Out]

x^4/(6*Sqrt[2 + x^6]) - Sqrt[2 + x^6]/(6*(2^(1/3)*(1 + Sqrt[3]) + x^2)) + (Sqrt[2 - Sqrt[3]]*(2^(1/3) + x^2)*S
qrt[(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 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(2*2^(5/6)*3^(3/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqr
t[3]) + x^2)^2]*Sqrt[2 + x^6]) - ((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[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(3*2^(
1/3)*3^(1/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

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 290

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

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 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 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}{\left (2+x^6\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{\left (2+x^3\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac{x^4}{6 \sqrt{2+x^6}}-\frac{1}{12} \operatorname{Subst}\left (\int \frac{x}{\sqrt{2+x^3}} \, dx,x,x^2\right )\\ &=\frac{x^4}{6 \sqrt{2+x^6}}-\frac{1}{12} \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 )}{6 \sqrt [6]{2} \sqrt{2+\sqrt{3}}}\\ &=\frac{x^4}{6 \sqrt{2+x^6}}-\frac{\sqrt{2+x^6}}{6 \left (\sqrt [3]{2} \left (1+\sqrt{3}\right )+x^2\right )}+\frac{\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\ 2^{5/6} 3^{3/4} \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{\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 )}{3 \sqrt [3]{2} \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*}

Mathematica [C]  time = 0.0053138, size = 29, normalized size = 0.08 \[ \frac{x^4 \, _2F_1\left (\frac{2}{3},\frac{3}{2};\frac{5}{3};-\frac{x^6}{2}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(2 + x^6)^(3/2),x]

[Out]

(x^4*Hypergeometric2F1[2/3, 3/2, 5/3, -x^6/2])/(8*Sqrt[2])

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Maple [C]  time = 0.022, size = 33, normalized size = 0.1 \begin{align*}{\frac{{x}^{4}}{6}{\frac{1}{\sqrt{{x}^{6}+2}}}}-{\frac{{x}^{4}\sqrt{2}}{48}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{2}{3}};\,{\frac{5}{3}};\,-{\frac{{x}^{6}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^4/(x^6+2)^(1/2)-1/48*2^(1/2)*x^4*hypergeom([1/2,2/3],[5/3],-1/2*x^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/(x^6 + 2)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{6} + 2} x^{3}}{x^{12} + 4 \, x^{6} + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^6 + 2)*x^3/(x^12 + 4*x^6 + 4), x)

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Sympy [A]  time = 0.583651, size = 36, normalized size = 0.1 \begin{align*} \frac{\sqrt{2} x^{4} \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{3}{2} \\ \frac{5}{3} \end{matrix}\middle |{\frac{x^{6} e^{i \pi }}{2}} \right )}}{24 \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*x**4*gamma(2/3)*hyper((2/3, 3/2), (5/3,), x**6*exp_polar(I*pi)/2)/(24*gamma(5/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/(x^6 + 2)^(3/2), x)