3.1534 \(\int \frac{x^2}{\sqrt{1+x^8}} \, dx\)

Optimal. Leaf size=223 \[ -\frac{\sqrt{\frac{\left (x^2+1\right )^2}{x^2}} \sqrt{-\frac{x^8+1}{x^4}} x^3 \text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{-\frac{\sqrt{2} x^4-2 x^2+\sqrt{2}}{x^2}}\right ),-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (x^2+1\right ) \sqrt{x^8+1}}-\frac{\sqrt{-\frac{\left (1-x^2\right )^2}{x^2}} \sqrt{-\frac{x^8+1}{x^4}} x^3 \text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{\frac{\sqrt{2} x^4+2 x^2+\sqrt{2}}{x^2}}\right ),-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (1-x^2\right ) \sqrt{x^8+1}} \]

[Out]

-(x^3*Sqrt[(1 + x^2)^2/x^2]*Sqrt[-((1 + x^8)/x^4)]*EllipticF[ArcSin[Sqrt[-((Sqrt[2] - 2*x^2 + Sqrt[2]*x^4)/x^2
)]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*(1 + x^2)*Sqrt[1 + x^8]) - (x^3*Sqrt[-((1 - x^2)^2/x^2)]*Sqrt[-
((1 + x^8)/x^4)]*EllipticF[ArcSin[Sqrt[(Sqrt[2] + 2*x^2 + Sqrt[2]*x^4)/x^2]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 +
Sqrt[2]]*(1 - x^2)*Sqrt[1 + x^8])

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Rubi [A]  time = 0.0476937, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {309, 1883} \[ -\frac{\sqrt{\frac{\left (x^2+1\right )^2}{x^2}} \sqrt{-\frac{x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{-\frac{\sqrt{2} x^4-2 x^2+\sqrt{2}}{x^2}}\right )|-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (x^2+1\right ) \sqrt{x^8+1}}-\frac{\sqrt{-\frac{\left (1-x^2\right )^2}{x^2}} \sqrt{-\frac{x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{\frac{\sqrt{2} x^4+2 x^2+\sqrt{2}}{x^2}}\right )|-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (1-x^2\right ) \sqrt{x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[1 + x^8],x]

[Out]

-(x^3*Sqrt[(1 + x^2)^2/x^2]*Sqrt[-((1 + x^8)/x^4)]*EllipticF[ArcSin[Sqrt[-((Sqrt[2] - 2*x^2 + Sqrt[2]*x^4)/x^2
)]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*(1 + x^2)*Sqrt[1 + x^8]) - (x^3*Sqrt[-((1 - x^2)^2/x^2)]*Sqrt[-
((1 + x^8)/x^4)]*EllipticF[ArcSin[Sqrt[(Sqrt[2] + 2*x^2 + Sqrt[2]*x^4)/x^2]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 +
Sqrt[2]]*(1 - x^2)*Sqrt[1 + x^8])

Rule 309

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Dist[1/(2*Rt[b/a, 4]), Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a + b*
x^8], x], x] - Dist[1/(2*Rt[b/a, 4]), Int[(1 - Rt[b/a, 4]*x^2)/Sqrt[a + b*x^8], x], x] /; FreeQ[{a, b}, x]

Rule 1883

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> -Simp[(c*d*x^3*Sqrt[-((c - d*x^2)^2/(c*d*x^2
))]*Sqrt[-((d^2*(a + b*x^8))/(b*c^2*x^4))]*EllipticF[ArcSin[(1*Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4
)/(c*d*x^2)])/2], -2*(1 - Sqrt[2])])/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]), x] /; FreeQ[{a, b, c, d}
, x] && EqQ[b*c^4 - a*d^4, 0]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{1+x^8}} \, dx &=-\left (\frac{1}{2} \int \frac{1-x^2}{\sqrt{1+x^8}} \, dx\right )+\frac{1}{2} \int \frac{1+x^2}{\sqrt{1+x^8}} \, dx\\ &=-\frac{x^3 \sqrt{\frac{\left (1+x^2\right )^2}{x^2}} \sqrt{-\frac{1+x^8}{x^4}} F\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{-\frac{\sqrt{2}-2 x^2+\sqrt{2} x^4}{x^2}}\right )|-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (1+x^2\right ) \sqrt{1+x^8}}-\frac{x^3 \sqrt{-\frac{\left (1-x^2\right )^2}{x^2}} \sqrt{-\frac{1+x^8}{x^4}} F\left (\sin ^{-1}\left (\frac{1}{2} \sqrt{\frac{\sqrt{2}+2 x^2+\sqrt{2} x^4}{x^2}}\right )|-2 \left (1-\sqrt{2}\right )\right )}{2 \sqrt{2+\sqrt{2}} \left (1-x^2\right ) \sqrt{1+x^8}}\\ \end{align*}

Mathematica [C]  time = 0.0026363, size = 22, normalized size = 0.1 \[ \frac{1}{3} x^3 \, _2F_1\left (\frac{3}{8},\frac{1}{2};\frac{11}{8};-x^8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[1 + x^8],x]

[Out]

(x^3*Hypergeometric2F1[3/8, 1/2, 11/8, -x^8])/3

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Maple [C]  time = 0.018, size = 17, normalized size = 0.1 \begin{align*}{\frac{{x}^{3}}{3}{\mbox{$_2$F$_1$}({\frac{3}{8}},{\frac{1}{2}};\,{\frac{11}{8}};\,-{x}^{8})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^8+1)^(1/2),x)

[Out]

1/3*x^3*hypergeom([3/8,1/2],[11/8],-x^8)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{x^{8} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(x^8 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{\sqrt{x^{8} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^2/sqrt(x^8 + 1), x)

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Sympy [C]  time = 0.52335, size = 29, normalized size = 0.13 \begin{align*} \frac{x^{3} \Gamma \left (\frac{3}{8}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{8}, \frac{1}{2} \\ \frac{11}{8} \end{matrix}\middle |{x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac{11}{8}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**8+1)**(1/2),x)

[Out]

x**3*gamma(3/8)*hyper((3/8, 1/2), (11/8,), x**8*exp_polar(I*pi))/(8*gamma(11/8))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{x^{8} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(x^8 + 1), x)