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

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

[Out]

-1/2*x^3*EllipticF(1/2*((2*x^2+2^(1/2)+x^4*2^(1/2))/x^2)^(1/2),(-2+2*2^(1/2))^(1/2))*(-(-x^2+1)^2/x^2)^(1/2)*(
(-x^8-1)/x^4)^(1/2)/(-x^2+1)/(x^8+1)^(1/2)/(2+2^(1/2))^(1/2)-1/2*x^3*EllipticF(1/2*((2*x^2-2^(1/2)-x^4*2^(1/2)
)/x^2)^(1/2),(-2+2*2^(1/2))^(1/2))*((x^2+1)^2/x^2)^(1/2)*((-x^8-1)/x^4)^(1/2)/(x^2+1)/(x^8+1)^(1/2)/(2+2^(1/2)
)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 223, normalized size of antiderivative = 1.00, 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*}

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Mathematica [C]  time = 0.00, size = 22, normalized size = 0.10 \[ \frac {1}{3} x^3 \, _2F_1\left (\frac {3}{8},\frac {1}{2};\frac {11}{8};-x^8\right ) \]

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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fricas [F]  time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\sqrt {x^{8} + 1}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {x^{8} + 1}}\,{d x} \]

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)

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maple [C]  time = 0.14, size = 17, normalized size = 0.08 \[ \frac {x^{3} \hypergeom \left (\left [\frac {3}{8}, \frac {1}{2}\right ], \left [\frac {11}{8}\right ], -x^{8}\right )}{3} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {x^{8} + 1}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{\sqrt {x^8+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^8 + 1)^(1/2),x)

[Out]

int(x^2/(x^8 + 1)^(1/2), x)

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sympy [C]  time = 0.68, size = 29, normalized size = 0.13 \[ \frac {x^{3} \Gamma \left (\frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, \frac {1}{2} \\ \frac {11}{8} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac {11}{8}\right )} \]

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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