3.1530 \(\int \frac {x^{10}}{\sqrt {1+x^8}} \, dx\)

Optimal. Leaf size=239 \[ \frac {1}{7} \sqrt {x^8+1} x^3+\frac {3 \sqrt {\frac {\left (x^2+1\right )^2}{x^2}} \sqrt {-\frac {x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} x^4-2 x^2+\sqrt {2}}{x^2}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{14 \sqrt {2+\sqrt {2}} \left (x^2+1\right ) \sqrt {x^8+1}}+\frac {3 \sqrt {-\frac {\left (1-x^2\right )^2}{x^2}} \sqrt {-\frac {x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac {1}{2} \sqrt {\frac {\sqrt {2} x^4+2 x^2+\sqrt {2}}{x^2}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{14 \sqrt {2+\sqrt {2}} \left (1-x^2\right ) \sqrt {x^8+1}} \]

[Out]

1/7*x^3*(x^8+1)^(1/2)+3/14*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)+3/14*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 = 0.07, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {321, 309, 1883} \[ \frac {1}{7} \sqrt {x^8+1} x^3+\frac {3 \sqrt {\frac {\left (x^2+1\right )^2}{x^2}} \sqrt {-\frac {x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} x^4-2 x^2+\sqrt {2}}{x^2}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{14 \sqrt {2+\sqrt {2}} \left (x^2+1\right ) \sqrt {x^8+1}}+\frac {3 \sqrt {-\frac {\left (1-x^2\right )^2}{x^2}} \sqrt {-\frac {x^8+1}{x^4}} x^3 F\left (\sin ^{-1}\left (\frac {1}{2} \sqrt {\frac {\sqrt {2} x^4+2 x^2+\sqrt {2}}{x^2}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{14 \sqrt {2+\sqrt {2}} \left (1-x^2\right ) \sqrt {x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/Sqrt[1 + x^8],x]

[Out]

(x^3*Sqrt[1 + x^8])/7 + (3*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])])/(14*Sqrt[2 + Sqrt[2]]*(1 + x^2)*Sqrt[1 + x^8]) + (3*x^3*Sqr
t[-((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])])/(14*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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^{10}}{\sqrt {1+x^8}} \, dx &=\frac {1}{7} x^3 \sqrt {1+x^8}-\frac {3}{7} \int \frac {x^2}{\sqrt {1+x^8}} \, dx\\ &=\frac {1}{7} x^3 \sqrt {1+x^8}+\frac {3}{14} \int \frac {1-x^2}{\sqrt {1+x^8}} \, dx-\frac {3}{14} \int \frac {1+x^2}{\sqrt {1+x^8}} \, dx\\ &=\frac {1}{7} x^3 \sqrt {1+x^8}+\frac {3 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 )}{14 \sqrt {2+\sqrt {2}} \left (1+x^2\right ) \sqrt {1+x^8}}+\frac {3 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 )}{14 \sqrt {2+\sqrt {2}} \left (1-x^2\right ) \sqrt {1+x^8}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.14 \[ \frac {1}{7} x^3 \left (\sqrt {x^8+1}-\, _2F_1\left (\frac {3}{8},\frac {1}{2};\frac {11}{8};-x^8\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/Sqrt[1 + x^8],x]

[Out]

(x^3*(Sqrt[1 + x^8] - Hypergeometric2F1[3/8, 1/2, 11/8, -x^8]))/7

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{10}}{\sqrt {x^{8} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^10/sqrt(x^8 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\sqrt {x^{8} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^10/sqrt(x^8 + 1), x)

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maple [C]  time = 0.15, size = 30, normalized size = 0.13 \[ -\frac {x^{3} \hypergeom \left (\left [\frac {3}{8}, \frac {1}{2}\right ], \left [\frac {11}{8}\right ], -x^{8}\right )}{7}+\frac {\sqrt {x^{8}+1}\, x^{3}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^8+1)^(1/2),x)

[Out]

1/7*x^3*(x^8+1)^(1/2)-1/7*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 \[ \int \frac {x^{10}}{\sqrt {x^{8} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^10/sqrt(x^8 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^{10}}{\sqrt {x^8+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^8 + 1)^(1/2),x)

[Out]

int(x^10/(x^8 + 1)^(1/2), x)

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sympy [C]  time = 1.05, size = 29, normalized size = 0.12 \[ \frac {x^{11} \Gamma \left (\frac {11}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{8} \\ \frac {19}{8} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac {19}{8}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(x**8+1)**(1/2),x)

[Out]

x**11*gamma(11/8)*hyper((1/2, 11/8), (19/8,), x**8*exp_polar(I*pi))/(8*gamma(19/8))

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