3.889 \(\int \frac {x^{10}}{\sqrt {1-x^4}} \, dx\)

Optimal. Leaf size=53 \[ -\frac {1}{9} \sqrt {1-x^4} x^7-\frac {7}{45} \sqrt {1-x^4} x^3-\frac {7}{15} F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {7}{15} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

7/15*EllipticE(x,I)-7/15*EllipticF(x,I)-7/45*x^3*(-x^4+1)^(1/2)-1/9*x^7*(-x^4+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 307, 221, 1181, 424} \[ -\frac {1}{9} \sqrt {1-x^4} x^7-\frac {7}{45} \sqrt {1-x^4} x^3-\frac {7}{15} F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {7}{15} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^10/Sqrt[1 - x^4],x]

[Out]

(-7*x^3*Sqrt[1 - x^4])/45 - (x^7*Sqrt[1 - x^4])/9 + (7*EllipticE[ArcSin[x], -1])/15 - (7*EllipticF[ArcSin[x],
-1])/15

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 307

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

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 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1181

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx &=-\frac {1}{9} x^7 \sqrt {1-x^4}+\frac {7}{9} \int \frac {x^6}{\sqrt {1-x^4}} \, dx\\ &=-\frac {7}{45} x^3 \sqrt {1-x^4}-\frac {1}{9} x^7 \sqrt {1-x^4}+\frac {7}{15} \int \frac {x^2}{\sqrt {1-x^4}} \, dx\\ &=-\frac {7}{45} x^3 \sqrt {1-x^4}-\frac {1}{9} x^7 \sqrt {1-x^4}-\frac {7}{15} \int \frac {1}{\sqrt {1-x^4}} \, dx+\frac {7}{15} \int \frac {1+x^2}{\sqrt {1-x^4}} \, dx\\ &=-\frac {7}{45} x^3 \sqrt {1-x^4}-\frac {1}{9} x^7 \sqrt {1-x^4}-\frac {7}{15} F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {7}{15} \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx\\ &=-\frac {7}{45} x^3 \sqrt {1-x^4}-\frac {1}{9} x^7 \sqrt {1-x^4}+\frac {7}{15} E\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {7}{15} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 43, normalized size = 0.81 \[ \frac {1}{45} x^3 \left (7 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};x^4\right )-\sqrt {1-x^4} \left (5 x^4+7\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/Sqrt[1 - x^4],x]

[Out]

(x^3*(-(Sqrt[1 - x^4]*(7 + 5*x^4)) + 7*Hypergeometric2F1[1/2, 3/4, 7/4, x^4]))/45

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 1} x^{10}}{x^{4} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^4 + 1)*x^10/(x^4 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\sqrt {-x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^10/sqrt(-x^4 + 1), x)

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maple [A]  time = 0.01, size = 68, normalized size = 1.28 \[ -\frac {\sqrt {-x^{4}+1}\, x^{7}}{9}-\frac {7 \sqrt {-x^{4}+1}\, x^{3}}{45}-\frac {7 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (x , i\right )+\EllipticF \left (x , i\right )\right )}{15 \sqrt {-x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(-x^4+1)^(1/2),x)

[Out]

-1/9*x^7*(-x^4+1)^(1/2)-7/45*x^3*(-x^4+1)^(1/2)-7/15*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/2)*(EllipticF(x,
I)-EllipticE(x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\sqrt {-x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^10/sqrt(-x^4 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(1 - x^4)^(1/2),x)

[Out]

int(x^10/(1 - x^4)^(1/2), x)

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sympy [A]  time = 1.21, size = 31, normalized size = 0.58 \[ \frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(-x**4+1)**(1/2),x)

[Out]

x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), x**4*exp_polar(2*I*pi))/(4*gamma(15/4))

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