3.911 \(\int \frac{x^{10}}{(1-x^4)^{3/2}} \, dx\)

Optimal. Leaf size=53 \[ \frac{21}{10} \text{EllipticF}\left (\sin ^{-1}(x),-1\right )+\frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} \sqrt{1-x^4} x^3-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

x^7/(2*Sqrt[1 - x^4]) + (7*x^3*Sqrt[1 - x^4])/10 - (21*EllipticE[ArcSin[x], -1])/10 + (21*EllipticF[ArcSin[x],
-1])/10

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Rubi [A]  time = 0.0210173, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {288, 321, 307, 221, 1181, 424} \[ \frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} \sqrt{1-x^4} x^3+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^10/(1 - x^4)^(3/2),x]

[Out]

x^7/(2*Sqrt[1 - x^4]) + (7*x^3*Sqrt[1 - x^4])/10 - (21*EllipticE[ArcSin[x], -1])/10 + (21*EllipticF[ArcSin[x],
-1])/10

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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 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 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 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]

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]

Rubi steps

\begin{align*} \int \frac{x^{10}}{\left (1-x^4\right )^{3/2}} \, dx &=\frac{x^7}{2 \sqrt{1-x^4}}-\frac{7}{2} \int \frac{x^6}{\sqrt{1-x^4}} \, dx\\ &=\frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} x^3 \sqrt{1-x^4}-\frac{21}{10} \int \frac{x^2}{\sqrt{1-x^4}} \, dx\\ &=\frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} x^3 \sqrt{1-x^4}+\frac{21}{10} \int \frac{1}{\sqrt{1-x^4}} \, dx-\frac{21}{10} \int \frac{1+x^2}{\sqrt{1-x^4}} \, dx\\ &=\frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} x^3 \sqrt{1-x^4}+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{21}{10} \int \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} \, dx\\ &=\frac{x^7}{2 \sqrt{1-x^4}}+\frac{7}{10} x^3 \sqrt{1-x^4}-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end{align*}

Mathematica [C]  time = 0.0092496, size = 49, normalized size = 0.92 \[ -\frac{x^3 \left (-7 \sqrt{1-x^4} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};x^4\right )+x^4+7\right )}{5 \sqrt{1-x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(1 - x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.01, size = 68, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}}{2}{\frac{1}{\sqrt{-{x}^{4}+1}}}}+{\frac{{x}^{3}}{5}\sqrt{-{x}^{4}+1}}+{\frac{21\,{\it EllipticF} \left ( x,i \right ) -21\,{\it EllipticE} \left ( x,i \right ) }{10}\sqrt{-{x}^{2}+1}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{-{x}^{4}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(-x^4+1)^(3/2),x)

[Out]

1/2*x^3/(-x^4+1)^(1/2)+1/5*x^3*(-x^4+1)^(1/2)+21/10*(-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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{{\left (-x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^10/(-x^4 + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-x^{4} + 1} x^{10}}{x^{8} - 2 \, x^{4} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^4 + 1)*x^10/(x^8 - 2*x^4 + 1), x)

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Sympy [A]  time = 1.43059, size = 31, normalized size = 0.58 \begin{align*} \frac{x^{11} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac{15}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(-x**4+1)**(3/2),x)

[Out]

x**11*gamma(11/4)*hyper((3/2, 11/4), (15/4,), x**4*exp_polar(2*I*pi))/(4*gamma(15/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{{\left (-x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^10/(-x^4 + 1)^(3/2), x)