3.980 \(\int \frac{x^6}{\sqrt{-1+x^4}} \, dx\)

Optimal. Leaf size=150 \[ \frac{3 \sqrt{x^2-1} \sqrt{x^2+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right ),\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{x^4-1}}+\frac{1}{5} \sqrt{x^4-1} x^3+\frac{3 \left (x^2+1\right ) x}{5 \sqrt{x^4-1}}-\frac{3 \sqrt{2} \sqrt{x^2-1} \sqrt{x^2+1} E\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{5 \sqrt{x^4-1}} \]

[Out]

(3*x*(1 + x^2))/(5*Sqrt[-1 + x^4]) + (x^3*Sqrt[-1 + x^4])/5 - (3*Sqrt[2]*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*Elliptic
E[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[-1 + x^4]) + (3*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[Arc
Sin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[2]*Sqrt[-1 + x^4])

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Rubi [A]  time = 0.0232515, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {321, 306, 222, 1185} \[ \frac{1}{5} \sqrt{x^4-1} x^3+\frac{3 \left (x^2+1\right ) x}{5 \sqrt{x^4-1}}+\frac{3 \sqrt{x^2-1} \sqrt{x^2+1} F\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{x^4-1}}-\frac{3 \sqrt{2} \sqrt{x^2-1} \sqrt{x^2+1} E\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{5 \sqrt{x^4-1}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[-1 + x^4],x]

[Out]

(3*x*(1 + x^2))/(5*Sqrt[-1 + x^4]) + (x^3*Sqrt[-1 + x^4])/5 - (3*Sqrt[2]*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*Elliptic
E[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[-1 + x^4]) + (3*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[Arc
Sin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[2]*Sqrt[-1 + x^4])

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 306

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, Dist[1/q, 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] && LtQ[a, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 1185

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Simp[(e*x*(q + c*x
^2))/(c*Sqrt[a + c*x^4]), x] - Simp[(Sqrt[2]*e*q*Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2)/q]*EllipticE[ArcSin[x/Sqrt[
(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[-a]*c*Sqrt[a + c*x^4]), x] /; EqQ[c*d + e*q, 0] && IntegerQ[q]] /; FreeQ[{a,
c, d, e}, x] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps

\begin{align*} \int \frac{x^6}{\sqrt{-1+x^4}} \, dx &=\frac{1}{5} x^3 \sqrt{-1+x^4}+\frac{3}{5} \int \frac{x^2}{\sqrt{-1+x^4}} \, dx\\ &=\frac{1}{5} x^3 \sqrt{-1+x^4}+\frac{3}{5} \int \frac{1}{\sqrt{-1+x^4}} \, dx-\frac{3}{5} \int \frac{1-x^2}{\sqrt{-1+x^4}} \, dx\\ &=\frac{3 x \left (1+x^2\right )}{5 \sqrt{-1+x^4}}+\frac{1}{5} x^3 \sqrt{-1+x^4}-\frac{3 \sqrt{2} \sqrt{-1+x^2} \sqrt{1+x^2} E\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-1+x^2}}\right )|\frac{1}{2}\right )}{5 \sqrt{-1+x^4}}+\frac{3 \sqrt{-1+x^2} \sqrt{1+x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-1+x^2}}\right )|\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{-1+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0100652, size = 46, normalized size = 0.31 \[ \frac{x^3 \left (\sqrt{1-x^4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};x^4\right )+x^4-1\right )}{5 \sqrt{x^4-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[-1 + x^4],x]

[Out]

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

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Maple [C]  time = 0.007, size = 57, normalized size = 0.4 \begin{align*}{\frac{{x}^{3}}{5}\sqrt{{x}^{4}-1}}-{{\frac{3\,i}{5}} \left ({\it EllipticF} \left ( ix,i \right ) -{\it EllipticE} \left ( ix,i \right ) \right ) \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^6/(x^4-1)^(1/2),x)

[Out]

1/5*x^3*(x^4-1)^(1/2)-3/5*I*(x^2+1)^(1/2)*(-x^2+1)^(1/2)/(x^4-1)^(1/2)*(EllipticF(I*x,I)-EllipticE(I*x,I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{x^{4} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4-1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(x^4 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{6}}{\sqrt{x^{4} - 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4-1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^6/sqrt(x^4 - 1), x)

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Sympy [C]  time = 0.915058, size = 27, normalized size = 0.18 \begin{align*} - \frac{i x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{x^{4}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(x**4-1)**(1/2),x)

[Out]

-I*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**4)/(4*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{x^{4} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4-1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(x^4 - 1), x)