3.10.13 \(\int \frac {x^2}{(1-x^4)^{3/2}} \, dx\) [913]

Optimal. Leaf size=35 \[ \frac {x^3}{2 \sqrt {1-x^4}}-\frac {1}{2} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {296, 313, 227, 1195, 435} \begin {gather*} \frac {1}{2} F(\text {ArcSin}(x)|-1)-\frac {1}{2} E(\text {ArcSin}(x)|-1)+\frac {x^3}{2 \sqrt {1-x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 - x^4)^(3/2),x]

[Out]

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

Rule 227

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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 313

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 435

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

Rule 1195

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 3.78, size = 20, normalized size = 0.57 \begin {gather*} \frac {1}{3} x^3 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 - x^4)^(3/2),x]

[Out]

(x^3*Hypergeometric2F1[3/4, 3/2, 7/4, x^4])/3

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Maple [A]
time = 0.16, size = 54, normalized size = 1.54

method	result	size
meijerg	\(\frac {x^{3} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{3}\)	\(15\)
default	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\EllipticF \left (x , i\right )-\EllipticE \left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)
risch	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\EllipticF \left (x , i\right )-\EllipticE \left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)
elliptic	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\EllipticF \left (x , i\right )-\EllipticE \left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*x^3/(-x^4+1)^(1/2)+1/2*(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-x^4 + 1)^(3/2), x)

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Fricas [A]
time = 0.08, size = 44, normalized size = 1.26 \begin {gather*} -\frac {\sqrt {-x^{4} + 1} x^{3} + {\left (x^{4} - 1\right )} E(\arcsin \left (x\right )\,|\,-1) - {\left (x^{4} - 1\right )} F(\arcsin \left (x\right )\,|\,-1)}{2 \, {\left (x^{4} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(-x^4 + 1)*x^3 + (x^4 - 1)*elliptic_e(arcsin(x), -1) - (x^4 - 1)*elliptic_f(arcsin(x), -1))/(x^4 - 1
)

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Sympy [A]
time = 0.33, size = 31, normalized size = 0.89 \begin {gather*} \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(2*I*pi))/(4*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-x^4 + 1)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^2}{{\left (1-x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(1 - x^4)^(3/2),x)

[Out]

int(x^2/(1 - x^4)^(3/2), x)

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