3.9.48 \(\int \frac {x^2}{\sqrt {a-b x^4}} \, dx\) [848]

Optimal. Leaf size=108 \[ \frac {a^{3/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{b^{3/4} \sqrt {a-b x^4}}-\frac {a^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{b^{3/4} \sqrt {a-b x^4}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {313, 230, 227, 1214, 1213, 435} \begin {gather*} \frac {a^{3/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{b^{3/4} \sqrt {a-b x^4}}-\frac {a^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{b^{3/4} \sqrt {a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a - b*x^4],x]

[Out]

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

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 230

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

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 1213

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + e*(x^2/d)]/Sqrt
[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1214

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + c*(x^4/a)]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 52, normalized size = 0.48 \begin {gather*} \frac {x^3 \sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {b x^4}{a}\right )}{3 \sqrt {a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a - b*x^4],x]

[Out]

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

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Maple [A]
time = 0.14, size = 88, normalized size = 0.81

method	result	size
default	\(-\frac {\sqrt {a}\, \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, \sqrt {b}}\)	\(88\)
elliptic	\(-\frac {\sqrt {a}\, \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, \sqrt {b}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-a^(1/2)/(1/a^(1/2)*b^(1/2))^(1/2)*(1-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2*b^(1/2)/a^(1/2))^(1/2)/(-b*x^4+a)^(1/2
)/b^(1/2)*(EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(1/a^(1/2)*b^(1/2))^(1/2),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/(-b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(-b*x^4 + a), x)

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Fricas [A]
time = 0.08, size = 76, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {-b} x \left (\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - \sqrt {-b} x \left (\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-b x^{4} + a}}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.38, size = 39, normalized size = 0.36 \begin {gather*} \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*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/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(-b*x^4 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{\sqrt {a-b\,x^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a - b*x^4)^(1/2),x)

[Out]

int(x^2/(a - b*x^4)^(1/2), x)

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