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

Optimal. Leaf size=81 \[ \frac {4 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b} \]

[Out]

-2/5*x*(-b*x^2+a)^(3/4)/b+4/5*a^(3/2)*(1-b*x^2/a)^(1/4)*(cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*a
rcsin(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/2)/(-b*x^2+a)^(1/4)

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {321, 229, 228} \[ \frac {4 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a - b*x^2)^(1/4),x]

[Out]

(-2*x*(a - b*x^2)^(3/4))/(5*b) + (4*a^(3/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])
/(5*b^(3/2)*(a - b*x^2)^(1/4))

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 229

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

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt [4]{a-b x^2}} \, dx &=-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}+\frac {(2 a) \int \frac {1}{\sqrt [4]{a-b x^2}} \, dx}{5 b}\\ &=-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}+\frac {\left (2 a \sqrt [4]{1-\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx}{5 b \sqrt [4]{a-b x^2}}\\ &=-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}+\frac {4 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 64, normalized size = 0.79 \[ \frac {2 x \left (a \sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {b x^2}{a}\right )-a+b x^2\right )}{5 b \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a - b*x^2)^(1/4),x]

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} x^{2}}{b x^{2} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(-b*x^2 + a)^(3/4)*x^2/(b*x^2 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-b*x^2 + a)^(1/4), x)

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-b*x^2 + a)^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\left (a-b\,x^2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a - b*x^2)^(1/4),x)

[Out]

int(x^2/(a - b*x^2)^(1/4), x)

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sympy [C]  time = 0.82, size = 29, normalized size = 0.36 \[ \frac {x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{3 \sqrt [4]{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*hyper((1/4, 3/2), (5/2,), b*x**2*exp_polar(2*I*pi)/a)/(3*a**(1/4))

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