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

Optimal. Leaf size=101 \[ \frac {4 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 b^{3/2} \left (a-b x^2\right )^{3/4}}-\frac {2 a x \sqrt [4]{a-b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a-b x^2} \]

[Out]

-2/21*a*x*(-b*x^2+a)^(1/4)/b+2/7*x^3*(-b*x^2+a)^(1/4)+4/21*a^(5/2)*(1-b*x^2/a)^(3/4)*(cos(1/2*arcsin(x*b^(1/2)
/a^(1/2)))^2)^(1/2)/cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/b
^(3/2)/(-b*x^2+a)^(3/4)

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Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {279, 321, 233, 232} \[ \frac {4 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 b^{3/2} \left (a-b x^2\right )^{3/4}}+\frac {2}{7} x^3 \sqrt [4]{a-b x^2}-\frac {2 a x \sqrt [4]{a-b x^2}}{21 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a - b*x^2)^(1/4),x]

[Out]

(-2*a*x*(a - b*x^2)^(1/4))/(21*b) + (2*x^3*(a - b*x^2)^(1/4))/7 + (4*a^(5/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[A
rcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*b^(3/2)*(a - b*x^2)^(3/4))

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 279

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

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{a-b x^2} \, dx &=\frac {2}{7} x^3 \sqrt [4]{a-b x^2}+\frac {1}{7} a \int \frac {x^2}{\left (a-b x^2\right )^{3/4}} \, dx\\ &=-\frac {2 a x \sqrt [4]{a-b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a-b x^2}+\frac {\left (2 a^2\right ) \int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx}{21 b}\\ &=-\frac {2 a x \sqrt [4]{a-b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a-b x^2}+\frac {\left (2 a^2 \left (1-\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx}{21 b \left (a-b x^2\right )^{3/4}}\\ &=-\frac {2 a x \sqrt [4]{a-b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a-b x^2}+\frac {4 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 b^{3/2} \left (a-b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 64, normalized size = 0.63 \[ \frac {2 x \sqrt [4]{a-b x^2} \left (\frac {a \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {b x^2}{a}\right )}{\sqrt [4]{1-\frac {b x^2}{a}}}-a+b x^2\right )}{7 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a - b*x^2)^(1/4),x]

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b x^{2} + a\right )}^{\frac {1}{4}} x^{2}, x\right ) \]

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)^(1/4)*x^2, x)

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

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((-b*x^2 + a)^(1/4)*x^2, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (-b \,x^{2}+a \right )^{\frac {1}{4}} x^{2}\, dx \]

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 \[ \int {\left (-b x^{2} + a\right )}^{\frac {1}{4}} x^{2}\,{d x} \]

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((-b*x^2 + a)^(1/4)*x^2, x)

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

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 = 0.90, size = 31, normalized size = 0.31 \[ \frac {\sqrt [4]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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