3.8.95 \(\int \sqrt [4]{a-b x^2} \, dx\) [795]

Optimal. Leaf size=78 \[ \frac {2}{3} x \sqrt [4]{a-b x^2}+\frac {2 a^{3/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {b} \left (a-b x^2\right )^{3/4}} \]

[Out]

2/3*x*(-b*x^2+a)^(1/4)+2/3*a^(3/2)*(1-b*x^2/a)^(3/4)*(cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcs
in(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/(-b*x^2+a)^(3/4)/b^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {201, 239, 238} \begin {gather*} \frac {2 a^{3/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {b} \left (a-b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a-b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(1/4),x]

[Out]

(2*x*(a - b*x^2)^(1/4))/3 + (2*a^(3/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*S
qrt[b]*(a - b*x^2)^(3/4))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 238

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

Rule 239

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.89, size = 47, normalized size = 0.60 \begin {gather*} \frac {x \sqrt [4]{a-b x^2} \, _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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(1/4),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (-b \,x^{2}+a \right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((-b*x^2 + a)^(1/4), x)

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

[Out]

integral((-b*x^2 + a)^(1/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.44, size = 27, normalized size = 0.35 \begin {gather*} \sqrt [4]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((-b*x^2 + a)^(1/4), x)

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Mupad [B]
time = 4.80, size = 38, normalized size = 0.49 \begin {gather*} \frac {x\,{\left (a-b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (1-\frac {b\,x^2}{a}\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^2)^(1/4),x)

[Out]

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

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