3.1229 \(\int \frac{x^2}{\sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=86 \[ \frac{\sqrt{a} x \sqrt [4]{1-\frac{a}{b x^4}} E\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{b} \sqrt [4]{a-b x^4}}-\frac{\left (a-b x^4\right )^{3/4}}{2 b x} \]

[Out]

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

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Rubi [A]  time = 0.0375278, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {311, 313, 335, 275, 228} \[ \frac{\sqrt{a} x \sqrt [4]{1-\frac{a}{b x^4}} E\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{b} \sqrt [4]{a-b x^4}}-\frac{\left (a-b x^4\right )^{3/4}}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a - b*x^4)^(1/4),x]

[Out]

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

Rule 311

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

Rule 313

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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]

Rubi steps

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

Mathematica [C]  time = 0.0083938, size = 52, normalized size = 0.6 \[ \frac{x^3 \sqrt [4]{1-\frac{b x^4}{a}} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{b x^4}{a}\right )}{3 \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a - b*x^4)^(1/4),x]

[Out]

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

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2}{\frac{1}{\sqrt [4]{-b{x}^{4}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-b*x^4 + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (-b x^{4} + a\right )}^{\frac{3}{4}} x^{2}}{b x^{4} - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.880348, size = 39, normalized size = 0.45 \begin{align*} \frac{x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*gamma(3/4)*hyper((1/4, 3/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(1/4)*gamma(7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-b*x^4 + a)^(1/4), x)