3.1263 \(\int \frac{x^2}{(a-b x^4)^{5/4}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0387146, antiderivative size = 81, 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 = {287, 313, 335, 275, 228} \[ \frac{1}{b x \sqrt [4]{a-b x^4}}-\frac{x \sqrt [4]{1-\frac{a}{b x^4}} E\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt{b} \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a - b*x^4)^(5/4),x]

[Out]

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

Rule 287

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> -Simp[(b*x*(a + b*x^4)^(1/4))^(-1), x] - Dist[1/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}{\left (a-b x^4\right )^{5/4}} \, dx &=\frac{1}{b x \sqrt [4]{a-b x^4}}+\frac{\int \frac{1}{x^2 \sqrt [4]{a-b x^4}} \, dx}{b}\\ &=\frac{1}{b x \sqrt [4]{a-b x^4}}+\frac{\left (\sqrt [4]{1-\frac{a}{b x^4}} x\right ) \int \frac{1}{\sqrt [4]{1-\frac{a}{b x^4}} x^3} \, dx}{b \sqrt [4]{a-b x^4}}\\ &=\frac{1}{b x \sqrt [4]{a-b x^4}}-\frac{\left (\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 )}{b \sqrt [4]{a-b x^4}}\\ &=\frac{1}{b x \sqrt [4]{a-b x^4}}-\frac{\left (\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 )}{2 b \sqrt [4]{a-b x^4}}\\ &=\frac{1}{b x \sqrt [4]{a-b x^4}}-\frac{\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 )}{\sqrt{a} \sqrt{b} \sqrt [4]{a-b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0118417, size = 55, normalized size = 0.68 \[ \frac{x^3 \sqrt [4]{1-\frac{b x^4}{a}} \, _2F_1\left (\frac{3}{4},\frac{5}{4};\frac{7}{4};\frac{b x^4}{a}\right )}{3 a \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a - b*x^4)^(5/4),x]

[Out]

(x^3*(1 - (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, (b*x^4)/a])/(3*a*(a - b*x^4)^(1/4))

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( -b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-b*x^4+a)^(5/4),x)

[Out]

int(x^2/(-b*x^4+a)^(5/4),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{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^2/(-b*x^4 + a)^(5/4), 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^{2} x^{8} - 2 \, a b x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(3/4)*x^2/(b^2*x^8 - 2*a*b*x^4 + a^2), x)

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Sympy [C]  time = 1.04633, size = 39, normalized size = 0.48 \begin{align*} \frac{x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{4 a^{\frac{5}{4}} \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)**(5/4),x)

[Out]

x**3*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(5/4)*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{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^2/(-b*x^4 + a)^(5/4), x)