3.2086 \(\int \frac{x^2}{\sqrt{a+\frac{b}{x^4}}} \, dx\)

Optimal. Leaf size=110 \[ \frac{b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{6 a^{5/4} \sqrt{a+\frac{b}{x^4}}}+\frac{x^3 \sqrt{a+\frac{b}{x^4}}}{3 a} \]

[Out]

(Sqrt[a + b/x^4]*x^3)/(3*a) + (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ell
ipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(6*a^(5/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0451209, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {335, 325, 220} \[ \frac{b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{6 a^{5/4} \sqrt{a+\frac{b}{x^4}}}+\frac{x^3 \sqrt{a+\frac{b}{x^4}}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b/x^4],x]

[Out]

(Sqrt[a + b/x^4]*x^3)/(3*a) + (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ell
ipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(6*a^(5/4)*Sqrt[a + b/x^4])

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 325

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{a+\frac{b}{x^4}}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{a+\frac{b}{x^4}} x^3}{3 a}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{3 a}\\ &=\frac{\sqrt{a+\frac{b}{x^4}} x^3}{3 a}+\frac{b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{6 a^{5/4} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.0200952, size = 64, normalized size = 0.58 \[ \frac{-b \sqrt{\frac{a x^4}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{a x^4}{b}\right )+a x^4+b}{3 a x \sqrt{a+\frac{b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b/x^4],x]

[Out]

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

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Maple [C]  time = 0.009, size = 124, normalized size = 1.1 \begin{align*}{\frac{1}{3\,a{x}^{2}} \left ( \sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}{x}^{5}a-b\sqrt{-{ \left ( i\sqrt{a}{x}^{2}-\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}\sqrt{{ \left ( i\sqrt{a}{x}^{2}+\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}},i \right ) +\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}xb \right ){\frac{1}{\sqrt{{\frac{a{x}^{4}+b}{{x}^{4}}}}}}{\frac{1}{\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((I*a^(1/2)/b^(1/2))^(1/2)*x^5*a-b*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/
2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)+(I*a^(1/2)/b^(1/2))^(1/2)*x*b)/((a*x^4+b)/x^4)^(1/2)/x^2/a/
(I*a^(1/2)/b^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{6} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a x^{4} + b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6*sqrt((a*x^4 + b)/x^4)/(a*x^4 + b), x)

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Sympy [C]  time = 1.38849, size = 42, normalized size = 0.38 \begin{align*} - \frac{x^{3} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)