3.2062 \(\int \sqrt{a+\frac{b}{x^4}} x^2 \, dx\)

Optimal. Leaf size=107 \[ \frac{1}{3} x^3 \sqrt{a+\frac{b}{x^4}}-\frac{b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt{a+\frac{b}{x^4}}} \]

[Out]

(Sqrt[a + b/x^4]*x^3)/3 - (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ellipti
cF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*a^(1/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0479689, antiderivative size = 107, 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, 277, 220} \[ \frac{1}{3} x^3 \sqrt{a+\frac{b}{x^4}}-\frac{b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt{a+\frac{b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^4]*x^2,x]

[Out]

(Sqrt[a + b/x^4]*x^3)/3 - (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ellipti
cF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*a^(1/4)*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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && 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 \sqrt{a+\frac{b}{x^4}} x^2 \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^4}}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{3} \sqrt{a+\frac{b}{x^4}} x^3-\frac{1}{3} (2 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{3} \sqrt{a+\frac{b}{x^4}} x^3-\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 )}{3 \sqrt [4]{a} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.164785, size = 93, normalized size = 0.87 \[ \frac{1}{3} x^2 \sqrt{a+\frac{b}{x^4}} \left (x-\frac{2 i b \sqrt{\frac{a x^4}{b}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}\right ),-1\right )}{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}} \left (a x^4+b\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^4]*x^2,x]

[Out]

(Sqrt[a + b/x^4]*x^2*(x - ((2*I)*b*Sqrt[1 + (a*x^4)/b]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]*x], -1])/
(Sqrt[(I*Sqrt[a])/Sqrt[b]]*(b + a*x^4))))/3

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Maple [C]  time = 0.038, size = 130, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}}{3\,a{x}^{4}+3\,b}\sqrt{{\frac{a{x}^{4}+b}{{x}^{4}}}} \left ( \sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}{x}^{5}a+2\,b\sqrt{-{\frac{i\sqrt{a}{x}^{2}-\sqrt{b}}{\sqrt{b}}}}\sqrt{{\frac{i\sqrt{a}{x}^{2}+\sqrt{b}}{\sqrt{b}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}},i \right ) +\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}xb \right ){\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*((a*x^4+b)/x^4)^(1/2)*x^2*((I*a^(1/2)/b^(1/2))^(1/2)*x^5*a+2*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)/(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 \sqrt{a + \frac{b}{x^{4}}} x^{2}\,{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(sqrt(a + b/x^4)*x^2, x)

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

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Sympy [C]  time = 1.35776, size = 44, normalized size = 0.41 \begin{align*} - \frac{\sqrt{a} 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 \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]

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

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