3.2064 \(\int \frac{\sqrt{a+\frac{b}{x^4}}}{x^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.042532, 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, 195, 220} \[ -\frac{a^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}-\frac{\sqrt{a+\frac{b}{x^4}}}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^4]/x^2,x]

[Out]

-Sqrt[a + b/x^4]/(3*x) - (a^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Elliptic
F[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*b^(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 195

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

Mathematica [C]  time = 0.0096724, size = 51, normalized size = 0.48 \[ -\frac{\sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{4};-\frac{a x^4}{b}\right )}{3 x \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^4]/x^2,x]

[Out]

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

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Maple [C]  time = 0.011, size = 132, normalized size = 1.2 \begin{align*} -{\frac{1}{3\,x \left ( a{x}^{4}+b \right ) }\sqrt{{\frac{a{x}^{4}+b}{{x}^{4}}}} \left ( -2\,a\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 ){x}^{3}+\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}{x}^{4}a+\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}b \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((a+b/x^4)^(1/2)/x^2,x)

[Out]

-1/3*((a*x^4+b)/x^4)^(1/2)*(-2*a*(-(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)*x^3+(I*a^(1/2)/b^(1/2))^(1/2)*x^4*a+(I*a^(1/2)/b^(1/2))^(1/2)*b)/
x/(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 \frac{\sqrt{a + \frac{b}{x^{4}}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(1/2)/x^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 (\frac{\sqrt{\frac{a x^{4} + b}{x^{4}}}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.36352, size = 39, normalized size = 0.36 \begin{align*} - \frac{\sqrt{a} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 x \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*exp_polar(I*pi)/(a*x**4))/(4*x*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x^4)/x^2, x)