3.2063 \(\int \sqrt{a+\frac{b}{x^4}} \, dx\)

Optimal. Leaf size=224 \[ -\frac{\sqrt [4]{a} \sqrt [4]{b} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{\sqrt{a+\frac{b}{x^4}}}+x \sqrt{a+\frac{b}{x^4}}-\frac{2 \sqrt{b} \sqrt{a+\frac{b}{x^4}}}{x \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )}+\frac{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{\sqrt{a+\frac{b}{x^4}}} \]

[Out]

(-2*Sqrt[b]*Sqrt[a + b/x^4])/((Sqrt[a] + Sqrt[b]/x^2)*x) + Sqrt[a + b/x^4]*x + (2*a^(1/4)*b^(1/4)*Sqrt[(a + b/
x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/Sqrt[a
+ b/x^4] - (a^(1/4)*b^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/Sqrt[a + b/x^4]

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Rubi [A]  time = 0.107818, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {242, 277, 305, 220, 1196} \[ x \sqrt{a+\frac{b}{x^4}}-\frac{2 \sqrt{b} \sqrt{a+\frac{b}{x^4}}}{x \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )}-\frac{\sqrt [4]{a} \sqrt [4]{b} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{\sqrt{a+\frac{b}{x^4}}}+\frac{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{\sqrt{a+\frac{b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^4],x]

[Out]

(-2*Sqrt[b]*Sqrt[a + b/x^4])/((Sqrt[a] + Sqrt[b]/x^2)*x) + Sqrt[a + b/x^4]*x + (2*a^(1/4)*b^(1/4)*Sqrt[(a + b/
x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/Sqrt[a
+ b/x^4] - (a^(1/4)*b^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/Sqrt[a + b/x^4]

Rule 242

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

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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.0089302, size = 47, normalized size = 0.21 \[ -\frac{x \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-\frac{a x^4}{b}\right )}{\sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^4],x]

[Out]

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

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Maple [C]  time = 0.011, size = 201, normalized size = 0.9 \begin{align*} -{\frac{x}{a{x}^{4}+b}\sqrt{{\frac{a{x}^{4}+b}{{x}^{4}}}} \left ( -2\,i\sqrt{a}\sqrt{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}}}}}x{\it EllipticF} \left ( x\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}},i \right ) +2\,i\sqrt{a}\sqrt{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}}}}}x{\it EllipticE} \left ( x\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}},i \right ) +\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)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt((a*x^4 + b)/x^4), x)

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Sympy [C]  time = 0.96244, size = 42, normalized size = 0.19 \begin{align*} - \frac{\sqrt{a} x \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x^4), x)