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

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

[Out]

x*(a+b/x^4)^(1/2)-2*b^(1/2)*(a+b/x^4)^(1/2)/x/(a^(1/2)+b^(1/2)/x^2)+2*a^(1/4)*b^(1/4)*(cos(2*arccot(a^(1/4)*x/
b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4)))*EllipticE(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a
^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)/(a+b/x^4)^(1/2)-a^(1/4)*b^(1/4)*(cos(2*arccot(a^
(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(
1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 224, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 1.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\frac {a x^{4} + b}{x^{4}}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{x^{4}}}\,{d x} \]

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)

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maple [C]  time = 0.01, size = 202, normalized size = 0.90 \[ \frac {\sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, \left (-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{4}-2 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {a}\, \sqrt {b}\, x \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+2 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {a}\, \sqrt {b}\, x \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \right ) x}{\left (a \,x^{4}+b \right ) \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}} \]

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((I*a^(1/2)/b^(1/2))^(1/2)*x,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((I*a^(1/2)/b^(1/2))^(1/2)*x,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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{x^{4}}}\,{d x} \]

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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mupad [B]  time = 1.23, size = 38, normalized size = 0.17 \[ -\frac {x\,\sqrt {a+\frac {b}{x^4}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x^4}{b}\right )}{\sqrt {\frac {a\,x^4}{b}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^4)^(1/2),x)

[Out]

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

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sympy [C]  time = 1.17, size = 42, normalized size = 0.19 \[ - \frac {\sqrt {a} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

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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