3.2096 \(\int \frac {1}{(a+\frac {b}{x^4})^{3/2} x^2} \, dx\)

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

[Out]

-1/2/a/x/(a+b/x^4)^(1/2)-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)))*Ellip
ticF(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^(5/4)/b^(1/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {335, 199, 220} \[ -\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {1}{2 a x \sqrt {a+\frac {b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^4)^(3/2)*x^2),x]

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 57, normalized size = 0.52 \[ \frac {\sqrt {\frac {a x^4}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^4}{b}\right )-1}{2 a x \sqrt {a+\frac {b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a^{2} x^{8} + 2 \, a b x^{4} + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(3/2)/x^2,x, algorithm="fricas")

[Out]

integral(x^6*sqrt((a*x^4 + b)/x^4)/(a^2*x^8 + 2*a*b*x^4 + b^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(3/2)/x^2,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^4)^(3/2)*x^2), x)

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maple [C]  time = 0.01, size = 113, normalized size = 1.03 \[ -\frac {\left (a \,x^{4}+b \right ) \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x -\sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )\right )}{2 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^4)^(3/2)/x^2,x)

[Out]

-1/2*(a*x^4+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((I
*a^(1/2)/b^(1/2))^(1/2)*x,I)+(I*a^(1/2)/b^(1/2))^(1/2)*x)/((a*x^4+b)/x^4)^(3/2)/x^6/a/(I*a^(1/2)/b^(1/2))^(1/2
)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(3/2)/x^2,x, algorithm="maxima")

[Out]

integrate(1/((a + b/x^4)^(3/2)*x^2), x)

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mupad [B]  time = 1.37, size = 39, normalized size = 0.35 \[ -\frac {{\left (\frac {b}{a}+x^4\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{2};\ \frac {5}{4};\ -\frac {b}{a\,x^4}\right )}{x\,{\left (a\,x^4+b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b/x^4)^(3/2)),x)

[Out]

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

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sympy [C]  time = 1.50, size = 37, normalized size = 0.34 \[ - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} x \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**4)**(3/2)/x**2,x)

[Out]

-gamma(1/4)*hyper((1/4, 3/2), (5/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(3/2)*x*gamma(5/4))

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