3.1230 \(\int \frac {1}{x^2 \sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {\sqrt {b} x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a-b x^4}} \]

[Out]

-(1-a/b/x^4)^(1/4)*x*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))*Ellip
ticE(sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/(-b*x^4+a)^(1/4)/a^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {313, 335, 275, 228} \[ -\frac {\sqrt {b} x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^4)^(1/4)),x]

[Out]

-((Sqrt[b]*(1 - a/(b*x^4))^(1/4)*x*EllipticE[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(Sqrt[a]*(a - b*x^4)^(1/4)))

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 313

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Dist[(x*(1 + a/(b*x^4))^(1/4))/(a + b*x^4)^(1/4), Int
[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] /; FreeQ[{a, b}, x] && NegQ[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}{x^2 \sqrt [4]{a-b x^4}} \, dx &=\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \int \frac {1}{\sqrt [4]{1-\frac {a}{b x^4}} x^3} \, dx}{\sqrt [4]{a-b x^4}}\\ &=-\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{1-\frac {a x^4}{b}}} \, dx,x,\frac {1}{x}\right )}{\sqrt [4]{a-b x^4}}\\ &=-\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^2}{b}}} \, dx,x,\frac {1}{x^2}\right )}{2 \sqrt [4]{a-b x^4}}\\ &=-\frac {\sqrt {b} \sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.82 \[ -\frac {\sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\frac {b x^4}{a}\right )}{x \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{6} - a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.34, size = 41, normalized size = 0.67 \[ -\frac {{\left (1-\frac {a}{b\,x^4}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ \frac {a}{b\,x^4}\right )}{2\,x\,{\left (a-b\,x^4\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^4)^(1/4)),x)

[Out]

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

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sympy [C]  time = 1.66, size = 31, normalized size = 0.51 \[ \frac {i e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{2 \sqrt [4]{b} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*exp(I*pi/4)*hyper((1/4, 1/2), (3/2,), a/(b*x**4))/(2*b**(1/4)*x**2)

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