3.1106 \(\int \frac {1}{x^2 \sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=75 \[ \frac {\sqrt {b} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a+b x^4}}-\frac {1}{x \sqrt [4]{a+b x^4}} \]

[Out]

-1/x/(b*x^4+a)^(1/4)+(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(
1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(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 = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {312, 281, 335, 275, 196} \[ \frac {\sqrt {b} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a+b x^4}}-\frac {1}{x \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^4)^(1/4)),x]

[Out]

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

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[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 281

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[(x*(1 + a/(b*x^4))^(1/4))/(b*(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 312

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> -Simp[(x*(a + b*x^4)^(1/4))^(-1), x] - Dist[b, Int[x^
2/(a + b*x^4)^(5/4), x], 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}{x^2 \sqrt [4]{a+b x^4}} \, dx &=-\frac {1}{x \sqrt [4]{a+b x^4}}-b \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx\\ &=-\frac {1}{x \sqrt [4]{a+b x^4}}-\frac {\left (\sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{\sqrt [4]{a+b x^4}}\\ &=-\frac {1}{x \sqrt [4]{a+b x^4}}+\frac {\left (\sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{\sqrt [4]{a+b x^4}}\\ &=-\frac {1}{x \sqrt [4]{a+b x^4}}+\frac {\left (\sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{2 \sqrt [4]{a+b x^4}}\\ &=-\frac {1}{x \sqrt [4]{a+b x^4}}+\frac {\sqrt {b} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-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 = 49, normalized size = 0.65 \[ -\frac {\sqrt [4]{\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};-\frac {b x^4}{a}\right )}{x \sqrt [4]{a+b x^4}} \]

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 = 0.90, 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 = 0.14, 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 = 1.46, size = 40, normalized size = 0.53 \[ -\frac {{\left (\frac {a}{b\,x^4}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {a}{b\,x^4}\right )}{2\,x\,{\left (b\,x^4+a\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^4)^(1/4)),x)

[Out]

-((a/(b*x^4) + 1)^(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 = 1.55, size = 39, normalized size = 0.52 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} x \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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