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

Optimal. Leaf size=71 \[ \frac {2 x}{\sqrt [4]{a+b x^2}}-\frac {2 \sqrt {a} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {229, 227, 196} \[ \frac {2 x}{\sqrt [4]{a+b x^2}}-\frac {2 \sqrt {a} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(-1/4),x]

[Out]

(2*x)/(a + b*x^2)^(1/4) - (2*Sqrt[a]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[
b]*(a + b*x^2)^(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 227

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

Rule 229

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.65 \[ \frac {x \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(-1/4),x]

[Out]

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

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(-1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(-1/4), x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(-1/4), x)

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mupad [B]  time = 4.85, size = 37, normalized size = 0.52 \[ \frac {x\,{\left (\frac {b\,x^2}{a}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^2)^(1/4),x)

[Out]

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

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sympy [C]  time = 0.78, size = 24, normalized size = 0.34 \[ \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\sqrt [4]{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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