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)/(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))

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Rubi [A]  time = 0.0138555, antiderivative size = 71, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [C]  time = 0.0060906, 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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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}, x\right ) \end{align*}

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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Sympy [C]  time = 0.678049, size = 24, normalized size = 0.34 \begin{align*} \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}} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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)