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

Optimal. Leaf size=93 \[ \frac{b x}{a \sqrt [4]{a+b x^2}}-\frac{\left (a+b x^2\right )^{3/4}}{a x}-\frac{\sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt [4]{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0252727, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {325, 229, 227, 196} \[ \frac{b x}{a \sqrt [4]{a+b x^2}}-\frac{\left (a+b x^2\right )^{3/4}}{a x}-\frac{\sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(1/4)),x]

[Out]

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.0084933, size = 49, normalized size = 0.53 \[ -\frac{\sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/x^2/(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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}{b x^{4} + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.761748, size = 27, normalized size = 0.29 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{\sqrt [4]{a} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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