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

Optimal. Leaf size=33 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}-b x^2}}\right )}{2 \sqrt{b}} \]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a/x^2 - b*x^2]]/(2*Sqrt[b])

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Rubi [A]  time = 0.0149095, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1979, 2008, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}-b x^2}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a - b*x^4)/x^2],x]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a/x^2 - b*x^2]]/(2*Sqrt[b])

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{\frac{a-b x^4}{x^2}}} \, dx &=\int \frac{1}{\sqrt{\frac{a}{x^2}-b x^2}} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{x}{\sqrt{\frac{a}{x^2}-b x^2}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}-b x^2}}\right )}{2 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0219543, size = 62, normalized size = 1.88 \[ \frac{\sqrt{a-b x^4} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a-b x^4}}\right )}{2 \sqrt{b} x \sqrt{\frac{a-b x^4}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a - b*x^4)/x^2],x]

[Out]

(Sqrt[a - b*x^4]*ArcTan[(Sqrt[b]*x^2)/Sqrt[a - b*x^4]])/(2*Sqrt[b]*x*Sqrt[(a - b*x^4)/x^2])

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Maple [B]  time = 0.013, size = 53, normalized size = 1.6 \begin{align*}{\frac{1}{2\,x}\sqrt{-b{x}^{4}+a}\arctan \left ({{x}^{2}\sqrt{b}{\frac{1}{\sqrt{-b{x}^{4}+a}}}} \right ){\frac{1}{\sqrt{-{\frac{b{x}^{4}-a}{{x}^{2}}}}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{x^{5}}{{\left (b x^{4} - a\right )} \sqrt{-b x^{4} + a}}\,{d x} + \frac{x^{2}}{2 \, \sqrt{-b x^{4} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(x^5/((b*x^4 - a)*sqrt(-b*x^4 + a)), x) + 1/2*x^2/sqrt(-b*x^4 + a)

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Fricas [A]  time = 1.02574, size = 198, normalized size = 6. \begin{align*} \left [-\frac{\sqrt{-b} \log \left (2 \, b x^{4} - 2 \, \sqrt{-b} x^{3} \sqrt{-\frac{b x^{4} - a}{x^{2}}} - a\right )}{4 \, b}, -\frac{\arctan \left (\frac{\sqrt{b} x^{3} \sqrt{-\frac{b x^{4} - a}{x^{2}}}}{b x^{4} - a}\right )}{2 \, \sqrt{b}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError