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

Optimal. Leaf size=32 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}+b x^2}}\right )}{2 \sqrt{b}} \]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[a/x^2 + b*x^2]]/(2*Sqrt[b])

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Rubi [A]  time = 0.0159634, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1979, 2008, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}+b x^2}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a + b*x^4)/x^2],x]

[Out]

ArcTanh[(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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^2}+b x^2}}\right )}{2 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0207978, size = 59, normalized size = 1.84 \[ \frac{\sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b} x \sqrt{\frac{a+b x^4}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a + b*x^4)/x^2],x]

[Out]

(Sqrt[a + b*x^4]*ArcTanh[(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 [A]  time = 0.051, size = 49, normalized size = 1.5 \begin{align*}{\frac{1}{2\,x}\sqrt{b{x}^{4}+a}\ln \left ({x}^{2}\sqrt{b}+\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)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))/b^(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 )}^{\frac{3}{2}}}\,{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)^(3/2), x) + 1/2*x^2/sqrt(b*x^4 + a)

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Fricas [A]  time = 0.835654, size = 196, normalized size = 6.12 \begin{align*} \left [\frac{\log \left (-2 \, b x^{4} - 2 \, \sqrt{b} x^{3} \sqrt{\frac{b x^{4} + a}{x^{2}}} - a\right )}{4 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} x^{3} \sqrt{\frac{b x^{4} + a}{x^{2}}}}{b x^{4} + a}\right )}{2 \, 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*log(-2*b*x^4 - 2*sqrt(b)*x^3*sqrt((b*x^4 + a)/x^2) - a)/sqrt(b), -1/2*sqrt(-b)*arctan(sqrt(-b)*x^3*sqrt((
b*x^4 + a)/x^2)/(b*x^4 + a))/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