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

Optimal. Leaf size=42 \[ x \sqrt{\frac{a}{x^2}+b}-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+b}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0229586, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1972, 242, 277, 217, 206} \[ x \sqrt{\frac{a}{x^2}+b}-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+b}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(a + b*x^2)/x^2],x]

[Out]

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

Rule 1972

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

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 277

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A]  time = 0.0280174, size = 62, normalized size = 1.48 \[ x \sqrt{\frac{a}{x^2}+b}-\frac{\sqrt{a} x \sqrt{\frac{a}{x^2}+b} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(a + b*x^2)/x^2],x]

[Out]

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

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Maple [A]  time = 0.005, size = 61, normalized size = 1.5 \begin{align*}{x\sqrt{{\frac{b{x}^{2}+a}{{x}^{2}}}} \left ( \sqrt{b{x}^{2}+a}-\sqrt{a}\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ) \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.85997, size = 259, normalized size = 6.17 \begin{align*} \left [x \sqrt{\frac{b x^{2} + a}{x^{2}}} + \frac{1}{2} \, \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{a} x \sqrt{\frac{b x^{2} + a}{x^{2}}} + 2 \, a}{x^{2}}\right ), x \sqrt{\frac{b x^{2} + a}{x^{2}}} + \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x \sqrt{\frac{b x^{2} + a}{x^{2}}}}{b x^{2} + a}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((a + b*x**2)/x**2), x)

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Giac [A]  time = 1.24815, size = 92, normalized size = 2.19 \begin{align*}{\left (\frac{a \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \sqrt{b x^{2} + a}\right )} \mathrm{sgn}\left (x\right ) - \frac{{\left (a \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + \sqrt{-a} \sqrt{a}\right )} \mathrm{sgn}\left (x\right )}{\sqrt{-a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + sqrt(b*x^2 + a))*sgn(x) - (a*arctan(sqrt(a)/sqrt(-a)) + sqrt(-a
)*sqrt(a))*sgn(x)/sqrt(-a)