3.388 \(\int \sqrt{\frac{-a+b x^3}{x^2}} \, dx\)

Optimal. Leaf size=53 \[ \frac{2}{3} x \sqrt{b x-\frac{a}{x^2}}+\frac{2}{3} \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{b x-\frac{a}{x^2}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0679645, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1979, 2007, 2029, 203} \[ \frac{2}{3} x \sqrt{b x-\frac{a}{x^2}}+\frac{2}{3} \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{b x-\frac{a}{x^2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-a + b*x^3)/x^2],x]

[Out]

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

Rule 1979

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

Rule 2007

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(p*(n - j)), x] + Dist
[a, Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, j, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[
Simplify[j*p + 1], 0]

Rule 2029

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

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 \sqrt{\frac{-a+b x^3}{x^2}} \, dx &=\int \sqrt{-\frac{a}{x^2}+b x} \, dx\\ &=\frac{2}{3} x \sqrt{-\frac{a}{x^2}+b x}-a \int \frac{1}{x^2 \sqrt{-\frac{a}{x^2}+b x}} \, dx\\ &=\frac{2}{3} x \sqrt{-\frac{a}{x^2}+b x}+\frac{1}{3} (2 a) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{1}{x \sqrt{-\frac{a}{x^2}+b x}}\right )\\ &=\frac{2}{3} x \sqrt{-\frac{a}{x^2}+b x}+\frac{2}{3} \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{-\frac{a}{x^2}+b x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0403021, size = 73, normalized size = 1.38 \[ \frac{2 x \sqrt{b x-\frac{a}{x^2}} \left (\sqrt{b x^3-a}-\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b x^3-a}}{\sqrt{a}}\right )\right )}{3 \sqrt{b x^3-a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(-a + b*x^3)/x^2],x]

[Out]

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

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Maple [A]  time = 0.019, size = 73, normalized size = 1.4 \begin{align*}{\frac{2\,x}{3}\sqrt{{\frac{b{x}^{3}-a}{{x}^{2}}}} \left ( \sqrt{b{x}^{3}-a}\sqrt{-a}+a{\it Artanh} \left ({\sqrt{b{x}^{3}-a}{\frac{1}{\sqrt{-a}}}} \right ) \right ){\frac{1}{\sqrt{b{x}^{3}-a}}}{\frac{1}{\sqrt{-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^3-a)/x^2)^(1/2),x)

[Out]

2/3*((b*x^3-a)/x^2)^(1/2)*x*((b*x^3-a)^(1/2)*(-a)^(1/2)+a*arctanh((b*x^3-a)^(1/2)/(-a)^(1/2)))/(b*x^3-a)^(1/2)
/(-a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{b x^{3} - a}{x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3-a)/x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((b*x^3 - a)/x^2), x)

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Fricas [A]  time = 0.821793, size = 258, normalized size = 4.87 \begin{align*} \left [\frac{2}{3} \, x \sqrt{\frac{b x^{3} - a}{x^{2}}} + \frac{1}{3} \, \sqrt{-a} \log \left (\frac{b x^{3} - 2 \, \sqrt{-a} x \sqrt{\frac{b x^{3} - a}{x^{2}}} - 2 \, a}{x^{3}}\right ), \frac{2}{3} \, x \sqrt{\frac{b x^{3} - a}{x^{2}}} - \frac{2}{3} \, \sqrt{a} \arctan \left (\frac{x \sqrt{\frac{b x^{3} - a}{x^{2}}}}{\sqrt{a}}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3-a)/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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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(((b*x**3-a)/x**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.30164, size = 88, normalized size = 1.66 \begin{align*} -\frac{2}{3} \,{\left (\sqrt{a} \arctan \left (\frac{\sqrt{b x^{3} - a}}{\sqrt{a}}\right ) - \sqrt{b x^{3} - a}\right )} \mathrm{sgn}\left (x\right ) + \frac{2}{3} \,{\left (\sqrt{a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{a}}\right ) - \sqrt{-a}\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3-a)/x^2)^(1/2),x, algorithm="giac")

[Out]

-2/3*(sqrt(a)*arctan(sqrt(b*x^3 - a)/sqrt(a)) - sqrt(b*x^3 - a))*sgn(x) + 2/3*(sqrt(a)*arctan(sqrt(-a)/sqrt(a)
) - sqrt(-a))*sgn(x)