\(\int \sqrt {\frac {-a+b x^3}{x^2}} \, dx\) [388]

Optimal result

Integrand size = 17, antiderivative size = 53 \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\frac {2}{3} x \sqrt {-\frac {a}{x^2}+b x}+\frac {2}{3} \sqrt {a} \arctan \left (\frac {\sqrt {a}}{x \sqrt {-\frac {a}{x^2}+b x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2004, 2032, 2054, 209} \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\frac {2}{3} \sqrt {a} \arctan \left (\frac {\sqrt {a}}{x \sqrt {b x-\frac {a}{x^2}}}\right )+\frac {2}{3} x \sqrt {b x-\frac {a}{x^2}} \]

[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 209

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

Rule 2004

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

Rule 2032

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 2054

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]

Rubi steps \begin{align*} \text {integral}& = \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) \text {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] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\frac {2 x \sqrt {-\frac {a}{x^2}+b x} \left (\sqrt {-a+b x^3}-\sqrt {a} \arctan \left (\frac {\sqrt {-a+b x^3}}{\sqrt {a}}\right )\right )}{3 \sqrt {-a+b x^3}} \]

[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])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {2 \sqrt {-\frac {-b \,x^{3}+a}{x^{2}}}\, x \left (\sqrt {b \,x^{3}-a}\, \sqrt {-a}+a \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}-a}}{\sqrt {-a}}\right )\right )}{3 \sqrt {b \,x^{3}-a}\, \sqrt {-a}}\)	\(73\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.06 \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\left [\frac {2}{3} \, x \sqrt {\frac {b x^{3} - a}{x^{2}}} + \frac {1}{3} \, \sqrt {-a} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a} x \sqrt {\frac {b x^{3} - a}{x^{2}}} - 2 \, a}{x^{3}}\right ), \frac {2}{3} \, x \sqrt {\frac {b x^{3} - a}{x^{2}}} - \frac {2}{3} \, \sqrt {a} \arctan \left (\frac {x \sqrt {\frac {b x^{3} - a}{x^{2}}}}{\sqrt {a}}\right )\right ] \]

[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))]

Sympy [F(-1)]

Timed out. \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\int { \sqrt {\frac {b x^{3} - a}{x^{2}}} \,d x } \]

[In]

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

[Out]

integrate(sqrt((b*x^3 - a)/x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=-\frac {2}{3} \, \sqrt {a} \arctan \left (\frac {\sqrt {b x^{3} - a}}{\sqrt {a}}\right ) \mathrm {sgn}\left (x\right ) + \frac {2}{3} \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {a}}\right ) - \sqrt {-a}\right )} \mathrm {sgn}\left (x\right ) + \frac {2}{3} \, \sqrt {b x^{3} - a} \mathrm {sgn}\left (x\right ) \]

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

Mupad [B] (verification not implemented)

Time = 9.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \sqrt {\frac {-a+b x^3}{x^2}} \, dx=\frac {2\,x\,\sqrt {b\,x-\frac {a}{x^2}}}{3}+\frac {2\,\sqrt {a}\,\mathrm {asin}\left (\frac {\sqrt {a}}{\sqrt {b}\,x^{3/2}}\right )\,\sqrt {b\,x-\frac {a}{x^2}}}{3\,\sqrt {b}\,\sqrt {x}\,\sqrt {1-\frac {a}{b\,x^3}}} \]

[In]

int((-(a - b*x^3)/x^2)^(1/2),x)

[Out]

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