3.4.14 \(\int \frac {1}{\sqrt {\frac {a-b x^3}{x}}} \, dx\)

Optimal. Leaf size=33 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x}-b x^2}}\right )}{3 \sqrt {b}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x}-b x^2}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a - b*x^3)/x],x]

[Out]

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

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])

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 66, normalized size = 2.00 \begin {gather*} \frac {2 \sqrt {a-b x^3} \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a-b x^3}}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {\frac {a-b x^3}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a - b*x^3)/x],x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.38, size = 35, normalized size = 1.06 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {\frac {a-b x^3}{x}}}{\sqrt {b} x}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[(a - b*x^3)/x],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 111, normalized size = 3.36 \begin {gather*} \left [-\frac {\sqrt {-b} \log \left (-8 \, b^{2} x^{6} + 8 \, a b x^{3} - a^{2} + 4 \, {\left (2 \, b x^{5} - a x^{2}\right )} \sqrt {-b} \sqrt {-\frac {b x^{3} - a}{x}}\right )}{6 \, b}, -\frac {\arctan \left (\frac {2 \, \sqrt {b} x^{2} \sqrt {-\frac {b x^{3} - a}{x}}}{2 \, b x^{3} - a}\right )}{3 \, \sqrt {b}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*sqrt(-b)*log(-8*b^2*x^6 + 8*a*b*x^3 - a^2 + 4*(2*b*x^5 - a*x^2)*sqrt(-b)*sqrt(-(b*x^3 - a)/x))/b, -1/3*a
rctan(2*sqrt(b)*x^2*sqrt(-(b*x^3 - a)/x)/(2*b*x^3 - a))/sqrt(b)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Undef/Unsigned Inf encountered in limitLimit: Max order reached or unable to make series expansion Error:
Bad Argument Value

________________________________________________________________________________________

maple [C]  time = 6.22, size = 471, normalized size = 14.27 \begin {gather*} \frac {4 \left (b \,x^{3}-a \right ) \left (1+i \sqrt {3}\right ) \sqrt {-\frac {\left (i \sqrt {3}+3\right ) b x}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )^{2} \sqrt {\frac {-2 b x +i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}-\left (a \,b^{2}\right )^{\frac {1}{3}}}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {2 b x +i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (\EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}+3\right ) b x}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}-3\right ) \left (1+i \sqrt {3}\right )}{\left (i \sqrt {3}-1\right ) \left (i \sqrt {3}+3\right )}}\right )-\EllipticPi \left (\sqrt {-\frac {\left (i \sqrt {3}+3\right ) b x}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \frac {1+i \sqrt {3}}{i \sqrt {3}+3}, \sqrt {\frac {\left (i \sqrt {3}-3\right ) \left (1+i \sqrt {3}\right )}{\left (i \sqrt {3}-1\right ) \left (i \sqrt {3}+3\right )}}\right )\right )}{\sqrt {-\frac {b \,x^{3}-a}{x}}\, \sqrt {-\left (b \,x^{3}-a \right ) x}\, \left (i \sqrt {3}+3\right ) \sqrt {-\frac {\left (-b x +\left (a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (-2 b x +i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}-\left (a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (2 b x +i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}\right ) x}{b^{2}}}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(b*x^3-a)*(1+I*3^(1/2))*(-(I*3^(1/2)+3)*x*b/(1+I*3^(1/2))/(-b*x+(a*b^2)^(1/3)))^(1/2)*(-b*x+(a*b^2)^(1/3))^2
*((I*3^(1/2)*(a*b^2)^(1/3)-2*b*x-(a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(a*b^2)^
(1/3)+2*b*x+(a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(a*b^2)^(1/3)))^(1/2)/b^2*(EllipticF((-(I*3^(1/2)+3)*x*b/(1+I*3
^(1/2))/(-b*x+(a*b^2)^(1/3)))^(1/2),((I*3^(1/2)-3)*(1+I*3^(1/2))/(I*3^(1/2)-1)/(I*3^(1/2)+3))^(1/2))-EllipticP
i((-(I*3^(1/2)+3)*x*b/(1+I*3^(1/2))/(-b*x+(a*b^2)^(1/3)))^(1/2),(1+I*3^(1/2))/(I*3^(1/2)+3),((I*3^(1/2)-3)*(1+
I*3^(1/2))/(I*3^(1/2)-1)/(I*3^(1/2)+3))^(1/2)))/(-(b*x^3-a)/x)^(1/2)/(-(b*x^3-a)*x)^(1/2)/(I*3^(1/2)+3)/(-1/b^
2*x*(-b*x+(a*b^2)^(1/3))*(I*3^(1/2)*(a*b^2)^(1/3)-2*b*x-(a*b^2)^(1/3))*(I*3^(1/2)*(a*b^2)^(1/3)+2*b*x+(a*b^2)^
(1/3)))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-\frac {b x^{3} - a}{x}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(-(b*x^3 - a)/x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{\sqrt {\frac {a-b\,x^3}{x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - b*x^3)/x)^(1/2),x)

[Out]

int(1/((a - b*x^3)/x)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\frac {a - b x^{3}}{x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt((a - b*x**3)/x), x)

________________________________________________________________________________________