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

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

[Out]

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

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Rubi [A]  time = 0.0140644, antiderivative size = 33, normalized size of antiderivative = 1., 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} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x}-b x^2}}\right )}{3 \sqrt{b}} \]

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 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 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 \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 = 0.0259932, size = 66, normalized size = 2. \[ \frac{2 \sqrt{a-b x^3} \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a-b x^3}}\right )}{3 \sqrt{b} \sqrt{x} \sqrt{\frac{a-b x^3}{x}}} \]

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])

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Maple [C]  time = 0.566, size = 471, normalized size = 14.3 \begin{align*} 4\,{\frac{ \left ( b{x}^{3}-a \right ) \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) ^{2}}{{b}^{2}\sqrt{- \left ( b{x}^{3}-a \right ) x} \left ( i\sqrt{3}+3 \right ) }\sqrt{-{\frac{ \left ( i\sqrt{3}+3 \right ) xb}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{{b}^{2}a}-2\,bx-\sqrt [3]{{b}^{2}a}}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{{b}^{2}a}+2\,bx+\sqrt [3]{{b}^{2}a}}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) }}} \left ({\it EllipticF} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}+3 \right ) xb}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) }}},\sqrt{{\frac{ \left ( i\sqrt{3}-3 \right ) \left ( 1+i\sqrt{3} \right ) }{ \left ( i\sqrt{3}+3 \right ) \left ( -1+i\sqrt{3} \right ) }}} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}+3 \right ) xb}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{{b}^{2}a} \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}+3 \right ) \left ( -1+i\sqrt{3} \right ) }}} \right ) \right ){\frac{1}{\sqrt{-{\frac{b{x}^{3}-a}{x}}}}}{\frac{1}{\sqrt{-{\frac{x \left ( -bx+\sqrt [3]{{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{{b}^{2}a}-2\,bx-\sqrt [3]{{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{{b}^{2}a}+2\,bx+\sqrt [3]{{b}^{2}a} \right ) }{{b}^{2}}}}}}} \end{align*}

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+(b^2*a)^(1/3)))^(1/2)*(-b*x+(b^2*a)^(1/3))^2
*((I*3^(1/2)*(b^2*a)^(1/3)-2*b*x-(b^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(b^2*a)
^(1/3)+2*b*x+(b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(b^2*a)^(1/3)))^(1/2)/b^2*(EllipticF((-(I*3^(1/2)+3)*x*b/(1+I*
3^(1/2))/(-b*x+(b^2*a)^(1/3)))^(1/2),((I*3^(1/2)-3)*(1+I*3^(1/2))/(I*3^(1/2)+3)/(-1+I*3^(1/2)))^(1/2))-Ellipti
cPi((-(I*3^(1/2)+3)*x*b/(1+I*3^(1/2))/(-b*x+(b^2*a)^(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)+3)/(-1+I*3^(1/2)))^(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+(b^2*a)^(1/3))*(I*3^(1/2)*(b^2*a)^(1/3)-2*b*x-(b^2*a)^(1/3))*(I*3^(1/2)*(b^2*a)^(1/3)+2*b*x+(b^2*
a)^(1/3)))^(1/2)

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

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)

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Fricas [A]  time = 1.54122, size = 240, normalized size = 7.27 \begin{align*} \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{align*}

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)]

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

[Out]

Exception raised: TypeError