3.2027 \(\int \frac{1}{\sqrt{a+\frac{b}{x^3}} x^2} \, dx\)

Optimal. Leaf size=221 \[ -\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}} \]

[Out]

(-2*Sqrt[2 + Sqrt[3]]*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sqrt[3])*
a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/
x)], -7 - 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)/x))/((1 + Sqrt[3])*a^(
1/3) + b^(1/3)/x)^2])

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Rubi [A]  time = 0.0624405, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {335, 218} \[ -\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x^3]*x^2),x]

[Out]

(-2*Sqrt[2 + Sqrt[3]]*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sqrt[3])*
a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/
x)], -7 - 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)/x))/((1 + Sqrt[3])*a^(
1/3) + b^(1/3)/x)^2])

Rule 335

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

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+\frac{b}{x^3}} x^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}+\frac{b^{2/3}}{x^2}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.011577, size = 49, normalized size = 0.22 \[ \frac{2 \sqrt{\frac{a x^3}{b}+1} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{a x^3}{b}\right )}{x \sqrt{a+\frac{b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x^3]*x^2),x]

[Out]

(2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b)])/(Sqrt[a + b/x^3]*x)

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Maple [B]  time = 0.011, size = 437, normalized size = 2. \begin{align*} -4\,{\frac{ \left ( a{x}^{3}+b \right ) \left ( i\sqrt{3}{x}^{2}{a}^{2}-2\,i\sqrt [3]{-b{a}^{2}}\sqrt{3}xa+i \left ( -b{a}^{2} \right ) ^{2/3}\sqrt{3}-{a}^{2}{x}^{2}+2\,\sqrt [3]{-b{a}^{2}}xa- \left ( -b{a}^{2} \right ) ^{2/3} \right ) }{\sqrt [3]{-b{a}^{2}}xa\sqrt{x \left ( a{x}^{3}+b \right ) } \left ( i\sqrt{3}-3 \right ) }\sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xa}{ \left ( -1+i\sqrt{3} \right ) \left ( -ax+\sqrt [3]{-b{a}^{2}} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-b{a}^{2}}+2\,ax+\sqrt [3]{-b{a}^{2}}}{ \left ( 1+i\sqrt{3} \right ) \left ( -ax+\sqrt [3]{-b{a}^{2}} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-b{a}^{2}}-2\,ax-\sqrt [3]{-b{a}^{2}}}{ \left ( -1+i\sqrt{3} \right ) \left ( -ax+\sqrt [3]{-b{a}^{2}} \right ) }}}{\it EllipticF} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xa}{ \left ( -1+i\sqrt{3} \right ) \left ( -ax+\sqrt [3]{-b{a}^{2}} \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 ){\frac{1}{\sqrt{{\frac{a{x}^{3}+b}{{x}^{3}}}}}}{\frac{1}{\sqrt{{\frac{x \left ( -ax+\sqrt [3]{-b{a}^{2}} \right ) \left ( i\sqrt{3}\sqrt [3]{-b{a}^{2}}+2\,ax+\sqrt [3]{-b{a}^{2}} \right ) \left ( i\sqrt{3}\sqrt [3]{-b{a}^{2}}-2\,ax-\sqrt [3]{-b{a}^{2}} \right ) }{{a}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + \frac{b}{x^{3}}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a + b/x^3)*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \sqrt{\frac{a x^{3} + b}{x^{3}}}}{a x^{3} + b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*sqrt((a*x^3 + b)/x^3)/(a*x^3 + b), x)

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Sympy [A]  time = 1.41462, size = 37, normalized size = 0.17 \begin{align*} - \frac{\Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt{a} x \Gamma \left (\frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(1/3)*hyper((1/3, 1/2), (4/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt(a)*x*gamma(4/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + \frac{b}{x^{3}}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a + b/x^3)*x^2), x)