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

Optimal. Leaf size=228 \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{2 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{\sqrt{a+b x^3}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0417935, antiderivative size = 228, 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 = {277, 218} \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{2 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{\sqrt{a+b x^3}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3]/x^3,x]

[Out]

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.0085822, size = 51, normalized size = 0.22 \[ -\frac{\sqrt{a+b x^3} \, _2F_1\left (-\frac{2}{3},-\frac{1}{2};\frac{1}{3};-\frac{b x^3}{a}\right )}{2 x^2 \sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3]/x^3,x]

[Out]

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

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Maple [A]  time = 0.017, size = 295, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,{x}^{2}}\sqrt{b{x}^{3}+a}}-{{\frac{i}{2}}\sqrt{3}\sqrt [3]{-{b}^{2}a}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}\sqrt{{ \left ( x-{\frac{1}{b}\sqrt [3]{-{b}^{2}a}} \right ) \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}}\sqrt{{-i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}{\it EllipticF} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}},\sqrt{{\frac{i\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a} \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a)/x^3, x)

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Sympy [A]  time = 0.926185, size = 42, normalized size = 0.18 \begin{align*} \frac{\sqrt{a} \Gamma \left (- \frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, - \frac{1}{2} \\ \frac{1}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a)/x^3, x)