3.2002 \(\int \frac{\sqrt{a+\frac{b}{x^3}}}{x^5} \, dx\)

Optimal. Leaf size=267 \[ \frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt{3}\right )}{55 b^{4/3} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}-\frac{6 a \sqrt{a+\frac{b}{x^3}}}{55 b x}-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 x^4} \]

[Out]

(-2*Sqrt[a + b/x^3])/(11*x^4) - (6*a*Sqrt[a + b/x^3])/(55*b*x) + (4*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*(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]])/(55*b^(4/3)*
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 = 0.123234, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 279, 321, 218} \[ \frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt{3}\right )}{55 b^{4/3} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}-\frac{6 a \sqrt{a+\frac{b}{x^3}}}{55 b x}-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^3]/x^5,x]

[Out]

(-2*Sqrt[a + b/x^3])/(11*x^4) - (6*a*Sqrt[a + b/x^3])/(55*b*x) + (4*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*(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]])/(55*b^(4/3)*
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 279

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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+\frac{b}{x^3}}}{x^5} \, dx &=-\operatorname{Subst}\left (\int x^3 \sqrt{a+b x^3} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 x^4}-\frac{1}{11} (3 a) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 x^4}-\frac{6 a \sqrt{a+\frac{b}{x^3}}}{55 b x}+\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )}{55 b}\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 x^4}-\frac{6 a \sqrt{a+\frac{b}{x^3}}}{55 b x}+\frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 \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 )}{55 b^{4/3} \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 = 0.0128065, size = 51, normalized size = 0.19 \[ -\frac{2 \sqrt{a+\frac{b}{x^3}} \, _2F_1\left (-\frac{11}{6},-\frac{1}{2};-\frac{5}{6};-\frac{a x^3}{b}\right )}{11 x^4 \sqrt{\frac{a x^3}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^3]/x^5,x]

[Out]

(-2*Sqrt[a + b/x^3]*Hypergeometric2F1[-11/6, -1/2, -5/6, -((a*x^3)/b)])/(11*x^4*Sqrt[1 + (a*x^3)/b])

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Maple [B]  time = 0.025, size = 2002, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^3)^(1/2)/x^5,x)

[Out]

-2/55*((a*x^3+b)/x^3)^(1/2)/x^4/(-b*a^2)^(1/3)*(-12*I*(-(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
))*3^(1/2)*x^8*a^3+24*I*(-(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))*(-b*a^2)^(1/3)*3^(1/2)*x^7*
a^2-12*I*(-(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))*(-b*a^2)^(2/3)*3^(1/2)*x^6*a+12*(-(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))*x^8*a^3-24*(-(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))*(-b*a^2)^(1/3)*x^7*a^2+12*(-(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))*(-b*a^2)^(2/3)*x
^6*a+3*I*(a*x^4+b*x)^(1/2)*(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)*(-b*a^2)^(1/3)*3^(1/2)*x^3*a-9*a*(a*x^4+b*x)^(1/2)*x^3*(-b*
a^2)^(1/3)*(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)+5*I*(a*x^4+b*x)^(1/2)*(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)*(-b*a^2)^(1/3)*3^(1/2)*b-15*(
a*x^4+b*x)^(1/2)*b*(-b*a^2)^(1/3)*(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))/(x*(a*x^3+b))^(1/2)/b/(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{\sqrt{a + \frac{b}{x^{3}}}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^3)^(1/2)/x^5,x, algorithm="maxima")

[Out]

integrate(sqrt(a + b/x^3)/x^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\frac{a x^{3} + b}{x^{3}}}}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^3)^(1/2)/x^5,x, algorithm="fricas")

[Out]

integral(sqrt((a*x^3 + b)/x^3)/x^5, x)

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Sympy [A]  time = 1.2113, size = 41, normalized size = 0.15 \begin{align*} - \frac{\sqrt{a} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{3}}} \right )}}{3 x^{4} \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**3)**(1/2)/x**5,x)

[Out]

-sqrt(a)*gamma(4/3)*hyper((-1/2, 4/3), (7/3,), b*exp_polar(I*pi)/(a*x**3))/(3*x**4*gamma(7/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + \frac{b}{x^{3}}}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^3)^(1/2)/x^5,x, algorithm="giac")

[Out]

integrate(sqrt(a + b/x^3)/x^5, x)