3.2029 \(\int \frac{1}{\sqrt{a+\frac{b}{x^3}} x^8} \, dx\)

Optimal. Leaf size=270 \[ -\frac{32 \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}+\frac{16 a \sqrt{a+\frac{b}{x^3}}}{55 b^2 x}-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 b x^4} \]

[Out]

(-2*Sqrt[a + b/x^3])/(11*b*x^4) + (16*a*Sqrt[a + b/x^3])/(55*b^2*x) - (32*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[Ar
cSin[((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*3^(1/4)*b^
(7/3)*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 = 0.11993, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {335, 321, 218} \[ -\frac{32 \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}+\frac{16 a \sqrt{a+\frac{b}{x^3}}}{55 b^2 x}-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x^3]*x^8),x]

[Out]

(-2*Sqrt[a + b/x^3])/(11*b*x^4) + (16*a*Sqrt[a + b/x^3])/(55*b^2*x) - (32*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[Ar
cSin[((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*3^(1/4)*b^
(7/3)*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 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{1}{\sqrt{a+\frac{b}{x^3}} x^8} \, dx &=-\operatorname{Subst}\left (\int \frac{x^6}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 b x^4}+\frac{(8 a) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )}{11 b}\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 b x^4}+\frac{16 a \sqrt{a+\frac{b}{x^3}}}{55 b^2 x}-\frac{\left (16 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^3}} \, dx,x,\frac{1}{x}\right )}{55 b^2}\\ &=-\frac{2 \sqrt{a+\frac{b}{x^3}}}{11 b x^4}+\frac{16 a \sqrt{a+\frac{b}{x^3}}}{55 b^2 x}-\frac{32 \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 \sqrt [4]{3} b^{7/3} \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 = 0.01246, size = 51, normalized size = 0.19 \[ -\frac{2 \sqrt{\frac{a x^3}{b}+1} \, _2F_1\left (-\frac{11}{6},\frac{1}{2};-\frac{5}{6};-\frac{a x^3}{b}\right )}{11 x^7 \sqrt{a+\frac{b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x^3]*x^8),x]

[Out]

(-2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[-11/6, 1/2, -5/6, -((a*x^3)/b)])/(11*Sqrt[a + b/x^3]*x^7)

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Maple [B]  time = 0.013, size = 2009, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^8/(a+b/x^3)^(1/2),x)

[Out]

2/55/((a*x^3+b)/x^3)^(1/2)/x^7*(a*x^3+b)/(-b*a^2)^(1/3)/b^2*(-32*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+64*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^2-32*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+
32*(-(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^3-64*(-(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)*Elliptic
F((-(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+32*(-(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+8*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-24*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)/(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^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(a+b/x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a + b/x^3)*x^8), 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}}}}{a x^{8} + b x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(a+b/x^3)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt((a*x^3 + b)/x^3)/(a*x^8 + b*x^5), x)

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Sympy [A]  time = 2.77529, size = 39, normalized size = 0.14 \begin{align*} - \frac{\Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt{a} x^{7} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(a+b/x**3)**(1/2),x)

[Out]

-gamma(7/3)*hyper((1/2, 7/3), (10/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt(a)*x**7*gamma(10/3))

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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^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(a+b/x^3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(a + b/x^3)*x^8), x)