3.2029 \(\int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^8} \, dx\)
Optimal. Leaf size=270 \[ -\frac {32 \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {16 a \sqrt {a+\frac {b}{x^3}}}{55 b^2 x}-\frac {2 \sqrt {a+\frac {b}{x^3}}}{11 b x^4} \]
[Out]
-2/11*(a+b/x^3)^(1/2)/b/x^4+16/55*a*(a+b/x^3)^(1/2)/b^2/x-32/165*a^2*(a^(1/3)+b^(1/3)/x)*EllipticF((b^(1/3)/x+
a^(1/3)*(1-3^(1/2)))/(b^(1/3)/x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)+b^(2/3
)/x^2-a^(1/3)*b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/b^(7/3)/(a+b/x^3)^(1/2)/(a^(1/3)*(a^
(1/3)+b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)
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Rubi [A] time = 0.12, antiderivative size = 270, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]
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 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]
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*}
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Mathematica [C] time = 0.01, 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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fricas [F] time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a x^{3} + b}{x^{3}}}}{a x^{8} + b x^{5}}, x\right ) \]
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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + \frac {b}{x^{3}}} x^{8}}\,{d x} \]
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)
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maple [B] time = 0.02, size = 2009, normalized size = 7.44 \[ \text {result too large to display} \]
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)/(-a^2*b)^(1/3)/b^2*(-32*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b
)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2
)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*
3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(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)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(
1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/
3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-
a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/
2)*x^7*a^2-32*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3
)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))
/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)
^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*3^(1/2)*x^6*a+32*(-(I*3
^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I
*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(
-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*
(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^8*a^3-64*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3
))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2
*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2
)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^
(1/2))*(-a^2*b)^(1/3)*x^7*a^2+32*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1
/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3
)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a
^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*x^6*a+
8*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)
*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*x^3*a-24*(a*x^4+b*x)^(1/2)*(-a^2*b)^(1/3)*
((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*
b)^(1/3))/a^2*x)^(1/2)*a*x^3-5*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^
2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*b+15*(a*x^4+b
*x)^(1/2)*(-a^2*b)^(1/3)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1
/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*b)/((a*x^3+b)*x)^(1/2)/(I*3^(1/2)-3)/((-a*x+(-a^2*b)^(1/3))*(2
*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + \frac {b}{x^{3}}} x^{8}}\,{d x} \]
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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^8\,\sqrt {a+\frac {b}{x^3}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(x^8*(a + b/x^3)^(1/2)),x)
[Out]
int(1/(x^8*(a + b/x^3)^(1/2)), x)
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sympy [A] time = 1.68, size = 39, normalized size = 0.14 \[ - \frac {\Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt {a} x^{7} \Gamma \left (\frac {10}{3}\right )} \]
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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