3.2049 \(\int \frac {1}{(a+\frac {b}{x^3})^{3/2} x^8} \, dx\)

Optimal. Leaf size=267 \[ \frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}} \]

[Out]

2/3/b/x^4/(a+b/x^3)^(1/2)-16/15*(a+b/x^3)^(1/2)/b^2/x+32/45*a*(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 = 267, normalized size of antiderivative = 1.00, 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, 288, 321, 218} \[ \frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^3)^(3/2)*x^8),x]

[Out]

2/(3*b*Sqrt[a + b/x^3]*x^4) - (16*Sqrt[a + b/x^3])/(15*b^2*x) + (32*Sqrt[2 + Sqrt[3]]*a*(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]])/(15*3^(1/4)*b^(7/3)*Sq
rt[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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 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}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx &=-\operatorname {Subst}\left (\int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {8 \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{3 b}\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {(16 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{15 b^2}\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {32 \sqrt {2+\sqrt {3}} a \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 )}{15 \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 = 0.02, size = 54, normalized size = 0.20 \[ -\frac {2 \sqrt {\frac {a x^3}{b}+1} \, _2F_1\left (-\frac {5}{6},\frac {3}{2};\frac {1}{6};-\frac {a x^3}{b}\right )}{5 b x^4 \sqrt {a+\frac {b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^3)^(3/2)*x^8),x]

[Out]

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

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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a x^{3} + b}{x^{3}}}}{a^{2} x^{8} + 2 \, a b x^{5} + b^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt((a*x^3 + b)/x^3)/(a^2*x^8 + 2*a*b*x^5 + b^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 2056, normalized size = 7.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/((a*x^3+b)/x^3)^(3/2)/x^8*(a*x^3+b)/(-a^2*b)^(1/3)/b^2*(32*I*3^(1/2)*(-(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)*Elliptic
F((-(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*x^3+b)*x)^(1/2)*x^5*a^2-64*I*(-a^2*b)^(1/3)*3^(1/2)*(-(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)*Ellipt
icF((-(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*x^3+b)*x)^(1/2)*x^4*a+32*I*(-a^2*b)^(2/3)*3^(1/2)*(-(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)*Ellipt
icF((-(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*x^3+b)*x)^(1/2)*x^3-32*(-(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*x
^3+b)*x)^(1/2)*x^5*a^2+64*(-a^2*b)^(1/3)*(-(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*x^3+b)*x)
^(1/2)*x^4*a-32*(-a^2*b)^(2/3)*(-(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*x^3+b)*x)^(1/2)*x^3
-5*I*(-a^2*b)^(1/3)*3^(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)*x^4*a+15*(-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)*x^4*a-3*I*(-
a^2*b)^(1/3)*3^(1/2)*(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*x^3+b)*x)^(1/2)+9*(-a^2*b)^(1/3)*(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*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 \[ \int \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^8\,{\left (a+\frac {b}{x^3}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(a + b/x^3)^(3/2)),x)

[Out]

int(1/(x^8*(a + b/x^3)^(3/2)), x)

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sympy [A]  time = 2.25, size = 39, normalized size = 0.15 \[ - \frac {\Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} x^{7} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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