3.481 \(\int \frac{1}{x^{10} \sqrt{-1+x^3}} \, dx\)

Optimal. Leaf size=63 \[ \frac{5 \sqrt{x^3-1}}{24 x^3}+\frac{5 \sqrt{x^3-1}}{36 x^6}+\frac{\sqrt{x^3-1}}{9 x^9}+\frac{5}{24} \tan ^{-1}\left (\sqrt{x^3-1}\right ) \]

[Out]

Sqrt[-1 + x^3]/(9*x^9) + (5*Sqrt[-1 + x^3])/(36*x^6) + (5*Sqrt[-1 + x^3])/(24*x^3) + (5*ArcTan[Sqrt[-1 + x^3]]
)/24

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Rubi [A]  time = 0.0194304, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 51, 63, 203} \[ \frac{5 \sqrt{x^3-1}}{24 x^3}+\frac{5 \sqrt{x^3-1}}{36 x^6}+\frac{\sqrt{x^3-1}}{9 x^9}+\frac{5}{24} \tan ^{-1}\left (\sqrt{x^3-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^10*Sqrt[-1 + x^3]),x]

[Out]

Sqrt[-1 + x^3]/(9*x^9) + (5*Sqrt[-1 + x^3])/(36*x^6) + (5*Sqrt[-1 + x^3])/(24*x^3) + (5*ArcTan[Sqrt[-1 + x^3]]
)/24

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^{10} \sqrt{-1+x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} x^4} \, dx,x,x^3\right )\\ &=\frac{\sqrt{-1+x^3}}{9 x^9}+\frac{5}{18} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} x^3} \, dx,x,x^3\right )\\ &=\frac{\sqrt{-1+x^3}}{9 x^9}+\frac{5 \sqrt{-1+x^3}}{36 x^6}+\frac{5}{24} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} x^2} \, dx,x,x^3\right )\\ &=\frac{\sqrt{-1+x^3}}{9 x^9}+\frac{5 \sqrt{-1+x^3}}{36 x^6}+\frac{5 \sqrt{-1+x^3}}{24 x^3}+\frac{5}{48} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} x} \, dx,x,x^3\right )\\ &=\frac{\sqrt{-1+x^3}}{9 x^9}+\frac{5 \sqrt{-1+x^3}}{36 x^6}+\frac{5 \sqrt{-1+x^3}}{24 x^3}+\frac{5}{24} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-1+x^3}\right )\\ &=\frac{\sqrt{-1+x^3}}{9 x^9}+\frac{5 \sqrt{-1+x^3}}{36 x^6}+\frac{5 \sqrt{-1+x^3}}{24 x^3}+\frac{5}{24} \tan ^{-1}\left (\sqrt{-1+x^3}\right )\\ \end{align*}

Mathematica [C]  time = 0.004775, size = 28, normalized size = 0.44 \[ \frac{2}{3} \sqrt{x^3-1} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^10*Sqrt[-1 + x^3]),x]

[Out]

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

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Maple [A]  time = 0.021, size = 48, normalized size = 0.8 \begin{align*}{\frac{5}{24}\arctan \left ( \sqrt{{x}^{3}-1} \right ) }+{\frac{1}{9\,{x}^{9}}\sqrt{{x}^{3}-1}}+{\frac{5}{36\,{x}^{6}}\sqrt{{x}^{3}-1}}+{\frac{5}{24\,{x}^{3}}\sqrt{{x}^{3}-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^10/(x^3-1)^(1/2),x)

[Out]

5/24*arctan((x^3-1)^(1/2))+1/9*(x^3-1)^(1/2)/x^9+5/36*(x^3-1)^(1/2)/x^6+5/24*(x^3-1)^(1/2)/x^3

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Maxima [A]  time = 1.4751, size = 89, normalized size = 1.41 \begin{align*} \frac{15 \,{\left (x^{3} - 1\right )}^{\frac{5}{2}} + 40 \,{\left (x^{3} - 1\right )}^{\frac{3}{2}} + 33 \, \sqrt{x^{3} - 1}}{72 \,{\left ({\left (x^{3} - 1\right )}^{3} + 3 \, x^{3} + 3 \,{\left (x^{3} - 1\right )}^{2} - 2\right )}} + \frac{5}{24} \, \arctan \left (\sqrt{x^{3} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(x^3-1)^(1/2),x, algorithm="maxima")

[Out]

1/72*(15*(x^3 - 1)^(5/2) + 40*(x^3 - 1)^(3/2) + 33*sqrt(x^3 - 1))/((x^3 - 1)^3 + 3*x^3 + 3*(x^3 - 1)^2 - 2) +
5/24*arctan(sqrt(x^3 - 1))

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Fricas [A]  time = 1.72123, size = 107, normalized size = 1.7 \begin{align*} \frac{15 \, x^{9} \arctan \left (\sqrt{x^{3} - 1}\right ) +{\left (15 \, x^{6} + 10 \, x^{3} + 8\right )} \sqrt{x^{3} - 1}}{72 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(x^3-1)^(1/2),x, algorithm="fricas")

[Out]

1/72*(15*x^9*arctan(sqrt(x^3 - 1)) + (15*x^6 + 10*x^3 + 8)*sqrt(x^3 - 1))/x^9

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Sympy [A]  time = 6.48934, size = 182, normalized size = 2.89 \begin{align*} \begin{cases} \frac{5 i \operatorname{acosh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{24} - \frac{5 i}{24 x^{\frac{3}{2}} \sqrt{-1 + \frac{1}{x^{3}}}} + \frac{5 i}{72 x^{\frac{9}{2}} \sqrt{-1 + \frac{1}{x^{3}}}} + \frac{i}{36 x^{\frac{15}{2}} \sqrt{-1 + \frac{1}{x^{3}}}} + \frac{i}{9 x^{\frac{21}{2}} \sqrt{-1 + \frac{1}{x^{3}}}} & \text{for}\: \frac{1}{\left |{x^{3}}\right |} > 1 \\- \frac{5 \operatorname{asin}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{24} + \frac{5}{24 x^{\frac{3}{2}} \sqrt{1 - \frac{1}{x^{3}}}} - \frac{5}{72 x^{\frac{9}{2}} \sqrt{1 - \frac{1}{x^{3}}}} - \frac{1}{36 x^{\frac{15}{2}} \sqrt{1 - \frac{1}{x^{3}}}} - \frac{1}{9 x^{\frac{21}{2}} \sqrt{1 - \frac{1}{x^{3}}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**10/(x**3-1)**(1/2),x)

[Out]

Piecewise((5*I*acosh(x**(-3/2))/24 - 5*I/(24*x**(3/2)*sqrt(-1 + x**(-3))) + 5*I/(72*x**(9/2)*sqrt(-1 + x**(-3)
)) + I/(36*x**(15/2)*sqrt(-1 + x**(-3))) + I/(9*x**(21/2)*sqrt(-1 + x**(-3))), 1/Abs(x**3) > 1), (-5*asin(x**(
-3/2))/24 + 5/(24*x**(3/2)*sqrt(1 - 1/x**3)) - 5/(72*x**(9/2)*sqrt(1 - 1/x**3)) - 1/(36*x**(15/2)*sqrt(1 - 1/x
**3)) - 1/(9*x**(21/2)*sqrt(1 - 1/x**3)), True))

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Giac [A]  time = 1.12292, size = 59, normalized size = 0.94 \begin{align*} \frac{15 \,{\left (x^{3} - 1\right )}^{\frac{5}{2}} + 40 \,{\left (x^{3} - 1\right )}^{\frac{3}{2}} + 33 \, \sqrt{x^{3} - 1}}{72 \, x^{9}} + \frac{5}{24} \, \arctan \left (\sqrt{x^{3} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(x^3-1)^(1/2),x, algorithm="giac")

[Out]

1/72*(15*(x^3 - 1)^(5/2) + 40*(x^3 - 1)^(3/2) + 33*sqrt(x^3 - 1))/x^9 + 5/24*arctan(sqrt(x^3 - 1))