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

Optimal. Leaf size=71 \[ -\frac{5 \sqrt{1-x^3}}{24 x^3}-\frac{5 \sqrt{1-x^3}}{36 x^6}-\frac{\sqrt{1-x^3}}{9 x^9}-\frac{5}{24} \tanh ^{-1}\left (\sqrt{1-x^3}\right ) \]

[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*ArcTanh[Sqrt[1 - x^3]])/
24

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Rubi [A]  time = 0.0280834, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 206} \[ -\frac{5 \sqrt{1-x^3}}{24 x^3}-\frac{5 \sqrt{1-x^3}}{36 x^6}-\frac{\sqrt{1-x^3}}{9 x^9}-\frac{5}{24} \tanh ^{-1}\left (\sqrt{1-x^3}\right ) \]

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*ArcTanh[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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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} \tanh ^{-1}\left (\sqrt{1-x^3}\right )\\ \end{align*}

Mathematica [C]  time = 0.0048222, size = 30, normalized size = 0.42 \[ -\frac{2}{3} \sqrt{1-x^3} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};1-x^3\right ) \]

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 = 0.023, size = 56, normalized size = 0.8 \begin{align*} -{\frac{5}{24}{\it Artanh} \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*arctanh((-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 = 0.996317, size = 122, normalized size = 1.72 \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}{48} \, \log \left (\sqrt{-x^{3} + 1} + 1\right ) + \frac{5}{48} \, \log \left (\sqrt{-x^{3} + 1} - 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
) - 5/48*log(sqrt(-x^3 + 1) + 1) + 5/48*log(sqrt(-x^3 + 1) - 1)

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Fricas [A]  time = 1.44787, size = 161, normalized size = 2.27 \begin{align*} -\frac{15 \, x^{9} \log \left (\sqrt{-x^{3} + 1} + 1\right ) - 15 \, x^{9} \log \left (\sqrt{-x^{3} + 1} - 1\right ) + 2 \,{\left (15 \, x^{6} + 10 \, x^{3} + 8\right )} \sqrt{-x^{3} + 1}}{144 \, 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]

-1/144*(15*x^9*log(sqrt(-x^3 + 1) + 1) - 15*x^9*log(sqrt(-x^3 + 1) - 1) + 2*(15*x^6 + 10*x^3 + 8)*sqrt(-x^3 +
1))/x^9

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Sympy [A]  time = 5.22443, size = 182, normalized size = 2.56 \begin{align*} \begin{cases} - \frac{5 \operatorname{acosh}{\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{for}\: \frac{1}{\left |{x^{3}}\right |} > 1 \\\frac{5 i \operatorname{asin}{\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{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*acosh(x**(-3/2))/24 + 5/(24*x**(3/2)*sqrt(-1 + x**(-3))) - 5/(72*x**(9/2)*sqrt(-1 + x**(-3))) -
1/(36*x**(15/2)*sqrt(-1 + x**(-3))) - 1/(9*x**(21/2)*sqrt(-1 + x**(-3))), 1/Abs(x**3) > 1), (5*I*asin(x**(-3/2
))/24 - 5*I/(24*x**(3/2)*sqrt(1 - 1/x**3)) + 5*I/(72*x**(9/2)*sqrt(1 - 1/x**3)) + I/(36*x**(15/2)*sqrt(1 - 1/x
**3)) + I/(9*x**(21/2)*sqrt(1 - 1/x**3)), True))

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Giac [A]  time = 1.11308, size = 103, normalized size = 1.45 \begin{align*} -\frac{15 \,{\left (x^{3} - 1\right )}^{2} \sqrt{-x^{3} + 1} - 40 \,{\left (-x^{3} + 1\right )}^{\frac{3}{2}} + 33 \, \sqrt{-x^{3} + 1}}{72 \, x^{9}} - \frac{5}{48} \, \log \left (\sqrt{-x^{3} + 1} + 1\right ) + \frac{5}{48} \, \log \left ({\left | \sqrt{-x^{3} + 1} - 1 \right |}\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)^2*sqrt(-x^3 + 1) - 40*(-x^3 + 1)^(3/2) + 33*sqrt(-x^3 + 1))/x^9 - 5/48*log(sqrt(-x^3 + 1)
+ 1) + 5/48*log(abs(sqrt(-x^3 + 1) - 1))