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

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

[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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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, 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, 204} \[ \frac {5 \sqrt {-x^3-1}}{24 x^3}-\frac {5 \sqrt {-x^3-1}}{36 x^6}+\frac {\sqrt {-x^3-1}}{9 x^9}-\frac {5}{24} \tan ^{-1}\left (\sqrt {-x^3-1}\right ) \]

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 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 204

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

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]]

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*}

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Mathematica [C]  time = 0.01, size = 28, normalized size = 0.39 \[ -\frac {2}{3} \sqrt {-x^3-1} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};x^3+1\right ) \]

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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fricas [A]  time = 0.85, size = 44, normalized size = 0.62 \[ -\frac {15 \, x^{9} \arctan \left (\sqrt {-x^{3} - 1}\right ) - {\left (15 \, x^{6} - 10 \, x^{3} + 8\right )} \sqrt {-x^{3} - 1}}{72 \, x^{9}} \]

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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giac [A]  time = 0.17, size = 59, normalized size = 0.83 \[ \frac {15 \, {\left (x^{3} + 1\right )}^{2} \sqrt {-x^{3} - 1} + 40 \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {-x^{3} - 1}}{72 \, x^{9}} - \frac {5}{24} \, \arctan \left (\sqrt {-x^{3} - 1}\right ) \]

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 - 5/24*arctan(sqrt(-x^3 - 1
))

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maple [A]  time = 0.03, size = 56, normalized size = 0.79 \[ -\frac {5 \arctan \left (\sqrt {-x^{3}-1}\right )}{24}+\frac {5 \sqrt {-x^{3}-1}}{24 x^{3}}-\frac {5 \sqrt {-x^{3}-1}}{36 x^{6}}+\frac {\sqrt {-x^{3}-1}}{9 x^{9}} \]

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 = 3.00, size = 74, normalized size = 1.04 \[ \frac {15 \, {\left (-x^{3} - 1\right )}^{\frac {5}{2}} + 40 \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {-x^{3} - 1}}{72 \, {\left ({\left (x^{3} + 1\right )}^{3} + 3 \, x^{3} - 3 \, {\left (x^{3} + 1\right )}^{2} + 2\right )}} - \frac {5}{24} \, \arctan \left (\sqrt {-x^{3} - 1}\right ) \]

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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mupad [B]  time = 0.03, size = 223, normalized size = 3.14 \[ \frac {5\,\sqrt {-x^3-1}}{24\,x^3}-\frac {5\,\sqrt {-x^3-1}}{36\,x^6}+\frac {\sqrt {-x^3-1}}{9\,x^9}+\frac {5\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3+1}\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{8\,\sqrt {-x^3-1}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^10*(- x^3 - 1)^(1/2)),x)

[Out]

(5*(- x^3 - 1)^(1/2))/(24*x^3) - (5*(- x^3 - 1)^(1/2))/(36*x^6) + (- x^3 - 1)^(1/2)/(9*x^9) + (5*((3^(1/2)*1i)
/2 + 3/2)*(x^3 + 1)^(1/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 +
3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin((
(x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(8*(- x^3 - 1)^(1/2)*
(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^
(1/2))

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sympy [C]  time = 8.48, size = 90, normalized size = 1.27 \[ - \frac {5 i \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{24} + \frac {5 i}{24 x^{\frac {3}{2}} \sqrt {1 + \frac {1}{x^{3}}}} + \frac {5 i}{72 x^{\frac {9}{2}} \sqrt {1 + \frac {1}{x^{3}}}} - \frac {i}{36 x^{\frac {15}{2}} \sqrt {1 + \frac {1}{x^{3}}}} + \frac {i}{9 x^{\frac {21}{2}} \sqrt {1 + \frac {1}{x^{3}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*I*asinh(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)))

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