3.5.81 \(\int \frac {1}{x^{10} \sqrt {-1+x^3}} \, dx\) [481]

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

[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 = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 = {272, 44, 65, 209} \begin {gather*} \frac {5}{24} \text {ArcTan}\left (\sqrt {x^3-1}\right )+\frac {5 \sqrt {x^3-1}}{24 x^3}+\frac {\sqrt {x^3-1}}{9 x^9}+\frac {5 \sqrt {x^3-1}}{36 x^6} \end {gather*}

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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

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

Rule 272

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} \text {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} \text {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} \text {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} \text {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} \text {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 [A]
time = 0.04, size = 43, normalized size = 0.68 \begin {gather*} \frac {\sqrt {-1+x^3} \left (8+10 x^3+15 x^6\right )}{72 x^9}+\frac {5}{24} \tan ^{-1}\left (\sqrt {-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^10*Sqrt[-1 + x^3]),x]

[Out]

(Sqrt[-1 + x^3]*(8 + 10*x^3 + 15*x^6))/(72*x^9) + (5*ArcTan[Sqrt[-1 + x^3]])/24

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Maple [A]
time = 0.35, size = 48, normalized size = 0.76

method	result	size
risch	\(\frac {15 x^{9}-5 x^{6}-2 x^{3}-8}{72 x^{9} \sqrt {x^{3}-1}}+\frac {5 \arctan \left (\sqrt {x^{3}-1}\right )}{24}\)	\(41\)
default	\(\frac {5 \arctan \left (\sqrt {x^{3}-1}\right )}{24}+\frac {\sqrt {x^{3}-1}}{9 x^{9}}+\frac {5 \sqrt {x^{3}-1}}{36 x^{6}}+\frac {5 \sqrt {x^{3}-1}}{24 x^{3}}\)	\(48\)
elliptic	\(\frac {5 \arctan \left (\sqrt {x^{3}-1}\right )}{24}+\frac {\sqrt {x^{3}-1}}{9 x^{9}}+\frac {5 \sqrt {x^{3}-1}}{36 x^{6}}+\frac {5 \sqrt {x^{3}-1}}{24 x^{3}}\)	\(48\)
trager	\(\frac {\left (15 x^{6}+10 x^{3}+8\right ) \sqrt {x^{3}-1}}{72 x^{9}}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {x^{3} \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {x^{3}-1}}{x^{3}}\right )}{48}\)	\(68\)
meijerg	\(-\frac {\sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \left (-\frac {\sqrt {\pi }\, \left (-148 x^{9}+144 x^{6}+96 x^{3}+128\right )}{384 x^{9}}+\frac {\sqrt {\pi }\, \left (240 x^{6}+160 x^{3}+128\right ) \sqrt {-x^{3}+1}}{384 x^{9}}+\frac {5 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )}{8}-\frac {5 \left (\frac {37}{30}-2 \ln \left (2\right )+3 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{16}+\frac {\sqrt {\pi }}{3 x^{9}}+\frac {\sqrt {\pi }}{4 x^{6}}+\frac {3 \sqrt {\pi }}{8 x^{3}}\right )}{3 \sqrt {\pi }\, \sqrt {\mathrm {signum}\left (x^{3}-1\right )}}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, size = 66, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 = 0.36, size = 39, normalized size = 0.62 \begin {gather*} \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 {gather*}

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 [C] Result contains complex when optimal does not.
time = 6.79, size = 182, normalized size = 2.89 \begin {gather*} \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 {gather*}

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 = 1.83, size = 44, normalized size = 0.70 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 0.04, size = 201, normalized size = 3.19 \begin {gather*} \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 {-\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+\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}}}\,\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+\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 )}} \end {gather*}

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^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(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)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin((-(x - 1)/((3^(1/2)*1
i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(8*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)
/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2))

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