3.5.76 \(\int \frac {1}{x^7 \sqrt {-1+x^3}} \, dx\)

Optimal. Leaf size=38 \[ \frac {1}{4} \tan ^{-1}\left (\sqrt {x^3-1}\right )+\frac {\sqrt {x^3-1} \left (3 x^3+2\right )}{12 x^6} \]

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 203} \begin {gather*} \frac {\sqrt {x^3-1}}{4 x^3}+\frac {1}{4} \tan ^{-1}\left (\sqrt {x^3-1}\right )+\frac {\sqrt {x^3-1}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*Sqrt[-1 + x^3]),x]

[Out]

Sqrt[-1 + x^3]/(6*x^6) + Sqrt[-1 + x^3]/(4*x^3) + ArcTan[Sqrt[-1 + x^3]]/4

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])

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^7 \sqrt {-1+x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x^3} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^3}}{6 x^6}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x^2} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^3}}{6 x^6}+\frac {\sqrt {-1+x^3}}{4 x^3}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^3\right )\\ &=\frac {\sqrt {-1+x^3}}{6 x^6}+\frac {\sqrt {-1+x^3}}{4 x^3}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {\sqrt {-1+x^3}}{6 x^6}+\frac {\sqrt {-1+x^3}}{4 x^3}+\frac {1}{4} \tan ^{-1}\left (\sqrt {-1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 28, normalized size = 0.74 \begin {gather*} \frac {2}{3} \sqrt {x^3-1} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*Sqrt[-1 + x^3]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 38, normalized size = 1.00 \begin {gather*} \frac {\sqrt {-1+x^3} \left (2+3 x^3\right )}{12 x^6}+\frac {1}{4} \tan ^{-1}\left (\sqrt {-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*Sqrt[-1 + x^3]),x]

[Out]

(Sqrt[-1 + x^3]*(2 + 3*x^3))/(12*x^6) + ArcTan[Sqrt[-1 + x^3]]/4

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fricas [A]  time = 0.46, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, x^{6} \arctan \left (\sqrt {x^{3} - 1}\right ) + {\left (3 \, x^{3} + 2\right )} \sqrt {x^{3} - 1}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3-1)^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*x^6*arctan(sqrt(x^3 - 1)) + (3*x^3 + 2)*sqrt(x^3 - 1))/x^6

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giac [A]  time = 0.32, size = 35, normalized size = 0.92 \begin {gather*} \frac {3 \, {\left (x^{3} - 1\right )}^{\frac {3}{2}} + 5 \, \sqrt {x^{3} - 1}}{12 \, x^{6}} + \frac {1}{4} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3-1)^(1/2),x, algorithm="giac")

[Out]

1/12*(3*(x^3 - 1)^(3/2) + 5*sqrt(x^3 - 1))/x^6 + 1/4*arctan(sqrt(x^3 - 1))

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maple [A]  time = 0.34, size = 36, normalized size = 0.95

method	result	size
default	\(\frac {\sqrt {x^{3}-1}}{6 x^{6}}+\frac {\sqrt {x^{3}-1}}{4 x^{3}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{4}\)	\(36\)
risch	\(\frac {3 x^{6}-x^{3}-2}{12 x^{6} \sqrt {x^{3}-1}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{4}\)	\(36\)
elliptic	\(\frac {\sqrt {x^{3}-1}}{6 x^{6}}+\frac {\sqrt {x^{3}-1}}{4 x^{3}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{4}\)	\(36\)
trager	\(\frac {\sqrt {x^{3}-1}\, \left (3 x^{3}+2\right )}{12 x^{6}}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \sqrt {x^{3}-1}}{x^{3}}\right )}{8}\)	\(62\)
meijerg	\(\frac {\sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \left (-\frac {\sqrt {\pi }}{2 x^{6}}-\frac {\sqrt {\pi }}{2 x^{3}}+\frac {3 \left (\frac {7}{6}-2 \ln \relax (2)+3 \ln \relax (x )+i \pi \right ) \sqrt {\pi }}{8}+\frac {\sqrt {\pi }\, \left (-7 x^{6}+8 x^{3}+8\right )}{16 x^{6}}-\frac {\sqrt {\pi }\, \left (12 x^{3}+8\right ) \sqrt {-x^{3}+1}}{16 x^{6}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )}{4}\right )}{3 \sqrt {\mathrm {signum}\left (x^{3}-1\right )}\, \sqrt {\pi }}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^3-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(x^3-1)^(1/2)/x^6+1/4*(x^3-1)^(1/2)/x^3+1/4*arctan((x^3-1)^(1/2))

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maxima [A]  time = 0.42, size = 48, normalized size = 1.26 \begin {gather*} \frac {3 \, {\left (x^{3} - 1\right )}^{\frac {3}{2}} + 5 \, \sqrt {x^{3} - 1}}{12 \, {\left (2 \, x^{3} + {\left (x^{3} - 1\right )}^{2} - 1\right )}} + \frac {1}{4} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3-1)^(1/2),x, algorithm="maxima")

[Out]

1/12*(3*(x^3 - 1)^(3/2) + 5*sqrt(x^3 - 1))/(2*x^3 + (x^3 - 1)^2 - 1) + 1/4*arctan(sqrt(x^3 - 1))

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mupad [B]  time = 0.20, size = 189, normalized size = 4.97 \begin {gather*} \frac {\sqrt {x^3-1}}{4\,x^3}+\frac {\sqrt {x^3-1}}{6\,x^6}-\frac {3\,\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 )}{4\,\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^7*(x^3 - 1)^(1/2)),x)

[Out]

(x^3 - 1)^(1/2)/(4*x^3) + (x^3 - 1)^(1/2)/(6*x^6) - (3*((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)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/
2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(4*(((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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sympy [B]  time = 2.43, size = 138, normalized size = 3.63 \begin {gather*} \begin {cases} \frac {i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{4} - \frac {i}{4 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} + \frac {i}{12 x^{\frac {9}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} + \frac {i}{6 x^{\frac {15}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{4} + \frac {1}{4 x^{\frac {3}{2}} \sqrt {1 - \frac {1}{x^{3}}}} - \frac {1}{12 x^{\frac {9}{2}} \sqrt {1 - \frac {1}{x^{3}}}} - \frac {1}{6 x^{\frac {15}{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**7/(x**3-1)**(1/2),x)

[Out]

Piecewise((I*acosh(x**(-3/2))/4 - I/(4*x**(3/2)*sqrt(-1 + x**(-3))) + I/(12*x**(9/2)*sqrt(-1 + x**(-3))) + I/(
6*x**(15/2)*sqrt(-1 + x**(-3))), 1/Abs(x**3) > 1), (-asin(x**(-3/2))/4 + 1/(4*x**(3/2)*sqrt(1 - 1/x**3)) - 1/(
12*x**(9/2)*sqrt(1 - 1/x**3)) - 1/(6*x**(15/2)*sqrt(1 - 1/x**3)), True))

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