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

Optimal. Leaf size=47 \[ \frac{\sqrt{x^3-1}}{4 x^3}+\frac{\sqrt{x^3-1}}{6 x^6}+\frac{1}{4} \tan ^{-1}\left (\sqrt{x^3-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.014413, antiderivative size = 47, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{x^3-1}}{4 x^3}+\frac{\sqrt{x^3-1}}{6 x^6}+\frac{1}{4} \tan ^{-1}\left (\sqrt{x^3-1}\right ) \]

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 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^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*}

Mathematica [C]  time = 0.0037645, size = 28, normalized size = 0.6 \[ \frac{2}{3} \sqrt{x^3-1} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};1-x^3\right ) \]

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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Maple [A]  time = 0.018, size = 36, normalized size = 0.8 \begin{align*}{\frac{1}{4}\arctan \left ( \sqrt{{x}^{3}-1} \right ) }+{\frac{1}{6\,{x}^{6}}\sqrt{{x}^{3}-1}}+{\frac{1}{4\,{x}^{3}}\sqrt{{x}^{3}-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57114, size = 65, normalized size = 1.38 \begin{align*} \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{align*}

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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Fricas [A]  time = 1.7159, size = 92, normalized size = 1.96 \begin{align*} \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{align*}

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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Sympy [A]  time = 3.8372, size = 138, normalized size = 2.94 \begin{align*} \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{align*}

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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Giac [A]  time = 1.11643, size = 47, normalized size = 1. \begin{align*} \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{align*}

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))