3.7.27 \(\int \frac {(-1+x^3) \sqrt {-1+x^6}}{x^{10}} \, dx\)

Optimal. Leaf size=49 \[ \frac {\left (-2 x^6-3 x^3+2\right ) \sqrt {x^6-1}}{18 x^9}-\frac {1}{3} \tan ^{-1}\left (\frac {x^3+1}{\sqrt {x^6-1}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1475, 807, 266, 47, 63, 203} \begin {gather*} -\frac {\sqrt {x^6-1}}{6 x^6}+\frac {1}{6} \tan ^{-1}\left (\sqrt {x^6-1}\right )-\frac {\left (x^6-1\right )^{3/2}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^3)*Sqrt[-1 + x^6])/x^10,x]

[Out]

-1/6*Sqrt[-1 + x^6]/x^6 - (-1 + x^6)^(3/2)/(9*x^9) + ArcTan[Sqrt[-1 + x^6]]/6

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^{10}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(-1+x) \sqrt {-1+x^2}}{x^4} \, dx,x,x^3\right )\\ &=-\frac {\left (-1+x^6\right )^{3/2}}{9 x^9}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2}}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\left (-1+x^6\right )^{3/2}}{9 x^9}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x^2} \, dx,x,x^6\right )\\ &=-\frac {\sqrt {-1+x^6}}{6 x^6}-\frac {\left (-1+x^6\right )^{3/2}}{9 x^9}+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^6\right )\\ &=-\frac {\sqrt {-1+x^6}}{6 x^6}-\frac {\left (-1+x^6\right )^{3/2}}{9 x^9}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^6}\right )\\ &=-\frac {\sqrt {-1+x^6}}{6 x^6}-\frac {\left (-1+x^6\right )^{3/2}}{9 x^9}+\frac {1}{6} \tan ^{-1}\left (\sqrt {-1+x^6}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 65, normalized size = 1.33 \begin {gather*} -\frac {x^6+\sqrt {1-x^6} x^6 \tanh ^{-1}\left (\sqrt {1-x^6}\right )-1}{6 x^6 \sqrt {x^6-1}}-\frac {\left (x^6-1\right )^{3/2}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^3)*Sqrt[-1 + x^6])/x^10,x]

[Out]

-1/9*(-1 + x^6)^(3/2)/x^9 - (-1 + x^6 + x^6*Sqrt[1 - x^6]*ArcTanh[Sqrt[1 - x^6]])/(6*x^6*Sqrt[-1 + x^6])

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IntegrateAlgebraic [A]  time = 0.15, size = 51, normalized size = 1.04 \begin {gather*} \frac {\left (2-3 x^3-2 x^6\right ) \sqrt {-1+x^6}}{18 x^9}-\frac {1}{3} \tan ^{-1}\left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^3)*Sqrt[-1 + x^6])/x^10,x]

[Out]

((2 - 3*x^3 - 2*x^6)*Sqrt[-1 + x^6])/(18*x^9) - ArcTan[Sqrt[-1 + x^6]/(-1 + x^3)]/3

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fricas [A]  time = 0.73, size = 51, normalized size = 1.04 \begin {gather*} \frac {6 \, x^{9} \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - 2 \, x^{9} - {\left (2 \, x^{6} + 3 \, x^{3} - 2\right )} \sqrt {x^{6} - 1}}{18 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)*(x^6-1)^(1/2)/x^10,x, algorithm="fricas")

[Out]

1/18*(6*x^9*arctan(-x^3 + sqrt(x^6 - 1)) - 2*x^9 - (2*x^6 + 3*x^3 - 2)*sqrt(x^6 - 1))/x^9

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{6} - 1} {\left (x^{3} - 1\right )}}{x^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)*(x^6-1)^(1/2)/x^10,x, algorithm="giac")

[Out]

integrate(sqrt(x^6 - 1)*(x^3 - 1)/x^10, x)

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maple [A]  time = 0.46, size = 42, normalized size = 0.86

method	result	size
risch	\(-\frac {2 x^{12}+3 x^{9}-4 x^{6}-3 x^{3}+2}{18 x^{9} \sqrt {x^{6}-1}}-\frac {\arcsin \left (\frac {1}{x^{3}}\right )}{6}\)	\(42\)
trager	\(-\frac {\left (2 x^{6}+3 x^{3}-2\right ) \sqrt {x^{6}-1}}{18 x^{9}}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right )-\sqrt {x^{6}-1}}{x^{3}}\right )}{6}\)	\(56\)
meijerg	\(\frac {\sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (-x^{6}+1\right )^{\frac {3}{2}}}{9 \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, x^{9}}+\frac {\sqrt {\mathrm {signum}\left (x^{6}-1\right )}\, \left (-\frac {2 \sqrt {\pi }}{x^{6}}-\left (-2 \ln \relax (2)-1+6 \ln \relax (x )+i \pi \right ) \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 x^{6}+8\right )}{4 x^{6}}-\frac {2 \sqrt {\pi }\, \sqrt {-x^{6}+1}}{x^{6}}+2 \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right ) \sqrt {\pi }\right )}{12 \sqrt {-\mathrm {signum}\left (x^{6}-1\right )}\, \sqrt {\pi }}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)*(x^6-1)^(1/2)/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/18*(2*x^12+3*x^9-4*x^6-3*x^3+2)/x^9/(x^6-1)^(1/2)-1/6*arcsin(1/x^3)

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maxima [A]  time = 0.53, size = 35, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {x^{6} - 1}}{6 \, x^{6}} - \frac {{\left (x^{6} - 1\right )}^{\frac {3}{2}}}{9 \, x^{9}} + \frac {1}{6} \, \arctan \left (\sqrt {x^{6} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)*(x^6-1)^(1/2)/x^10,x, algorithm="maxima")

[Out]

-1/6*sqrt(x^6 - 1)/x^6 - 1/9*(x^6 - 1)^(3/2)/x^9 + 1/6*arctan(sqrt(x^6 - 1))

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mupad [B]  time = 1.10, size = 35, normalized size = 0.71 \begin {gather*} \frac {\mathrm {atan}\left (\sqrt {x^6-1}\right )}{6}-\frac {\sqrt {x^6-1}}{6\,x^6}-\frac {{\left (x^6-1\right )}^{3/2}}{9\,x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 - 1)*(x^6 - 1)^(1/2))/x^10,x)

[Out]

atan((x^6 - 1)^(1/2))/6 - (x^6 - 1)^(1/2)/(6*x^6) - (x^6 - 1)^(3/2)/(9*x^9)

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sympy [A]  time = 4.25, size = 49, normalized size = 1.00 \begin {gather*} - \frac {\begin {cases} \frac {\left (x^{6} - 1\right )^{\frac {3}{2}}}{3 x^{9}} & \text {for}\: x > -1 \wedge x < 1 \end {cases}}{3} + \frac {\begin {cases} \frac {\operatorname {acos}{\left (\frac {1}{x^{3}} \right )}}{2} - \frac {\sqrt {1 - \frac {1}{x^{6}}}}{2 x^{3}} & \text {for}\: x > -1 \wedge x < 1 \end {cases}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)*(x**6-1)**(1/2)/x**10,x)

[Out]

-Piecewise(((x**6 - 1)**(3/2)/(3*x**9), (x > -1) & (x < 1)))/3 + Piecewise((acos(x**(-3))/2 - sqrt(1 - 1/x**6)
/(2*x**3), (x > -1) & (x < 1)))/3

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