3.6.33 \(\int \frac {\sqrt {-1+x^3} (-2+x^3+2 x^6)}{x^{10}} \, dx\)

Optimal. Leaf size=41 \[ \frac {2}{3} \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {2 \sqrt {x^3-1} \left (3 x^6+x^3-1\right )}{9 x^9} \]

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Rubi [A]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.54, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1474, 897, 1257, 1157, 12, 288, 203} \begin {gather*} -\frac {2 \sqrt {x^3-1}}{3 x^3}+\frac {2}{3} \tan ^{-1}\left (\sqrt {x^3-1}\right )+\frac {2 \sqrt {x^3-1}}{9 x^9}-\frac {2 \sqrt {x^3-1}}{9 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x^3]*(-2 + x^3 + 2*x^6))/x^10,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1474

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

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x^3} \left (-2+x^3+2 x^6\right )}{x^{10}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x} \left (-2+x+2 x^2\right )}{x^4} \, dx,x,x^3\right )\\ &=\frac {2}{3} \operatorname {Subst}\left (\int \frac {x^2 \left (1+5 x^2+2 x^4\right )}{\left (1+x^2\right )^4} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {2 \sqrt {-1+x^3}}{9 x^9}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {2-18 x^2-12 x^4}{\left (1+x^2\right )^3} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {2 \sqrt {-1+x^3}}{9 x^9}-\frac {2 \sqrt {-1+x^3}}{9 x^6}+\frac {1}{36} \operatorname {Subst}\left (\int \frac {48 x^2}{\left (1+x^2\right )^2} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {2 \sqrt {-1+x^3}}{9 x^9}-\frac {2 \sqrt {-1+x^3}}{9 x^6}+\frac {4}{3} \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {2 \sqrt {-1+x^3}}{9 x^9}-\frac {2 \sqrt {-1+x^3}}{9 x^6}-\frac {2 \sqrt {-1+x^3}}{3 x^3}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^3}\right )\\ &=\frac {2 \sqrt {-1+x^3}}{9 x^9}-\frac {2 \sqrt {-1+x^3}}{9 x^6}-\frac {2 \sqrt {-1+x^3}}{3 x^3}+\frac {2}{3} \tan ^{-1}\left (\sqrt {-1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 41, normalized size = 1.00 \begin {gather*} \frac {2}{3} \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {2 \sqrt {x^3-1} \left (3 x^6+x^3-1\right )}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-1 + x^3]*(-2 + x^3 + 2*x^6))/x^10,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 41, normalized size = 1.00 \begin {gather*} -\frac {2 \sqrt {-1+x^3} \left (-1+x^3+3 x^6\right )}{9 x^9}+\frac {2}{3} \tan ^{-1}\left (\sqrt {-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-1 + x^3]*(-2 + x^3 + 2*x^6))/x^10,x]

[Out]

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

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fricas [A]  time = 0.47, size = 38, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (3 \, x^{9} \arctan \left (\sqrt {x^{3} - 1}\right ) - {\left (3 \, x^{6} + x^{3} - 1\right )} \sqrt {x^{3} - 1}\right )}}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.64, size = 44, normalized size = 1.07 \begin {gather*} -\frac {2 \, {\left (3 \, {\left (x^{3} - 1\right )}^{\frac {5}{2}} + 7 \, {\left (x^{3} - 1\right )}^{\frac {3}{2}} + 3 \, \sqrt {x^{3} - 1}\right )}}{9 \, x^{9}} + \frac {2}{3} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(3*(x^3 - 1)^(5/2) + 7*(x^3 - 1)^(3/2) + 3*sqrt(x^3 - 1))/x^9 + 2/3*arctan(sqrt(x^3 - 1))

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maple [A]  time = 0.40, size = 41, normalized size = 1.00

method	result	size
risch	\(-\frac {2 \left (3 x^{9}-2 x^{6}-2 x^{3}+1\right )}{9 x^{9} \sqrt {x^{3}-1}}+\frac {2 \arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(41\)
default	\(\frac {2 \sqrt {x^{3}-1}}{9 x^{9}}-\frac {2 \sqrt {x^{3}-1}}{9 x^{6}}-\frac {2 \sqrt {x^{3}-1}}{3 x^{3}}+\frac {2 \arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(48\)
elliptic	\(\frac {2 \sqrt {x^{3}-1}}{9 x^{9}}-\frac {2 \sqrt {x^{3}-1}}{9 x^{6}}-\frac {2 \sqrt {x^{3}-1}}{3 x^{3}}+\frac {2 \arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(48\)
trager	\(-\frac {2 \sqrt {x^{3}-1}\, \left (3 x^{6}+x^{3}-1\right )}{9 x^{9}}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \sqrt {x^{3}-1}}{x^{3}}\right )}{3}\)	\(65\)
meijerg	\(\frac {\sqrt {\mathrm {signum}\left (x^{3}-1\right )}\, \left (-\frac {2 \sqrt {\pi }}{x^{3}}-\left (-2 \ln \relax (2)-1+3 \ln \relax (x )+i \pi \right ) \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 x^{3}+8\right )}{4 x^{3}}-\frac {2 \sqrt {\pi }\, \sqrt {-x^{3}+1}}{x^{3}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )\right )}{3 \sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \sqrt {\pi }}-\frac {\sqrt {\mathrm {signum}\left (x^{3}-1\right )}\, \left (\frac {\sqrt {\pi }}{x^{6}}-\frac {\sqrt {\pi }}{x^{3}}+\frac {\left (\frac {1}{2}-2 \ln \relax (2)+3 \ln \relax (x )+i \pi \right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (x^{6}-8 x^{3}+8\right )}{8 x^{6}}+\frac {\sqrt {\pi }\, \left (-4 x^{3}+8\right ) \sqrt {-x^{3}+1}}{8 x^{6}}-\frac {\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )}{2}\right )}{6 \sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \sqrt {\pi }}-\frac {\sqrt {\mathrm {signum}\left (x^{3}-1\right )}\, \left (-\frac {2 \sqrt {\pi }}{3 x^{9}}+\frac {\sqrt {\pi }}{2 x^{6}}+\frac {\sqrt {\pi }}{4 x^{3}}-\frac {\left (\frac {5}{6}-2 \ln \relax (2)+3 \ln \relax (x )+i \pi \right ) \sqrt {\pi }}{8}+\frac {\sqrt {\pi }\, \left (20 x^{9}-48 x^{6}-96 x^{3}+128\right )}{192 x^{9}}-\frac {\sqrt {\pi }\, \left (-48 x^{6}-32 x^{3}+128\right ) \sqrt {-x^{3}+1}}{192 x^{9}}+\frac {\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )}{4}\right )}{3 \sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \sqrt {\pi }}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.68, size = 113, normalized size = 2.76 \begin {gather*} -\frac {3 \, {\left (x^{3} - 1\right )}^{\frac {5}{2}} + 8 \, {\left (x^{3} - 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {x^{3} - 1}}{36 \, {\left ({\left (x^{3} - 1\right )}^{3} + 3 \, x^{3} + 3 \, {\left (x^{3} - 1\right )}^{2} - 2\right )}} + \frac {{\left (x^{3} - 1\right )}^{\frac {3}{2}} - \sqrt {x^{3} - 1}}{12 \, {\left (2 \, x^{3} + {\left (x^{3} - 1\right )}^{2} - 1\right )}} - \frac {2 \, \sqrt {x^{3} - 1}}{3 \, x^{3}} + \frac {2}{3} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(3*(x^3 - 1)^(5/2) + 8*(x^3 - 1)^(3/2) - 3*sqrt(x^3 - 1))/((x^3 - 1)^3 + 3*x^3 + 3*(x^3 - 1)^2 - 2) + 1/
12*((x^3 - 1)^(3/2) - sqrt(x^3 - 1))/(2*x^3 + (x^3 - 1)^2 - 1) - 2/3*sqrt(x^3 - 1)/x^3 + 2/3*arctan(sqrt(x^3 -
1))

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mupad [B]  time = 0.22, size = 201, normalized size = 4.90 \begin {gather*} \frac {2\,\sqrt {x^3-1}}{9\,x^9}-\frac {2\,\sqrt {x^3-1}}{9\,x^6}-\frac {2\,\sqrt {x^3-1}}{3\,x^3}-\frac {2\,\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 )}{\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(((x^3 - 1)^(1/2)*(x^3 + 2*x^6 - 2))/x^10,x)

[Out]

(2*(x^3 - 1)^(1/2))/(9*x^9) - (2*(x^3 - 1)^(1/2))/(9*x^6) - (2*(x^3 - 1)^(1/2))/(3*x^3) - (2*((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)))/(((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 [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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