3.19.79 \(\int \frac {(-2+x^3) (-1+x^3)^{2/3}}{x^3 (-1+2 x^3)} \, dx\)

Optimal. Leaf size=174 \[ -\frac {1}{6} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}+x\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}-x}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{2 \sqrt {3}}-\frac {\left (x^3-1\right )^{2/3}}{x^2}+\frac {1}{4} \log \left (-\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right )+\frac {1}{12} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 143, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {580, 530, 239, 377, 200, 31, 634, 618, 204, 628} \begin {gather*} -\frac {1}{2} \log \left (\frac {x}{\sqrt [3]{x^3-1}}+1\right )-\frac {1}{4} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{x^3-1}}}{\sqrt {3}}\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\left (x^3-1\right )^{2/3}}{x^2}+\frac {1}{4} \log \left (-\frac {x}{\sqrt [3]{x^3-1}}+\frac {x^2}{\left (x^3-1\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 204

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

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 530

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [C]  time = 0.24, size = 146, normalized size = 0.84 \begin {gather*} \frac {1}{6} \left (-2 \log \left (\frac {x}{\sqrt [3]{1-x^3}}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{1-x^3}}-1}{\sqrt {3}}\right )-\frac {6 \left (x^3-1\right )^{2/3}}{x^2}+\log \left (-\frac {x}{\sqrt [3]{1-x^3}}+\frac {x^2}{\left (1-x^3\right )^{2/3}}+1\right )\right )-\frac {x^4 \sqrt [3]{1-x^3} F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};x^3,2 x^3\right )}{4 \sqrt [3]{x^3-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*(x^4*(1 - x^3)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, x^3, 2*x^3])/(-1 + x^3)^(1/3) + ((-6*(-1 + x^3)^(2/3))/x^
2 - 2*Sqrt[3]*ArcTan[(-1 + (2*x)/(1 - x^3)^(1/3))/Sqrt[3]] + Log[1 + x^2/(1 - x^3)^(2/3) - x/(1 - x^3)^(1/3)]
- 2*Log[1 + x/(1 - x^3)^(1/3)])/6

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IntegrateAlgebraic [A]  time = 0.27, size = 174, normalized size = 1.00 \begin {gather*} -\frac {1}{6} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}+x\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}-x}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{2 \sqrt {3}}-\frac {\left (x^3-1\right )^{2/3}}{x^2}+\frac {1}{4} \log \left (-\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right )+\frac {1}{12} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 6.68, size = 214, normalized size = 1.23 \begin {gather*} \frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {383838 \, \sqrt {3} {\left (x^{10} - 3 \, x^{4} - 2 \, x\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} + 13468 \, \sqrt {3} {\left (x^{11} - 3 \, x^{8} + 4 \, x^{2}\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (198653 \, x^{12} + 393594 \, x^{9} + 5568 \, x^{6} - 400090 \, x^{3} - 198189\right )}}{3 \, {\left (185185 \, x^{12} + 370434 \, x^{9} - 96 \, x^{6} - 370322 \, x^{3} - 185193\right )}}\right ) - x^{2} \log \left (\frac {8 \, x^{9} - 12 \, x^{6} + 6 \, x^{3} - 3 \, {\left (x^{10} - 3 \, x^{4} - 2 \, x\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} + 3 \, {\left (x^{11} - 3 \, x^{8} + 4 \, x^{2}\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1}{8 \, x^{9} - 12 \, x^{6} + 6 \, x^{3} - 1}\right ) - 12 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{12 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*sqrt(3)*x^2*arctan(1/3*(383838*sqrt(3)*(x^10 - 3*x^4 - 2*x)*(x^3 - 1)^(2/3) + 13468*sqrt(3)*(x^11 - 3*
x^8 + 4*x^2)*(x^3 - 1)^(1/3) + sqrt(3)*(198653*x^12 + 393594*x^9 + 5568*x^6 - 400090*x^3 - 198189))/(185185*x^
12 + 370434*x^9 - 96*x^6 - 370322*x^3 - 185193)) - x^2*log((8*x^9 - 12*x^6 + 6*x^3 - 3*(x^10 - 3*x^4 - 2*x)*(x
^3 - 1)^(2/3) + 3*(x^11 - 3*x^8 + 4*x^2)*(x^3 - 1)^(1/3) - 1)/(8*x^9 - 12*x^6 + 6*x^3 - 1)) - 12*(x^3 - 1)^(2/
3))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x^{3} - 2\right )}}{{\left (2 \, x^{3} - 1\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - 1)^(2/3)*(x^3 - 2)/((2*x^3 - 1)*x^3), x)

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maple [C]  time = 8.05, size = 801, normalized size = 4.60

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^3-1)^(2/3)/x^2-1/6*ln((-11+60*x*(x^3-1)^(2/3)+108*x^2*(x^3-1)^(1/3)+22*x^12-44*x^9+27*x^11*(x^3-1)^(1/3)-8
1*x^8*(x^3-1)^(1/3)+132*x^6-110*x^3+76*RootOf(4*_Z^2-2*_Z+1)^2*x^12-456*RootOf(4*_Z^2-2*_Z+1)^2*x^9+912*RootOf
(4*_Z^2-2*_Z+1)^2*x^6-608*RootOf(4*_Z^2-2*_Z+1)^2*x^3+38*RootOf(4*_Z^2-2*_Z+1)-720*RootOf(4*_Z^2-2*_Z+1)*x^6+5
56*RootOf(4*_Z^2-2*_Z+1)*x^3-98*RootOf(4*_Z^2-2*_Z+1)*x^12+284*x^9*RootOf(4*_Z^2-2*_Z+1)+54*RootOf(4*_Z^2-2*_Z
+1)*(x^3-1)^(2/3)*x^10+6*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(1/3)*x^11-18*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(1/3)*x^8-1
62*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(2/3)*x^4-108*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(2/3)*x+24*RootOf(4*_Z^2-2*_Z+1)*
(x^3-1)^(1/3)*x^2-30*(x^3-1)^(2/3)*x^10+90*(x^3-1)^(2/3)*x^4)/(2*x^3-1)^3)+1/3*RootOf(4*_Z^2-2*_Z+1)*ln(-(19-6
*x*(x^3-1)^(2/3)+108*x^2*(x^3-1)^(1/3)-19*x^12-38*x^9+27*x^11*(x^3-1)^(1/3)-81*x^8*(x^3-1)^(1/3)+38*x^3+44*Roo
tOf(4*_Z^2-2*_Z+1)^2*x^12-264*RootOf(4*_Z^2-2*_Z+1)^2*x^9+528*RootOf(4*_Z^2-2*_Z+1)^2*x^6-352*RootOf(4*_Z^2-2*
_Z+1)^2*x^3-22*RootOf(4*_Z^2-2*_Z+1)-456*RootOf(4*_Z^2-2*_Z+1)*x^6+260*RootOf(4*_Z^2-2*_Z+1)*x^3-16*RootOf(4*_
Z^2-2*_Z+1)*x^12+272*x^9*RootOf(4*_Z^2-2*_Z+1)+54*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(2/3)*x^10-60*RootOf(4*_Z^2-2*
_Z+1)*(x^3-1)^(1/3)*x^11+180*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(1/3)*x^8-162*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(2/3)*x
^4-108*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(2/3)*x-240*RootOf(4*_Z^2-2*_Z+1)*(x^3-1)^(1/3)*x^2+3*(x^3-1)^(2/3)*x^10-
9*(x^3-1)^(2/3)*x^4)/(2*x^3-1)^3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x^{3} - 2\right )}}{{\left (2 \, x^{3} - 1\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - 1)^(2/3)*(x^3 - 2)/((2*x^3 - 1)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3-1\right )}^{2/3}\,\left (x^3-2\right )}{x^3\,\left (2\,x^3-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} - 2\right )}{x^{3} \left (2 x^{3} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((x - 1)*(x**2 + x + 1))**(2/3)*(x**3 - 2)/(x**3*(2*x**3 - 1)), x)

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