3.14.27 \(\int \frac {\sqrt [3]{-x+x^3}}{x^2} \, dx\)

Optimal. Leaf size=106 \[ -\frac {3 \sqrt [3]{x^3-x}}{2 x}-\frac {1}{2} \log \left (\sqrt [3]{x^3-x}-x\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x}+x}\right )+\frac {1}{4} \log \left (\sqrt [3]{x^3-x} x+\left (x^3-x\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 188, normalized size of antiderivative = 1.77, number of steps used = 11, number of rules used = 11, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {2020, 2032, 329, 275, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} -\frac {3 \sqrt [3]{x^3-x}}{2 x}-\frac {x^{2/3} \left (x^2-1\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2-1}}\right )}{2 \left (x^3-x\right )^{2/3}}+\frac {x^{2/3} \left (x^2-1\right )^{2/3} \log \left (\frac {x^{4/3}}{\left (x^2-1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2-1}}+1\right )}{4 \left (x^3-x\right )^{2/3}}-\frac {\sqrt {3} x^{2/3} \left (x^2-1\right )^{2/3} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1}{\sqrt {3}}\right )}{2 \left (x^3-x\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + x^3)^(1/3)/x^2,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/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]

Rule 2020

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

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.40 \begin {gather*} -\frac {3 \sqrt [3]{x \left (x^2-1\right )} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};x^2\right )}{2 x \sqrt [3]{1-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^3)^(1/3)/x^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 106, normalized size = 1.00 \begin {gather*} -\frac {3 \sqrt [3]{x^3-x}}{2 x}-\frac {1}{2} \log \left (\sqrt [3]{x^3-x}-x\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x}+x}\right )+\frac {1}{4} \log \left (\sqrt [3]{x^3-x} x+\left (x^3-x\right )^{2/3}+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 104, normalized size = 0.98 \begin {gather*} -\frac {2 \, \sqrt {3} x \arctan \left (-\frac {44032959556 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {1}{3}} x + \sqrt {3} {\left (16754327161 \, x^{2} - 2707204793\right )} - 10524305234 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {2}{3}}}{81835897185 \, x^{2} - 1102302937}\right ) + x \log \left (-3 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} x + 3 \, {\left (x^{3} - x\right )}^{\frac {2}{3}} + 1\right ) + 6 \, {\left (x^{3} - x\right )}^{\frac {1}{3}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*x*arctan(-(44032959556*sqrt(3)*(x^3 - x)^(1/3)*x + sqrt(3)*(16754327161*x^2 - 2707204793) - 10
524305234*sqrt(3)*(x^3 - x)^(2/3))/(81835897185*x^2 - 1102302937)) + x*log(-3*(x^3 - x)^(1/3)*x + 3*(x^3 - x)^
(2/3) + 1) + 6*(x^3 - x)^(1/3))/x

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giac [A]  time = 0.19, size = 74, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {3}{2} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + \frac {1}{4} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-1/x^2 + 1)^(1/3) + 1)) - 3/2*(-1/x^2 + 1)^(1/3) + 1/4*log((-1/x^2 + 1)^(2/
3) + (-1/x^2 + 1)^(1/3) + 1) - 1/2*log(abs((-1/x^2 + 1)^(1/3) - 1))

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maple [C]  time = 1.93, size = 787, normalized size = 7.42

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*(x*(x^2-1))^(1/3)/x+(1/12*RootOf(_Z^2-6*_Z+36)*ln((-47*RootOf(_Z^2-6*_Z+36)^2*x^4+3207*RootOf(_Z^2-6*_Z+3
6)*x^4+2925*(x^6-2*x^4+x^2)^(1/3)*RootOf(_Z^2-6*_Z+36)*x^2+235*RootOf(_Z^2-6*_Z+36)^2*x^2-6930*x^4+2925*RootOf
(_Z^2-6*_Z+36)*(x^6-2*x^4+x^2)^(2/3)-5238*(x^6-2*x^4+x^2)^(1/3)*x^2-5601*RootOf(_Z^2-6*_Z+36)*x^2-5238*(x^6-2*
x^4+x^2)^(2/3)-2925*RootOf(_Z^2-6*_Z+36)*(x^6-2*x^4+x^2)^(1/3)-188*RootOf(_Z^2-6*_Z+36)^2+11340*x^2+5238*(x^6-
2*x^4+x^2)^(1/3)+2394*RootOf(_Z^2-6*_Z+36)-4410)/(-1+x)/(1+x))-1/12*ln(-(47*RootOf(_Z^2-6*_Z+36)^2*x^4+2643*Ro
otOf(_Z^2-6*_Z+36)*x^4+2925*(x^6-2*x^4+x^2)^(1/3)*RootOf(_Z^2-6*_Z+36)*x^2-235*RootOf(_Z^2-6*_Z+36)^2*x^2-1062
0*x^4+2925*RootOf(_Z^2-6*_Z+36)*(x^6-2*x^4+x^2)^(2/3)-12312*(x^6-2*x^4+x^2)^(1/3)*x^2-2781*RootOf(_Z^2-6*_Z+36
)*x^2-12312*(x^6-2*x^4+x^2)^(2/3)-2925*RootOf(_Z^2-6*_Z+36)*(x^6-2*x^4+x^2)^(1/3)+188*RootOf(_Z^2-6*_Z+36)^2+1
3806*x^2+12312*(x^6-2*x^4+x^2)^(1/3)+138*RootOf(_Z^2-6*_Z+36)-3186)/(-1+x)/(1+x))*RootOf(_Z^2-6*_Z+36)+1/2*ln(
-(47*RootOf(_Z^2-6*_Z+36)^2*x^4+2643*RootOf(_Z^2-6*_Z+36)*x^4+2925*(x^6-2*x^4+x^2)^(1/3)*RootOf(_Z^2-6*_Z+36)*
x^2-235*RootOf(_Z^2-6*_Z+36)^2*x^2-10620*x^4+2925*RootOf(_Z^2-6*_Z+36)*(x^6-2*x^4+x^2)^(2/3)-12312*(x^6-2*x^4+
x^2)^(1/3)*x^2-2781*RootOf(_Z^2-6*_Z+36)*x^2-12312*(x^6-2*x^4+x^2)^(2/3)-2925*RootOf(_Z^2-6*_Z+36)*(x^6-2*x^4+
x^2)^(1/3)+188*RootOf(_Z^2-6*_Z+36)^2+13806*x^2+12312*(x^6-2*x^4+x^2)^(1/3)+138*RootOf(_Z^2-6*_Z+36)-3186)/(-1
+x)/(1+x)))*(x*(x^2-1))^(1/3)/x*(x^2*(x^2-1)^2)^(1/3)/(x^2-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - x)^(1/3)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - x)^(1/3)/x^2,x)

[Out]

int((x^3 - x)^(1/3)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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