3.18.29 \(\int x^4 \sqrt [3]{-x+x^3} \, dx\) [1729]

Optimal. Leaf size=117 \[ \frac {1}{108} \sqrt [3]{-x+x^3} \left (-5 x-3 x^3+18 x^5\right )+\frac {5 \text {ArcTan}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x+x^3}}\right )}{54 \sqrt {3}}+\frac {5}{162} \log \left (-x+\sqrt [3]{-x+x^3}\right )-\frac {5}{324} \log \left (x^2+x \sqrt [3]{-x+x^3}+\left (-x+x^3\right )^{2/3}\right ) \]

[Out]

1/108*(x^3-x)^(1/3)*(18*x^5-3*x^3-5*x)+5/162*arctan(3^(1/2)*x/(x+2*(x^3-x)^(1/3)))*3^(1/2)+5/162*ln(-x+(x^3-x)
^(1/3))-5/324*ln(x^2+x*(x^3-x)^(1/3)+(x^3-x)^(2/3))

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Rubi [A]
time = 0.10, antiderivative size = 159, normalized size of antiderivative = 1.36, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2046, 2049, 2057, 335, 281, 337} \begin {gather*} \frac {5 \left (x^2-1\right )^{2/3} x^{2/3} \text {ArcTan}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1}{\sqrt {3}}\right )}{54 \sqrt {3} \left (x^3-x\right )^{2/3}}-\frac {1}{36} \sqrt [3]{x^3-x} x^3-\frac {5}{108} \sqrt [3]{x^3-x} x+\frac {1}{6} \sqrt [3]{x^3-x} x^5+\frac {5 \left (x^2-1\right )^{2/3} x^{2/3} \log \left (x^{2/3}-\sqrt [3]{x^2-1}\right )}{108 \left (x^3-x\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(-x + x^3)^(1/3),x]

[Out]

(-5*x*(-x + x^3)^(1/3))/108 - (x^3*(-x + x^3)^(1/3))/36 + (x^5*(-x + x^3)^(1/3))/6 + (5*x^(2/3)*(-1 + x^2)^(2/
3)*ArcTan[(1 + (2*x^(2/3))/(-1 + x^2)^(1/3))/Sqrt[3]])/(54*Sqrt[3]*(-x + x^3)^(2/3)) + (5*x^(2/3)*(-1 + x^2)^(
2/3)*Log[x^(2/3) - (-1 + x^2)^(1/3)])/(108*(-x + x^3)^(2/3))

Rule 281

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 335

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 337

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

Rule 2046

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 + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 x^4 \sqrt [3]{-x+x^3} \, dx &=\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {1}{9} \int \frac {x^5}{\left (-x+x^3\right )^{2/3}} \, dx\\ &=-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {5}{54} \int \frac {x^3}{\left (-x+x^3\right )^{2/3}} \, dx\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {5}{81} \int \frac {x}{\left (-x+x^3\right )^{2/3}} \, dx\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (-1+x^2\right )^{2/3}} \, dx}{81 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {x^3}{\left (-1+x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{27 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {x}{\left (-1+x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{54 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{54 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{162 \left (-x+x^3\right )^{2/3}}+\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{162 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}+\frac {5 x^{2/3} \left (-1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{162 \left (-x+x^3\right )^{2/3}}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{324 \left (-x+x^3\right )^{2/3}}+\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{108 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}+\frac {5 x^{2/3} \left (-1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{162 \left (-x+x^3\right )^{2/3}}-\frac {5 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 )}{324 \left (-x+x^3\right )^{2/3}}-\frac {\left (5 x^{2/3} \left (-1+x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{54 \left (-x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{-x+x^3}-\frac {1}{36} x^3 \sqrt [3]{-x+x^3}+\frac {1}{6} x^5 \sqrt [3]{-x+x^3}+\frac {5 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 )}{54 \sqrt {3} \left (-x+x^3\right )^{2/3}}+\frac {5 x^{2/3} \left (-1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{162 \left (-x+x^3\right )^{2/3}}-\frac {5 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 )}{324 \left (-x+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 1.13, size = 169, normalized size = 1.44 \begin {gather*} \frac {15 x^2-6 x^4-63 x^6+54 x^8+10 \sqrt {3} x^{2/3} \left (-1+x^2\right )^{2/3} \text {ArcTan}\left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{-1+x^2}}\right )+10 x^{2/3} \left (-1+x^2\right )^{2/3} \log \left (-x^{2/3}+\sqrt [3]{-1+x^2}\right )-5 x^{2/3} \left (-1+x^2\right )^{2/3} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right )}{324 \left (x \left (-1+x^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(-x + x^3)^(1/3),x]

[Out]

(15*x^2 - 6*x^4 - 63*x^6 + 54*x^8 + 10*Sqrt[3]*x^(2/3)*(-1 + x^2)^(2/3)*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*
(-1 + x^2)^(1/3))] + 10*x^(2/3)*(-1 + x^2)^(2/3)*Log[-x^(2/3) + (-1 + x^2)^(1/3)] - 5*x^(2/3)*(-1 + x^2)^(2/3)
*Log[x^(4/3) + x^(2/3)*(-1 + x^2)^(1/3) + (-1 + x^2)^(2/3)])/(324*(x*(-1 + x^2))^(2/3))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 3.01, size = 33, normalized size = 0.28

method	result	size
meijerg	\(\frac {3 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {16}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {8}{3}\right ], \left [\frac {11}{3}\right ], x^{2}\right )}{16 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{3}}}\)	\(33\)
trager	\(\frac {x \left (18 x^{4}-3 x^{2}-5\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{108}-\frac {5 \ln \left (1395 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}-6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {2}{3}}-6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}} x -6303 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-5580 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+5355 \left (x^{3}-x \right )^{\frac {2}{3}}+5355 x \left (x^{3}-x \right )^{\frac {1}{3}}+5510 x^{2}+8007 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-1653\right )}{162}-\frac {5 \ln \left (1395 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}-6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {2}{3}}-6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}} x -6303 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-5580 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+5355 \left (x^{3}-x \right )^{\frac {2}{3}}+5355 x \left (x^{3}-x \right )^{\frac {1}{3}}+5510 x^{2}+8007 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-1653\right ) \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{54}+\frac {5 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (1395 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {2}{3}}+6768 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}} x +7233 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-5580 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+7611 \left (x^{3}-x \right )^{\frac {2}{3}}+7611 x \left (x^{3}-x \right )^{\frac {1}{3}}+7766 x^{2}-11727 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-4942\right )}{54}\)	\(468\)
risch	\(\text {Expression too large to display}\)	\(796\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x^3-x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/16*signum(x^2-1)^(1/3)/(-signum(x^2-1))^(1/3)*x^(16/3)*hypergeom([-1/3,8/3],[11/3],x^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^3-x)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^3 - x)^(1/3)*x^4, x)

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Fricas [A]
time = 0.50, size = 112, normalized size = 0.96 \begin {gather*} \frac {5}{162} \, \sqrt {3} \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 ) + \frac {1}{108} \, {\left (18 \, x^{5} - 3 \, x^{3} - 5 \, x\right )} {\left (x^{3} - x\right )}^{\frac {1}{3}} + \frac {5}{324} \, \log \left (-3 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} x + 3 \, {\left (x^{3} - x\right )}^{\frac {2}{3}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^3-x)^(1/3),x, algorithm="fricas")

[Out]

5/162*sqrt(3)*arctan(-(44032959556*sqrt(3)*(x^3 - x)^(1/3)*x + sqrt(3)*(16754327161*x^2 - 2707204793) - 105243
05234*sqrt(3)*(x^3 - x)^(2/3))/(81835897185*x^2 - 1102302937)) + 1/108*(18*x^5 - 3*x^3 - 5*x)*(x^3 - x)^(1/3)
+ 5/324*log(-3*(x^3 - x)^(1/3)*x + 3*(x^3 - x)^(2/3) + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \sqrt [3]{x \left (x - 1\right ) \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(x**3-x)**(1/3),x)

[Out]

Integral(x**4*(x*(x - 1)*(x + 1))**(1/3), x)

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Giac [A]
time = 0.43, size = 109, normalized size = 0.93 \begin {gather*} -\frac {1}{108} \, {\left (5 \, {\left (\frac {1}{x^{2}} - 1\right )}^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 13 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} - 10 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )} x^{6} - \frac {5}{162} \, \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 {5}{324} \, \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 {5}{162} \, \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^4*(x^3-x)^(1/3),x, algorithm="giac")

[Out]

-1/108*(5*(1/x^2 - 1)^2*(-1/x^2 + 1)^(1/3) - 13*(-1/x^2 + 1)^(4/3) - 10*(-1/x^2 + 1)^(1/3))*x^6 - 5/162*sqrt(3
)*arctan(1/3*sqrt(3)*(2*(-1/x^2 + 1)^(1/3) + 1)) - 5/324*log((-1/x^2 + 1)^(2/3) + (-1/x^2 + 1)^(1/3) + 1) + 5/
162*log(abs((-1/x^2 + 1)^(1/3) - 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\left (x^3-x\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x^3 - x)^(1/3),x)

[Out]

int(x^4*(x^3 - x)^(1/3), x)

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