3.4.6 \(\int \sqrt [3]{x (1-x^2)} \, dx\) [306]

3.4.6.1 Optimal result
3.4.6.2 Mathematica [A] (verified)
3.4.6.3 Rubi [A] (warning: unable to verify)
3.4.6.4 Maple [C] (verified)
3.4.6.5 Fricas [A] (verification not implemented)
3.4.6.6 Sympy [F]
3.4.6.7 Maxima [F]
3.4.6.8 Giac [A] (verification not implemented)
3.4.6.9 Mupad [B] (verification not implemented)
3.4.6.10 Reduce [F]

3.4.6.1 Optimal result

Integrand size = 13, antiderivative size = 93 \[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\frac {1}{2} x \sqrt [3]{x \left (1-x^2\right )}+\frac {\arctan \left (\frac {2 x-\sqrt [3]{x \left (1-x^2\right )}}{\sqrt {3} \sqrt [3]{x \left (1-x^2\right )}}\right )}{2 \sqrt {3}}+\frac {\log (x)}{12}-\frac {1}{4} \log \left (x+\sqrt [3]{x \left (1-x^2\right )}\right ) \]

output
1/2*x*(x*(-x^2+1))^(1/3)+1/12*ln(x)-1/4*ln(x+(x*(-x^2+1))^(1/3))+1/6*arcta 
n(1/3*(2*x-(x*(-x^2+1))^(1/3))/(x*(-x^2+1))^(1/3)*3^(1/2))*3^(1/2)
 
3.4.6.2 Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.47 \[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\frac {\sqrt [3]{x-x^3} \left (6 x^{4/3} \sqrt [3]{-1+x^2}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{-1+x^2}}\right )+2 \log \left (-x^{2/3}+\sqrt [3]{-1+x^2}\right )-\log \left (x^{4/3}+x^{2/3} \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right )\right )}{12 \sqrt [3]{x} \sqrt [3]{-1+x^2}} \]

input
Integrate[(x*(1 - x^2))^(1/3),x]
 
output
((x - x^3)^(1/3)*(6*x^(4/3)*(-1 + x^2)^(1/3) + 2*Sqrt[3]*ArcTan[(Sqrt[3]*x 
^(2/3))/(x^(2/3) + 2*(-1 + x^2)^(1/3))] + 2*Log[-x^(2/3) + (-1 + x^2)^(1/3 
)] - Log[x^(4/3) + x^(2/3)*(-1 + x^2)^(1/3) + (-1 + x^2)^(2/3)]))/(12*x^(1 
/3)*(-1 + x^2)^(1/3))
 
3.4.6.3 Rubi [A] (warning: unable to verify)

Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2078, 1910, 1938, 266, 807, 853}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt [3]{x \left (1-x^2\right )} \, dx\)

\(\Big \downarrow \) 2078

\(\displaystyle \int \sqrt [3]{x-x^3}dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{3} \int \frac {x}{\left (x-x^3\right )^{2/3}}dx+\frac {1}{2} \sqrt [3]{x-x^3} x\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {\left (1-x^2\right )^{2/3} x^{2/3} \int \frac {\sqrt [3]{x}}{\left (1-x^2\right )^{2/3}}dx}{3 \left (x-x^3\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x-x^3} x\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\left (1-x^2\right )^{2/3} x^{2/3} \int \frac {x}{\left (1-x^2\right )^{2/3}}d\sqrt [3]{x}}{\left (x-x^3\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x-x^3} x\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\left (1-x^2\right )^{2/3} x^{2/3} \int \frac {x^{2/3}}{(1-x)^{2/3}}dx^{2/3}}{2 \left (x-x^3\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x-x^3} x\)

\(\Big \downarrow \) 853

\(\displaystyle \frac {\left (1-x^2\right )^{2/3} x^{2/3} \left (-\frac {\arctan \left (\frac {1-\frac {2 x^{2/3}}{\sqrt [3]{1-x}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (-x^{2/3}-\sqrt [3]{1-x}\right )\right )}{2 \left (x-x^3\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x-x^3} x\)

input
Int[(x*(1 - x^2))^(1/3),x]
 
output
(x*(x - x^3)^(1/3))/2 + (x^(2/3)*(1 - x^2)^(2/3)*(-(ArcTan[(1 - (2*x^(2/3) 
)/(1 - x)^(1/3))/Sqrt[3]]/Sqrt[3]) - Log[-(1 - x)^(1/3) - x^(2/3)]/2))/(2* 
(x - x^3)^(2/3))
 

3.4.6.3.1 Defintions of rubi rules used

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 807
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[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 853
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-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 1910
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1))   Int[x^j*(a*x^j 
+ b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, 
n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
 

rule 1938
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[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] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 2078
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && G 
eneralizedBinomialQ[u, x] &&  !GeneralizedBinomialMatchQ[u, x]
 
3.4.6.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 2.62 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.16

method result size
meijerg \(\frac {3 x^{\frac {4}{3}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};x^{2}\right )}{4}\) \(15\)
pseudoelliptic \(\frac {x \left (2 \sqrt {3}\, \arctan \left (\frac {\left (-2 \left (-x^{3}+x \right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+6 \left (-x^{3}+x \right )^{\frac {1}{3}} x -2 \ln \left (\frac {\left (-x^{3}+x \right )^{\frac {1}{3}}+x}{x}\right )+\ln \left (\frac {\left (-x^{3}+x \right )^{\frac {2}{3}}-\left (-x^{3}+x \right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )\right )}{12 \left (\left (-x^{3}+x \right )^{\frac {1}{3}}+x \right ) \left (\left (-x^{3}+x \right )^{\frac {2}{3}}-\left (-x^{3}+x \right )^{\frac {1}{3}} x +x^{2}\right )}\) \(132\)
trager \(\frac {\left (-x^{3}+x \right )^{\frac {1}{3}} x}{2}-\frac {\ln \left (4959 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+6768 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {2}{3}}+22833 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {1}{3}} x +14412 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+5355 \left (-x^{3}+x \right )^{\frac {2}{3}}-2256 \left (-x^{3}+x \right )^{\frac {1}{3}} x -19836 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-7060 x^{2}+3513 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+2118\right )}{6}+\frac {\ln \left (-6354 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-6768 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {2}{3}}+16065 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {1}{3}} x +24951 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+7611 \left (-x^{3}+x \right )^{\frac {2}{3}}+2256 \left (-x^{3}+x \right )^{\frac {1}{3}} x +25416 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-6061 x^{2}-21438 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+3857\right )}{6}-\frac {\ln \left (-6354 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-6768 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {2}{3}}+16065 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-x^{3}+x \right )^{\frac {1}{3}} x +24951 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+7611 \left (-x^{3}+x \right )^{\frac {2}{3}}+2256 \left (-x^{3}+x \right )^{\frac {1}{3}} x +25416 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-6061 x^{2}-21438 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+3857\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )}{2}\) \(445\)
risch \(\text {Expression too large to display}\) \(788\)

input
int((x*(-x^2+1))^(1/3),x,method=_RETURNVERBOSE)
 
output
3/4*x^(4/3)*hypergeom([-1/3,2/3],[5/3],x^2)
 
3.4.6.5 Fricas [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {44032959556 \, \sqrt {3} {\left (-x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (16754327161 \, x^{2} - 2707204793\right )} + 10524305234 \, \sqrt {3} {\left (-x^{3} + x\right )}^{\frac {2}{3}}}{81835897185 \, x^{2} - 1102302937}\right ) + \frac {1}{2} \, {\left (-x^{3} + x\right )}^{\frac {1}{3}} x - \frac {1}{12} \, \log \left (3 \, {\left (-x^{3} + x\right )}^{\frac {1}{3}} x + 3 \, {\left (-x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \]

input
integrate((x*(-x^2+1))^(1/3),x, algorithm="fricas")
 
output
-1/6*sqrt(3)*arctan((44032959556*sqrt(3)*(-x^3 + x)^(1/3)*x - sqrt(3)*(167 
54327161*x^2 - 2707204793) + 10524305234*sqrt(3)*(-x^3 + x)^(2/3))/(818358 
97185*x^2 - 1102302937)) + 1/2*(-x^3 + x)^(1/3)*x - 1/12*log(3*(-x^3 + x)^ 
(1/3)*x + 3*(-x^3 + x)^(2/3) + 1)
 
3.4.6.6 Sympy [F]

\[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\int \sqrt [3]{x \left (1 - x^{2}\right )}\, dx \]

input
integrate((x*(-x**2+1))**(1/3),x)
 
output
Integral((x*(1 - x**2))**(1/3), x)
 
3.4.6.7 Maxima [F]

\[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\int { \left (-{\left (x^{2} - 1\right )} x\right )^{\frac {1}{3}} \,d x } \]

input
integrate((x*(-x^2+1))^(1/3),x, algorithm="maxima")
 
output
integrate((-(x^2 - 1)*x)^(1/3), x)
 
3.4.6.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\frac {1}{2} \, x^{2} {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) + \frac {1}{12} \, \log \left ({\left (\frac {1}{x^{2}} - 1\right )}^{\frac {2}{3}} - {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{6} \, \log \left ({\left | {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) \]

input
integrate((x*(-x^2+1))^(1/3),x, algorithm="giac")
 
output
1/2*x^2*(1/x^2 - 1)^(1/3) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 - 1)^ 
(1/3) - 1)) + 1/12*log((1/x^2 - 1)^(2/3) - (1/x^2 - 1)^(1/3) + 1) - 1/6*lo 
g(abs((1/x^2 - 1)^(1/3) + 1))
 
3.4.6.9 Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.31 \[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\frac {3\,x\,{\left (x-x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ x^2\right )}{4\,{\left (1-x^2\right )}^{1/3}} \]

input
int((-x*(x^2 - 1))^(1/3),x)
 
output
(3*x*(x - x^3)^(1/3)*hypergeom([-1/3, 2/3], 5/3, x^2))/(4*(1 - x^2)^(1/3))
 
3.4.6.10 Reduce [F]

\[ \int \sqrt [3]{x \left (1-x^2\right )} \, dx=\frac {x^{\frac {4}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}}{2}-\frac {\left (\int \frac {x^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}}{x^{2}-1}d x \right )}{3} \]

input
int(( - x**3 + x)**(1/3),x)
 
output
(3*x**(1/3)*( - x**2 + 1)**(1/3)*x - 2*int((x**(1/3)*( - x**2 + 1)**(1/3)) 
/(x**2 - 1),x))/6