\(\int \frac {\sqrt [3]{-1+x^3} (-1+2 x^3)}{x^5} \, dx\) [1415]

Optimal result

Integrand size = 20, antiderivative size = 101 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\frac {\left (1-9 x^3\right ) \sqrt [3]{-1+x^3}}{4 x^4}-\frac {2 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}-\frac {2}{3} \log \left (-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{3} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \] Output:

1/4*(-9*x^3+1)*(x^3-1)^(1/3)/x^4-2/3*arctan(3^(1/2)*x/(x+2*(x^3-1)^(1/3))) 
*3^(1/2)-2/3*ln(-x+(x^3-1)^(1/3))+1/3*ln(x^2+x*(x^3-1)^(1/3)+(x^3-1)^(2/3) 
)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\frac {\left (1-9 x^3\right ) \sqrt [3]{-1+x^3}}{4 x^4}-\frac {2 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}-\frac {2}{3} \log \left (-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{3} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \] Input:

Integrate[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^5,x]
 

Output:

((1 - 9*x^3)*(-1 + x^3)^(1/3))/(4*x^4) - (2*ArcTan[(Sqrt[3]*x)/(x + 2*(-1 
+ x^3)^(1/3))])/Sqrt[3] - (2*Log[-x + (-1 + x^3)^(1/3)])/3 + Log[x^2 + x*( 
-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]/3
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {953, 809, 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.

\(\displaystyle \int \frac {\sqrt [3]{x^3-1} \left (2 x^3-1\right )}{x^5} \, dx\)

\(\Big \downarrow \) 953

\(\displaystyle 2 \int \frac {\sqrt [3]{x^3-1}}{x^2}dx-\frac {\left (x^3-1\right )^{4/3}}{4 x^4}\)

\(\Big \downarrow \) 809

\(\displaystyle 2 \left (\int \frac {x}{\left (x^3-1\right )^{2/3}}dx-\frac {\sqrt [3]{x^3-1}}{x}\right )-\frac {\left (x^3-1\right )^{4/3}}{4 x^4}\)

\(\Big \downarrow \) 853

\(\displaystyle 2 \left (-\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\sqrt [3]{x^3-1}}{x}-\frac {1}{2} \log \left (x-\sqrt [3]{x^3-1}\right )\right )-\frac {\left (x^3-1\right )^{4/3}}{4 x^4}\)

Input:

Int[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^5,x]
 

Output:

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

Defintions of rubi rules used

rule 809

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

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 953

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[d/e^n   Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, 
c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && 
(IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G 
tQ[m + n, -1]))
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.84 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.56

method	result	size
risch	\(-\frac {9 x^{6}-10 x^{3}+1}{4 x^{4} \left (x^{3}-1\right )^{\frac {2}{3}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\)	\(57\)
meijerg	\(-\frac {2 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}{{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x}+\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {4}{3}}}{4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x^{4}}\)	\(66\)
pseudoelliptic	\(\frac {8 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}-1\right )^{\frac {1}{3}}\right )}{3 x}\right ) x^{4}+4 \ln \left (\frac {x^{2}+x \left (x^{3}-1\right )^{\frac {1}{3}}+\left (x^{3}-1\right )^{\frac {2}{3}}}{x^{2}}\right ) x^{4}-8 \ln \left (\frac {-x +\left (x^{3}-1\right )^{\frac {1}{3}}}{x}\right ) x^{4}-27 x^{3} \left (x^{3}-1\right )^{\frac {1}{3}}+3 \left (x^{3}-1\right )^{\frac {1}{3}}}{12 x^{4}}\)	\(107\)
trager	\(-\frac {\left (9 x^{3}-1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{4 x^{4}}-\frac {2 \ln \left (-17421502720 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x^{3}-3103405824 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -21033432144 \left (x^{3}-1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{2}[phone] \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{3}-1314589509 x \left (x^{3}-1\right )^{\frac {2}{3}}[phone] x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-262015609 x^{3}[phone] \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}-17790832624 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )[phone]\right )}{3}+\frac {32 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \ln \left (14505155072 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x^{3}[phone] \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -24136837968 \left (x^{3}-1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{2}[phone] \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{3}-1508552373 x \left (x^{3}-1\right )^{\frac {2}{3}}[phone] x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}[phone] x^{3}-116041240576 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}-7804179168 \operatorname {RootOf}\left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-79743881\right )}{3}\)	\(308\)

Input:

int((x^3-1)^(1/3)*(2*x^3-1)/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(9*x^6-10*x^3+1)/x^4/(x^3-1)^(2/3)+1/signum(x^3-1)^(2/3)*(-signum(x^3 
-1))^(2/3)*x^2*hypergeom([2/3,2/3],[5/3],x^3)
 

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=-\frac {8 \, \sqrt {3} x^{4} \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} - 7200\right )}}{58653 \, x^{3} - 8000}\right ) + 4 \, x^{4} \log \left (-3 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + 1\right ) + 3 \, {\left (9 \, x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{12 \, x^{4}} \] Input:

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="fricas")
 

Output:

-1/12*(8*sqrt(3)*x^4*arctan(-(25382*sqrt(3)*(x^3 - 1)^(1/3)*x^2 - 13720*sq 
rt(3)*(x^3 - 1)^(2/3)*x + sqrt(3)*(5831*x^3 - 7200))/(58653*x^3 - 8000)) + 
 4*x^4*log(-3*(x^3 - 1)^(1/3)*x^2 + 3*(x^3 - 1)^(2/3)*x + 1) + 3*(9*x^3 - 
1)*(x^3 - 1)^(1/3))/x^4
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.38 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=- \begin {cases} \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} - \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} + \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {otherwise} \end {cases} + \frac {2 e^{\frac {i \pi }{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \] Input:

integrate((x**3-1)**(1/3)*(2*x**3-1)/x**5,x)
 

Output:

-Piecewise(((-1 + x**(-3))**(1/3)*exp(-2*I*pi/3)*gamma(-4/3)/(3*gamma(-1/3 
)) - (-1 + x**(-3))**(1/3)*exp(-2*I*pi/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)) 
, 1/Abs(x**3) > 1), (-(1 - 1/x**3)**(1/3)*gamma(-4/3)/(3*gamma(-1/3)) + (1 
 - 1/x**3)**(1/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)), True)) + 2*exp(I*pi/3) 
*gamma(-1/3)*hyper((-1/3, -1/3), (2/3,), x**3)/(3*x*gamma(2/3))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - \frac {{\left (x^{3} - 1\right )}^{\frac {4}{3}}}{4 \, x^{4}} + \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {2}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \] Input:

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\int { \frac {{\left (2 \, x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x^{5}} \,d x } \] Input:

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="giac")
 

Output:

integrate((2*x^3 - 1)*(x^3 - 1)^(1/3)/x^5, x)
 

Mupad [B] (verification not implemented)

Time = 8.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\frac {{\left (x^3-1\right )}^{1/3}-x^3\,{\left (x^3-1\right )}^{1/3}}{4\,x^4}-\frac {2\,{\left (x^3-1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ x^3\right )}{x\,{\left (1-x^3\right )}^{1/3}} \] Input:

int(((x^3 - 1)^(1/3)*(2*x^3 - 1))/x^5,x)
 

Output:

((x^3 - 1)^(1/3) - x^3*(x^3 - 1)^(1/3))/(4*x^4) - (2*(x^3 - 1)^(1/3)*hyper 
geom([-1/3, -1/3], 2/3, x^3))/(x*(1 - x^3)^(1/3))
 

Reduce [F]

\[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx=\frac {-\left (x^{3}-1\right )^{\frac {1}{3}} x^{3}+\left (x^{3}-1\right )^{\frac {1}{3}}+8 \left (\int \frac {\left (x^{3}-1\right )^{\frac {1}{3}}}{x^{2}}d x \right ) x^{4}}{4 x^{4}} \] Input:

int((x^3-1)^(1/3)*(2*x^3-1)/x^5,x)
 

Output:

( - (x**3 - 1)**(1/3)*x**3 + (x**3 - 1)**(1/3) + 8*int((x**3 - 1)**(1/3)/x 
**2,x)*x**4)/(4*x**4)