\(\int \frac {(-1+x^5)^{2/3}}{x^6} \, dx\) [1273]

Optimal result

Integrand size = 13, antiderivative size = 92 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=-\frac {\left (-1+x^5\right )^{2/3}}{5 x^5}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^5}}{\sqrt {3}}\right )}{5 \sqrt {3}}-\frac {2}{15} \log \left (1+\sqrt [3]{-1+x^5}\right )+\frac {1}{15} \log \left (1-\sqrt [3]{-1+x^5}+\left (-1+x^5\right )^{2/3}\right ) \] Output:

-1/5*(x^5-1)^(2/3)/x^5+2/15*3^(1/2)*arctan(-1/3*3^(1/2)+2/3*(x^5-1)^(1/3)* 
3^(1/2))-2/15*ln(1+(x^5-1)^(1/3))+1/15*ln(1-(x^5-1)^(1/3)+(x^5-1)^(2/3))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=\frac {1}{15} \left (-\frac {3 \left (-1+x^5\right )^{2/3}}{x^5}-2 \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{-1+x^5}}{\sqrt {3}}\right )-2 \log \left (1+\sqrt [3]{-1+x^5}\right )+\log \left (1-\sqrt [3]{-1+x^5}+\left (-1+x^5\right )^{2/3}\right )\right ) \] Input:

Integrate[(-1 + x^5)^(2/3)/x^6,x]
 

Output:

((-3*(-1 + x^5)^(2/3))/x^5 - 2*Sqrt[3]*ArcTan[(1 - 2*(-1 + x^5)^(1/3))/Sqr 
t[3]] - 2*Log[1 + (-1 + x^5)^(1/3)] + Log[1 - (-1 + x^5)^(1/3) + (-1 + x^5 
)^(2/3)])/15
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 51, 68, 16, 1083, 217}

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.

\(\displaystyle \int \frac {\left (x^5-1\right )^{2/3}}{x^6} \, dx\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {1}{5} \int \frac {\left (x^5-1\right )^{2/3}}{x^{10}}dx^5\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 68

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (-\frac {3}{2} \int \frac {1}{\sqrt [3]{x^5-1}+1}d\sqrt [3]{x^5-1}+\frac {3}{2} \int \frac {1}{x^{10}-\sqrt [3]{x^5-1}+1}d\sqrt [3]{x^5-1}+\frac {\log \left (x^5\right )}{2}\right )-\frac {\left (x^5-1\right )^{2/3}}{x^5}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\frac {3}{2} \int \frac {1}{x^{10}-\sqrt [3]{x^5-1}+1}d\sqrt [3]{x^5-1}+\frac {\log \left (x^5\right )}{2}-\frac {3}{2} \log \left (\sqrt [3]{x^5-1}+1\right )\right )-\frac {\left (x^5-1\right )^{2/3}}{x^5}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (-3 \int \frac {1}{-x^{10}-3}d\left (2 \sqrt [3]{x^5-1}-1\right )+\frac {\log \left (x^5\right )}{2}-\frac {3}{2} \log \left (\sqrt [3]{x^5-1}+1\right )\right )-\frac {\left (x^5-1\right )^{2/3}}{x^5}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^5-1}-1}{\sqrt {3}}\right )+\frac {\log \left (x^5\right )}{2}-\frac {3}{2} \log \left (\sqrt [3]{x^5-1}+1\right )\right )-\frac {\left (x^5-1\right )^{2/3}}{x^5}\right )\)

Input:

Int[(-1 + x^5)^(2/3)/x^6,x]
 

Output:

(-((-1 + x^5)^(2/3)/x^5) + (2*(Sqrt[3]*ArcTan[(-1 + 2*(-1 + x^5)^(1/3))/Sq 
rt[3]] + Log[x^5]/2 - (3*Log[1 + (-1 + x^5)^(1/3)])/2))/3)/5
 

Defintions of rubi rules used

rule 16

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

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 68

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 217

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

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1083

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

Maple [C] (warning: unable to verify)

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

Time = 4.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {\left (x^{5}-1\right )^{\frac {2}{3}}}{5 x^{5}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {1}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{5}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+5 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{15 \pi \operatorname {signum}\left (x^{5}-1\right )^{\frac {1}{3}}}\)	\(96\)
meijerg	\(\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} \left (-\frac {\pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], x^{5}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+5 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{5}}\right )}{15 \pi {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}}}\)	\(97\)
pseudoelliptic	\(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{5}-1\right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right ) x^{5}+\ln \left (1-\left (x^{5}-1\right )^{\frac {1}{3}}+\left (x^{5}-1\right )^{\frac {2}{3}}\right ) x^{5}-2 \ln \left (1+\left (x^{5}-1\right )^{\frac {1}{3}}\right ) x^{5}-3 \left (x^{5}-1\right )^{\frac {2}{3}}}{15 \left (1-\left (x^{5}-1\right )^{\frac {1}{3}}+\left (x^{5}-1\right )^{\frac {2}{3}}\right ) \left (1+\left (x^{5}-1\right )^{\frac {1}{3}}\right )}\)	\(107\)
trager	\(-\frac {\left (x^{5}-1\right )^{\frac {2}{3}}}{5 x^{5}}+\frac {64 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \ln \left (-\frac {5879915417871360 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2} x^{5}-2220442568272704 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) x^{5}[phone] x^{5}+12071260649204448 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {2}{3}}-188157293371883520 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2}-12071260649204448 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {1}{3}}-184009801566659 \left (x^{5}-1\right )^{\frac {2}{3}}+14031232455161568 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )[phone] \left (x^{5}-1\right )^{\frac {1}{3}}-204426174545379}{x^{5}}\right )}{5}+\frac {2 \ln \left (-\frac {2939957708935680 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2} x^{5}+1048972165200192 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) x^{5}[phone] x^{5}-6035630324602224 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {2}{3}}-94078646685941760 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2}+6035630324602224 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {1}{3}}-29133751568723 \left (x^{5}-1\right )^{\frac {2}{3}}-5055644421623664 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )[phone] \left (x^{5}-1\right )^{\frac {1}{3}}-39341938058083}{x^{5}}\right )}{15}-\frac {64 \ln \left (-\frac {2939957708935680 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2} x^{5}+1048972165200192 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) x^{5}[phone] x^{5}-6035630324602224 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {2}{3}}-94078646685941760 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )^{2}+6035630324602224 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right ) \left (x^{5}-1\right )^{\frac {1}{3}}-29133751568723 \left (x^{5}-1\right )^{\frac {2}{3}}-5055644421623664 \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )[phone] \left (x^{5}-1\right )^{\frac {1}{3}}-39341938058083}{x^{5}}\right ) \operatorname {RootOf}\left (9216 \textit {\_Z}^{2}-96 \textit {\_Z} +1\right )}{5}\)	\(441\)

Input:

int((x^5-1)^(2/3)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/5*(x^5-1)^(2/3)/x^5+1/15/Pi*3^(1/2)*GAMMA(2/3)/signum(x^5-1)^(1/3)*(-si 
gnum(x^5-1))^(1/3)*(2/9*Pi*3^(1/2)/GAMMA(2/3)*x^5*hypergeom([1,1,4/3],[2,2 
],x^5)+2/3*(-1/6*Pi*3^(1/2)-3/2*ln(3)+5*ln(x)+I*Pi)*Pi*3^(1/2)/GAMMA(2/3))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=\frac {2 \, \sqrt {3} x^{5} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{5} - 1\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + x^{5} \log \left ({\left (x^{5} - 1\right )}^{\frac {2}{3}} - {\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1\right ) - 2 \, x^{5} \log \left ({\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1\right ) - 3 \, {\left (x^{5} - 1\right )}^{\frac {2}{3}}}{15 \, x^{5}} \] Input:

integrate((x^5-1)^(2/3)/x^6,x, algorithm="fricas")
 

Output:

1/15*(2*sqrt(3)*x^5*arctan(2/3*sqrt(3)*(x^5 - 1)^(1/3) - 1/3*sqrt(3)) + x^ 
5*log((x^5 - 1)^(2/3) - (x^5 - 1)^(1/3) + 1) - 2*x^5*log((x^5 - 1)^(1/3) + 
 1) - 3*(x^5 - 1)^(2/3))/x^5
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{5}}} \right )}}{5 x^{\frac {5}{3}} \Gamma \left (\frac {4}{3}\right )} \] Input:

integrate((x**5-1)**(2/3)/x**6,x)
 

Output:

-gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(2*I*pi)/x**5)/(5*x**(5/3) 
*gamma(4/3))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=\frac {2}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{5} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {{\left (x^{5} - 1\right )}^{\frac {2}{3}}}{5 \, x^{5}} + \frac {1}{15} \, \log \left ({\left (x^{5} - 1\right )}^{\frac {2}{3}} - {\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {2}{15} \, \log \left ({\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1\right ) \] Input:

integrate((x^5-1)^(2/3)/x^6,x, algorithm="maxima")
 

Output:

2/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^5 - 1)^(1/3) - 1)) - 1/5*(x^5 - 1)^( 
2/3)/x^5 + 1/15*log((x^5 - 1)^(2/3) - (x^5 - 1)^(1/3) + 1) - 2/15*log((x^5 
 - 1)^(1/3) + 1)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=\frac {2}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{5} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {{\left (x^{5} - 1\right )}^{\frac {2}{3}}}{5 \, x^{5}} + \frac {1}{15} \, \log \left ({\left (x^{5} - 1\right )}^{\frac {2}{3}} - {\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {2}{15} \, \log \left ({\left | {\left (x^{5} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) \] Input:

integrate((x^5-1)^(2/3)/x^6,x, algorithm="giac")
 

Output:

2/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^5 - 1)^(1/3) - 1)) - 1/5*(x^5 - 1)^( 
2/3)/x^5 + 1/15*log((x^5 - 1)^(2/3) - (x^5 - 1)^(1/3) + 1) - 2/15*log(abs( 
(x^5 - 1)^(1/3) + 1))
 

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=-\frac {2\,\ln \left (\frac {4\,{\left (x^5-1\right )}^{1/3}}{25}+\frac {4}{25}\right )}{15}-\ln \left (9\,{\left (-\frac {1}{15}+\frac {\sqrt {3}\,1{}\mathrm {i}}{15}\right )}^2+\frac {4\,{\left (x^5-1\right )}^{1/3}}{25}\right )\,\left (-\frac {1}{15}+\frac {\sqrt {3}\,1{}\mathrm {i}}{15}\right )+\ln \left (9\,{\left (\frac {1}{15}+\frac {\sqrt {3}\,1{}\mathrm {i}}{15}\right )}^2+\frac {4\,{\left (x^5-1\right )}^{1/3}}{25}\right )\,\left (\frac {1}{15}+\frac {\sqrt {3}\,1{}\mathrm {i}}{15}\right )-\frac {{\left (x^5-1\right )}^{2/3}}{5\,x^5} \] Input:

int((x^5 - 1)^(2/3)/x^6,x)
                                                                                    
                                                                                    
 

Output:

log(9*((3^(1/2)*1i)/15 + 1/15)^2 + (4*(x^5 - 1)^(1/3))/25)*((3^(1/2)*1i)/1 
5 + 1/15) - log(9*((3^(1/2)*1i)/15 - 1/15)^2 + (4*(x^5 - 1)^(1/3))/25)*((3 
^(1/2)*1i)/15 - 1/15) - (2*log((4*(x^5 - 1)^(1/3))/25 + 4/25))/15 - (x^5 - 
 1)^(2/3)/(5*x^5)
 

Reduce [F]

\[ \int \frac {\left (-1+x^5\right )^{2/3}}{x^6} \, dx=\frac {-3 \left (x^{5}-1\right )^{\frac {2}{3}}+10 \left (\int \frac {\left (x^{5}-1\right )^{\frac {2}{3}}}{x^{6}-x}d x \right ) x^{5}}{15 x^{5}} \] Input:

int((x^5-1)^(2/3)/x^6,x)
 

Output:

( - 3*(x**5 - 1)**(2/3) + 10*int((x**5 - 1)**(2/3)/(x**6 - x),x)*x**5)/(15 
*x**5)