\(\int \frac {(1+x^3)^{2/3}}{x^4} \, dx\) [1229]

Optimal result

Integrand size = 13, antiderivative size = 90 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=-\frac {\left (1+x^3\right )^{2/3}}{3 x^3}+\frac {2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{9} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{9} \log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=\frac {1}{9} \left (-\frac {3 \left (1+x^3\right )^{2/3}}{x^3}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^3}\right )-\log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )\right ) \] Input:

Integrate[(1 + x^3)^(2/3)/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 51, 67, 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.

\(\displaystyle \int \frac {\left (x^3+1\right )^{2/3}}{x^4} \, dx\)

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{x^3+1}}d\sqrt [3]{x^3+1}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}-\frac {1}{2} \log \left (x^3\right )\right )-\frac {\left (x^3+1\right )^{2/3}}{x^3}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )-\frac {\left (x^3+1\right )^{2/3}}{x^3}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (-3 \int \frac {1}{-x^6-3}d\left (2 \sqrt [3]{x^3+1}+1\right )-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )-\frac {\left (x^3+1\right )^{2/3}}{x^3}\right )\)

\(\Big \downarrow \) 217

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

Input:

Int[(1 + x^3)^(2/3)/x^4,x]
 

Output:

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

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 67

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] && PosQ[(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] (verified)

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

Time = 1.92 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {\pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+3 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{3}}\right )}{9 \pi }\)	\(76\)
risch	\(-\frac {\left (x^{3}+1\right )^{\frac {2}{3}}}{3 x^{3}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], -x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{9 \pi }\)	\(76\)
pseudoelliptic	\(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{3}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right ) x^{3}-\ln \left (1+\left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}\right ) x^{3}+2 \ln \left (-1+\left (x^{3}+1\right )^{\frac {1}{3}}\right ) x^{3}-3 \left (x^{3}+1\right )^{\frac {2}{3}}}{9 \left (1+\left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}\right ) \left (-1+\left (x^{3}+1\right )^{\frac {1}{3}}\right )}\)	\(104\)
trager	\(-\frac {\left (x^{3}+1\right )^{\frac {2}{3}}}{3 x^{3}}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \left (x^{3}+1\right )^{\frac {1}{3}}+19 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-5}{x^{3}}\right )}{9}-\frac {2 \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+24 \left (x^{3}+1\right )^{\frac {1}{3}}+11 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+20}{x^{3}}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{9}-\frac {2 \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+24 \left (x^{3}+1\right )^{\frac {1}{3}}+11 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+20}{x^{3}}\right )}{9}\)	\(360\)

Input:

int((x^3+1)^(2/3)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/9/Pi*3^(1/2)*GAMMA(2/3)*(1/9*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1,1,4 
/3],[2,3],-x^3)-2/3*(-1/6*Pi*3^(1/2)-3/2*ln(3)-1+3*ln(x))*Pi*3^(1/2)/GAMMA 
(2/3)+Pi*3^(1/2)/GAMMA(2/3)/x^3)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=\frac {2 \, \sqrt {3} x^{3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{9 \, x^{3}} \] Input:

integrate((x^3+1)^(2/3)/x^4,x, algorithm="fricas")
 

Output:

1/9*(2*sqrt(3)*x^3*arctan(2/3*sqrt(3)*(x^3 + 1)^(1/3) + 1/3*sqrt(3)) - x^3 
*log((x^3 + 1)^(2/3) + (x^3 + 1)^(1/3) + 1) + 2*x^3*log((x^3 + 1)^(1/3) - 
1) - 3*(x^3 + 1)^(2/3))/x^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.34 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \] Input:

integrate((x**3+1)**(2/3)/x**4,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{3 \, x^{3}} - \frac {1}{9} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {2}{9} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \] Input:

integrate((x^3+1)^(2/3)/x^4,x, algorithm="maxima")
 

Output:

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/3*(x^3 + 1)^(2 
/3)/x^3 - 1/9*log((x^3 + 1)^(2/3) + (x^3 + 1)^(1/3) + 1) + 2/9*log((x^3 + 
1)^(1/3) - 1)
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{3 \, x^{3}} - \frac {1}{9} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {2}{9} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:

integrate((x^3+1)^(2/3)/x^4,x, algorithm="giac")
 

Output:

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/3*(x^3 + 1)^(2 
/3)/x^3 - 1/9*log((x^3 + 1)^(2/3) + (x^3 + 1)^(1/3) + 1) + 2/9*log(abs((x^ 
3 + 1)^(1/3) - 1))
 

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^4} \, dx=\frac {2\,\ln \left (\frac {4\,{\left (x^3+1\right )}^{1/3}}{9}-\frac {4}{9}\right )}{9}+\ln \left (\frac {4\,{\left (x^3+1\right )}^{1/3}}{9}-9\,{\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )-\ln \left (\frac {4\,{\left (x^3+1\right )}^{1/3}}{9}-9\,{\left (\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2\right )\,\left (\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )-\frac {{\left (x^3+1\right )}^{2/3}}{3\,x^3} \] Input:

int((x^3 + 1)^(2/3)/x^4,x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [F]

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

int((x^3+1)^(2/3)/x^4,x)
 

Output:

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