\(\int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx\) [1333]

Optimal result

Integrand size = 13, antiderivative size = 96 \[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{x^2+x^6}}\right )-\frac {1}{4} \log \left (-x^2+\sqrt [3]{x^2+x^6}\right )+\frac {1}{8} \log \left (x^4+x^2 \sqrt [3]{x^2+x^6}+\left (x^2+x^6\right )^{2/3}\right ) \] Output:

1/4*3^(1/2)*arctan(3^(1/2)*x^2/(x^2+2*(x^6+x^2)^(1/3)))-1/4*ln(-x^2+(x^6+x 
^2)^(1/3))+1/8*ln(x^4+x^2*(x^6+x^2)^(1/3)+(x^6+x^2)^(2/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.53 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\frac {x^{2/3} \sqrt [3]{1+x^4} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{4/3}}{x^{4/3}+2 \sqrt [3]{1+x^4}}\right )-2 \log \left (-x^{4/3}+\sqrt [3]{1+x^4}\right )+\log \left (x^{8/3}+x^{4/3} \sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right )\right )}{8 \sqrt [3]{x^2+x^6}} \] Input:

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

Output:

(x^(2/3)*(1 + x^4)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(4/3))/(x^(4/3) + 2* 
(1 + x^4)^(1/3))] - 2*Log[-x^(4/3) + (1 + x^4)^(1/3)] + Log[x^(8/3) + x^(4 
/3)*(1 + x^4)^(1/3) + (1 + x^4)^(2/3)]))/(8*(x^2 + x^6)^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1919, 1917, 266, 807, 769}

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.

\(\displaystyle \int \frac {x}{\sqrt [3]{x^6+x^2}} \, dx\)

\(\Big \downarrow \) 1919

\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [3]{x^6+x^2}}dx^2\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {\sqrt [3]{x^2} \sqrt [3]{x^4+1} \int \frac {1}{\sqrt [3]{x^2} \sqrt [3]{x^4+1}}dx^2}{2 \sqrt [3]{x^6+x^2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 \sqrt [3]{x^2} \sqrt [3]{x^4+1} \int \frac {\sqrt [3]{x^2}}{\sqrt [3]{x^{12}+1}}d\sqrt [3]{x^2}}{2 \sqrt [3]{x^6+x^2}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 \sqrt [3]{x^2} \sqrt [3]{x^4+1} \int \frac {1}{\sqrt [3]{x^6+1}}dx^4}{4 \sqrt [3]{x^6+x^2}}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {3 \sqrt [3]{x^2} \sqrt [3]{x^4+1} \left (\frac {\arctan \left (\frac {\frac {2 x^4}{\sqrt [3]{x^6+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x^6+1}-x^4\right )\right )}{4 \sqrt [3]{x^6+x^2}}\)

Input:

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

Output:

(3*(x^2)^(1/3)*(1 + x^4)^(1/3)*(ArcTan[(1 + (2*x^4)/(1 + x^6)^(1/3))/Sqrt[ 
3]]/Sqrt[3] - Log[-x^4 + (1 + x^6)^(1/3)]/2))/(4*(x^2 + x^6)^(1/3))
 

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 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, 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 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1919

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

Maple [C] (verified)

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

Time = 3.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(\frac {3 x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{4}\right )}{4}\)	\(17\)
pseudoelliptic	\(\frac {\ln \left (\frac {x^{4}+\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}} x^{2}+\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {2}{3}}}{x^{4}}\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (x^{2}+2 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 x^{2}}\right )}{4}-\frac {\ln \left (\frac {-x^{2}+\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}}}{x^{2}}\right )}{4}\)	\(94\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{4}-734 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{4}-585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{3}} x^{2}-55 x^{4}-585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-204 x^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-204 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-309 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-22\right )}{4}-\frac {\ln \left (149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{4}+436 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{4}+585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{3}} x^{2}-640 x^{4}+585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-789 x^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-789 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}+607 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-480\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )}{4}+\frac {\ln \left (149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{4}+436 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{4}+585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{3}} x^{2}-640 x^{4}+585 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-789 x^{2} \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-789 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}+607 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-480\right )}{4}\)	\(416\)

Input:

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

Output:

3/4*x^(4/3)*hypergeom([1/3,1/3],[4/3],-x^4)
 

Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} x^{2} - \sqrt {3} {\left (539 \, x^{4} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {2}{3}}}{2205 \, x^{4} + 2197}\right ) - \frac {1}{8} \, \log \left (3 \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{6} + x^{2}\right )}^{\frac {2}{3}} + 1\right ) \] Input:

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

Output:

1/4*sqrt(3)*arctan(-(196*sqrt(3)*(x^6 + x^2)^(1/3)*x^2 - sqrt(3)*(539*x^4 
+ 507) - 1274*sqrt(3)*(x^6 + x^2)^(2/3))/(2205*x^4 + 2197)) - 1/8*log(3*(x 
^6 + x^2)^(1/3)*x^2 - 3*(x^6 + x^2)^(2/3) + 1)
 

Sympy [F]

\[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\int \frac {x}{\sqrt [3]{x^{2} \left (x^{4} + 1\right )}}\, dx \] Input:

integrate(x/(x**6+x**2)**(1/3),x)
 

Output:

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

Maxima [F]

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

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

Output:

integrate(x/(x^6 + x^2)^(1/3), x)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=-\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{8} \, \log \left ({\left (\frac {1}{x^{4}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{4} \, \log \left ({\left | {\left (\frac {1}{x^{4}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:

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

Output:

-1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^4 + 1)^(1/3) + 1)) + 1/8*log((1/x^ 
4 + 1)^(2/3) + (1/x^4 + 1)^(1/3) + 1) - 1/4*log(abs((1/x^4 + 1)^(1/3) - 1) 
)
 

Mupad [B] (verification not implemented)

Time = 8.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.32 \[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\frac {3\,x^2\,{\left (x^4+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x^4\right )}{4\,{\left (x^6+x^2\right )}^{1/3}} \] Input:

int(x/(x^2 + x^6)^(1/3),x)
 

Output:

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

Reduce [F]

\[ \int \frac {x}{\sqrt [3]{x^2+x^6}} \, dx=\int \frac {x^{\frac {1}{3}}}{\left (x^{4}+1\right )^{\frac {1}{3}}}d x \] Input:

int(x/(x^6+x^2)^(1/3),x)
 

Output:

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