3.17.52 \(\int x^6 \sqrt [3]{x+x^3} \, dx\) [1652]

3.17.52.1 Optimal result
3.17.52.2 Mathematica [A] (verified)
3.17.52.3 Rubi [A] (warning: unable to verify)
3.17.52.4 Maple [C] (verified)
3.17.52.5 Fricas [A] (verification not implemented)
3.17.52.6 Sympy [F]
3.17.52.7 Maxima [F]
3.17.52.8 Giac [A] (verification not implemented)
3.17.52.9 Mupad [F(-1)]

3.17.52.1 Optimal result

Integrand size = 13, antiderivative size = 112 \[ \int x^6 \sqrt [3]{x+x^3} \, dx=\frac {1}{648} \sqrt [3]{x+x^3} \left (20 x-12 x^3+9 x^5+81 x^7\right )+\frac {5 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{81 \sqrt {3}}+\frac {5}{243} \log \left (-x+\sqrt [3]{x+x^3}\right )-\frac {5}{486} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \]

output
1/648*(x^3+x)^(1/3)*(81*x^7+9*x^5-12*x^3+20*x)+5/243*arctan(3^(1/2)*x/(x+2 
*(x^3+x)^(1/3)))*3^(1/2)+5/243*ln(-x+(x^3+x)^(1/3))-5/486*ln(x^2+x*(x^3+x) 
^(1/3)+(x^3+x)^(2/3))
 
3.17.52.2 Mathematica [A] (verified)

Time = 3.23 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.63 \[ \int x^6 \sqrt [3]{x+x^3} \, dx=\frac {\sqrt [3]{x+x^3} \left (60 x^{4/3} \sqrt [3]{1+x^2}-36 x^{10/3} \sqrt [3]{1+x^2}+27 x^{16/3} \sqrt [3]{1+x^2}+243 x^{22/3} \sqrt [3]{1+x^2}+40 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )+40 \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )-20 \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{1944 \sqrt [3]{x} \sqrt [3]{1+x^2}} \]

input
Integrate[x^6*(x + x^3)^(1/3),x]
 
output
((x + x^3)^(1/3)*(60*x^(4/3)*(1 + x^2)^(1/3) - 36*x^(10/3)*(1 + x^2)^(1/3) 
 + 27*x^(16/3)*(1 + x^2)^(1/3) + 243*x^(22/3)*(1 + x^2)^(1/3) + 40*Sqrt[3] 
*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))] + 40*Log[-x^(2/3) 
 + (1 + x^2)^(1/3)] - 20*Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2) 
^(2/3)]))/(1944*x^(1/3)*(1 + x^2)^(1/3))
 
3.17.52.3 Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.38, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {1927, 1930, 1930, 1930, 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.

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

\(\Big \downarrow \) 1927

\(\displaystyle \frac {1}{12} \int \frac {x^7}{\left (x^3+x\right )^{2/3}}dx+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \int \frac {x^5}{\left (x^3+x\right )^{2/3}}dx\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \int \frac {x^3}{\left (x^3+x\right )^{2/3}}dx\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2}{3} \int \frac {x}{\left (x^3+x\right )^{2/3}}dx\right )\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {\sqrt [3]{x}}{\left (x^2+1\right )^{2/3}}dx}{3 \left (x^3+x\right )^{2/3}}\right )\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x}{\left (x^2+1\right )^{2/3}}d\sqrt [3]{x}}{\left (x^3+x\right )^{2/3}}\right )\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x^{2/3}}{(x+1)^{2/3}}dx^{2/3}}{\left (x^3+x\right )^{2/3}}\right )\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

\(\Big \downarrow \) 853

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} x^5 \sqrt [3]{x^3+x}-\frac {8}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^{2/3}-\sqrt [3]{x+1}\right )\right )}{\left (x^3+x\right )^{2/3}}\right )\right )\right )+\frac {1}{8} \sqrt [3]{x^3+x} x^7\)

input
Int[x^6*(x + x^3)^(1/3),x]
 
output
(x^7*(x + x^3)^(1/3))/8 + ((x^5*(x + x^3)^(1/3))/6 - (8*((x^3*(x + x^3)^(1 
/3))/4 - (5*((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[x^(2/3) - (1 + x)^(1 
/3)]/2))/(x + x^3)^(2/3)))/6))/9)/12
 

3.17.52.3.1 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 1927
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1930
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + 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]
 
3.17.52.4 Maple [C] (verified)

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

Time = 1.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.15

method result size
meijerg \(\frac {3 x^{\frac {22}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {11}{3}\right ], \left [\frac {14}{3}\right ], -x^{2}\right )}{22}\) \(17\)
pseudoelliptic \(-\frac {x^{4} \left (-243 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x^{7}-27 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x^{5}+36 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x^{3}+40 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-60 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +20 \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )-40 \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right )\right )}{1944 {\left ({\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}\right )}^{4} {\left ({\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x \right )}^{4}}\) \(180\)
trager \(\frac {x \left (81 x^{6}+9 x^{4}-12 x^{2}+20\right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{648}-\frac {5 \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+20 x^{2}+18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+8\right )}{243}-\frac {5 \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+20 x^{2}+18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+8\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{81}+\frac {5 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-4 x^{2}-48 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-3\right )}{81}\) \(447\)
risch \(\text {Expression too large to display}\) \(746\)

input
int(x^6*(x^3+x)^(1/3),x,method=_RETURNVERBOSE)
 
output
3/22*x^(22/3)*hypergeom([-1/3,11/3],[14/3],-x^2)
 
3.17.52.5 Fricas [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int x^6 \sqrt [3]{x+x^3} \, dx=\frac {5}{243} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{648} \, {\left (81 \, x^{7} + 9 \, x^{5} - 12 \, x^{3} + 20 \, x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}} + \frac {5}{486} \, \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \]

input
integrate(x^6*(x^3+x)^(1/3),x, algorithm="fricas")
 
output
5/243*sqrt(3)*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^2 + 
507) - 1274*sqrt(3)*(x^3 + x)^(2/3))/(2205*x^2 + 2197)) + 1/648*(81*x^7 + 
9*x^5 - 12*x^3 + 20*x)*(x^3 + x)^(1/3) + 5/486*log(3*(x^3 + x)^(1/3)*x - 3 
*(x^3 + x)^(2/3) + 1)
 
3.17.52.6 Sympy [F]

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

input
integrate(x**6*(x**3+x)**(1/3),x)
 
output
Integral(x**6*(x*(x**2 + 1))**(1/3), x)
 
3.17.52.7 Maxima [F]

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

input
integrate(x^6*(x^3+x)^(1/3),x, algorithm="maxima")
 
output
integrate((x^3 + x)^(1/3)*x^6, x)
 
3.17.52.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int x^6 \sqrt [3]{x+x^3} \, dx=\frac {1}{648} \, {\left (20 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {10}{3}} - 72 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {7}{3}} + 93 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} + 40 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )} x^{8} - \frac {5}{243} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {5}{486} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {5}{243} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

input
integrate(x^6*(x^3+x)^(1/3),x, algorithm="giac")
 
output
1/648*(20*(1/x^2 + 1)^(10/3) - 72*(1/x^2 + 1)^(7/3) + 93*(1/x^2 + 1)^(4/3) 
 + 40*(1/x^2 + 1)^(1/3))*x^8 - 5/243*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 
+ 1)^(1/3) + 1)) - 5/486*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) + 
5/243*log(abs((1/x^2 + 1)^(1/3) - 1))
 
3.17.52.9 Mupad [F(-1)]

Timed out. \[ \int x^6 \sqrt [3]{x+x^3} \, dx=\int x^6\,{\left (x^3+x\right )}^{1/3} \,d x \]

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