Integrand size = 13, antiderivative size = 107 \[ \int x^4 \sqrt [3]{x+x^3} \, dx=\frac {1}{108} \sqrt [3]{x+x^3} \left (-5 x+3 x^3+18 x^5\right )-\frac {5 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{54 \sqrt {3}}-\frac {5}{162} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {5}{324} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \] Output:
1/108*(x^3+x)^(1/3)*(18*x^5+3*x^3-5*x)-5/162*arctan(3^(1/2)*x/(x+2*(x^3+x) ^(1/3)))*3^(1/2)-5/162*ln(-x+(x^3+x)^(1/3))+5/324*ln(x^2+x*(x^3+x)^(1/3)+( x^3+x)^(2/3))
Time = 1.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.56 \[ \int x^4 \sqrt [3]{x+x^3} \, dx=\frac {\sqrt [3]{x+x^3} \left (-15 x^{4/3} \sqrt [3]{1+x^2}+9 x^{10/3} \sqrt [3]{1+x^2}+54 x^{16/3} \sqrt [3]{1+x^2}-10 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-10 \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )+5 \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{324 \sqrt [3]{x} \sqrt [3]{1+x^2}} \] Input:
Integrate[x^4*(x + x^3)^(1/3),x]
Output:
((x + x^3)^(1/3)*(-15*x^(4/3)*(1 + x^2)^(1/3) + 9*x^(10/3)*(1 + x^2)^(1/3) + 54*x^(16/3)*(1 + x^2)^(1/3) - 10*Sqrt[3]*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2 /3) + 2*(1 + x^2)^(1/3))] - 10*Log[-x^(2/3) + (1 + x^2)^(1/3)] + 5*Log[x^( 4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)]))/(324*x^(1/3)*(1 + x^2) ^(1/3))
Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {1927, 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.
\(\displaystyle \int x^4 \sqrt [3]{x^3+x} \, dx\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {1}{9} \int \frac {x^5}{\left (x^3+x\right )^{2/3}}dx+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \int \frac {x^3}{\left (x^3+x\right )^{2/3}}dx\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2}{3} \int \frac {x}{\left (x^3+x\right )^{2/3}}dx\right )\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {\sqrt [3]{x}}{\left (x^2+1\right )^{2/3}}dx}{3 \left (x^3+x\right )^{2/3}}\right )\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x}{\left (x^2+1\right )^{2/3}}d\sqrt [3]{x}}{\left (x^3+x\right )^{2/3}}\right )\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x^{2/3}}{(x+1)^{2/3}}dx^{2/3}}{\left (x^3+x\right )^{2/3}}\right )\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{4} x^3 \sqrt [3]{x^3+x}-\frac {5}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^{2/3}-\sqrt [3]{x+1}\right )\right )}{\left (x^3+x\right )^{2/3}}\right )\right )+\frac {1}{6} \sqrt [3]{x^3+x} x^5\) |
Input:
Int[x^4*(x + x^3)^(1/3),x]
Output:
(x^5*(x + x^3)^(1/3))/6 + ((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
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]
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]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 2.52 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.16
method | result | size |
meijerg | \(\frac {3 x^{\frac {16}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {8}{3}\right ], \left [\frac {11}{3}\right ], -x^{2}\right )}{16}\) | \(17\) |
pseudoelliptic | \(\frac {x^{3} \left (54 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x^{5}+9 x^{3} {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+10 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-15 x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+5 \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}}{x^{2}}\right )-10 \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x}{x}\right )\right )}{324 {\left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}\right )}^{3} {\left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x \right )}^{3}}\) | \(166\) |
trager | \(\frac {x \left (18 x^{4}+3 x^{2}-5\right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{108}-\frac {5 \ln \left (-36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-27 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -33 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-15 x \left (x^{3}+x \right )^{\frac {1}{3}}+36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+20 x^{2}-51 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+15\right )}{162}+\frac {5 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+45 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +30 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+9 x \left (x^{3}+x \right )^{\frac {1}{3}}+9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-25 x^{2}-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-10\right )}{54}\) | \(299\) |
risch | \(\text {Expression too large to display}\) | \(739\) |
Input:
int(x^4*(x^3+x)^(1/3),x,method=_RETURNVERBOSE)
Output:
3/16*x^(16/3)*hypergeom([-1/3,8/3],[11/3],-x^2)
Time = 0.24 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96 \[ \int x^4 \sqrt [3]{x+x^3} \, dx=-\frac {5}{162} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{108} \, {\left (18 \, x^{5} + 3 \, x^{3} - 5 \, x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}} - \frac {5}{324} \, \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \] Input:
integrate(x^4*(x^3+x)^(1/3),x, algorithm="fricas")
Output:
-5/162*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)) + 1/108*(18*x^5 + 3*x^3 - 5*x)*(x^3 + x)^(1/3) - 5/324*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x )^(2/3) + 1)
\[ \int x^4 \sqrt [3]{x+x^3} \, dx=\int x^{4} \sqrt [3]{x \left (x^{2} + 1\right )}\, dx \] Input:
integrate(x**4*(x**3+x)**(1/3),x)
Output:
Integral(x**4*(x*(x**2 + 1))**(1/3), x)
\[ \int x^4 \sqrt [3]{x+x^3} \, dx=\int { {\left (x^{3} + x\right )}^{\frac {1}{3}} x^{4} \,d x } \] Input:
integrate(x^4*(x^3+x)^(1/3),x, algorithm="maxima")
Output:
integrate((x^3 + x)^(1/3)*x^4, x)
Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.82 \[ \int x^4 \sqrt [3]{x+x^3} \, dx=-\frac {1}{108} \, {\left (5 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {7}{3}} - 13 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} - 10 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )} x^{6} + \frac {5}{162} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {5}{324} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {5}{162} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:
integrate(x^4*(x^3+x)^(1/3),x, algorithm="giac")
Output:
-1/108*(5*(1/x^2 + 1)^(7/3) - 13*(1/x^2 + 1)^(4/3) - 10*(1/x^2 + 1)^(1/3)) *x^6 + 5/162*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) + 5/324 *log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 5/162*log(abs((1/x^2 + 1 )^(1/3) - 1))
Timed out. \[ \int x^4 \sqrt [3]{x+x^3} \, dx=\int x^4\,{\left (x^3+x\right )}^{1/3} \,d x \] Input:
int(x^4*(x + x^3)^(1/3),x)
Output:
int(x^4*(x + x^3)^(1/3), x)
\[ \int x^4 \sqrt [3]{x+x^3} \, dx=\frac {x^{\frac {16}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{6}+\frac {x^{\frac {10}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{36}-\frac {5 x^{\frac {4}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{108}+\frac {5 \left (\int \frac {x^{\frac {1}{3}}}{\left (x^{2}+1\right )^{\frac {2}{3}}}d x \right )}{81} \] Input:
int(x^4*(x^3+x)^(1/3),x)
Output:
(54*x**(1/3)*(x**2 + 1)**(1/3)*x**5 + 9*x**(1/3)*(x**2 + 1)**(1/3)*x**3 - 15*x**(1/3)*(x**2 + 1)**(1/3)*x + 20*int((x**(1/3)*(x**2 + 1)**(1/3))/(x** 2 + 1),x))/324