Integrand size = 9, antiderivative size = 94 \[ \int \sqrt [3]{x+x^3} \, dx=\frac {1}{2} x \sqrt [3]{x+x^3}-\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{12} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \]
1/2*x*(x^3+x)^(1/3)-1/6*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))*3^(1/2)-1/6* ln(-x+(x^3+x)^(1/3))+1/12*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))
Time = 0.53 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.41 \[ \int \sqrt [3]{x+x^3} \, dx=\frac {\sqrt [3]{x+x^3} \left (6 x^{4/3} \sqrt [3]{1+x^2}-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-2 \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )+\log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{12 \sqrt [3]{x} \sqrt [3]{1+x^2}} \]
((x + x^3)^(1/3)*(6*x^(4/3)*(1 + x^2)^(1/3) - 2*Sqrt[3]*ArcTan[(Sqrt[3]*x^ (2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))] - 2*Log[-x^(2/3) + (1 + x^2)^(1/3)] + Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)]))/(12*x^(1/3)*( 1 + x^2)^(1/3))
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {1910, 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.
| \(\displaystyle \int \sqrt [3]{x^3+x} \, dx\) |
| \(\Big \downarrow \) 1910 |
| \(\displaystyle \frac {1}{3} \int \frac {x}{\left (x^3+x\right )^{2/3}}dx+\frac {1}{2} \sqrt [3]{x^3+x} x\) |
| \(\Big \downarrow \) 1938 |
| \(\displaystyle \frac {\left (x^2+1\right )^{2/3} x^{2/3} \int \frac {\sqrt [3]{x}}{\left (x^2+1\right )^{2/3}}dx}{3 \left (x^3+x\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x^3+x} x\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\left (x^2+1\right )^{2/3} x^{2/3} \int \frac {x}{\left (x^2+1\right )^{2/3}}d\sqrt [3]{x}}{\left (x^3+x\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x^3+x} x\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {\left (x^2+1\right )^{2/3} x^{2/3} \int \frac {x^{2/3}}{(x+1)^{2/3}}dx^{2/3}}{2 \left (x^3+x\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x^3+x} x\) |
| \(\Big \downarrow \) 853 |
| \(\displaystyle \frac {\left (x^2+1\right )^{2/3} x^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^{2/3}-\sqrt [3]{x+1}\right )\right )}{2 \left (x^3+x\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x^3+x} x\) |
(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))/(2*( x + x^3)^(2/3))
3.13.97.3.1 Defintions of rubi rules used
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[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1)) Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && !IntegerQ[p] && LtQ[0, j, n] && GtQ[p, 0] && NeQ[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 = 6.50 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.18
| method | result | size |
| meijerg | \(\frac {3 x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{2}\right )}{4}\) | \(17\) |
| pseudoelliptic | \(\frac {x \left (6 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x +2 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-2 \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x}{x}\right )+\ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )\right )}{12 \left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x \right ) \left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x +x^{2}\right )}\) | \(134\) |
| trager | \(\frac {x \left (x^{3}+x \right )^{\frac {1}{3}}}{2}-\frac {\ln \left (-36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-27 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -33 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-51 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+15\right )}{6}+\frac {\operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+45 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +30 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+9 x \left (x^{3}+x \right )^{\frac {1}{3}}-25 x^{2}-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-10\right )}{2}\) | \(287\) |
| risch | \(\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x}{2}+\frac {\left (-\frac {\ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}-38 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+16 x^{4}+30 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}-70 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-96 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+18 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}+28 x^{2}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-32 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )+12}{x^{2}+1}\right )}{6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+20 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}-48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-100 x^{4}+30 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}+14 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+36 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}-48 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-140 x^{2}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-40}{x^{2}+1}\right )}{12}\right ) {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} {\left (\left (x^{2}+1\right )^{2} x^{2}\right )}^{\frac {1}{3}}}{\left (x^{2}+1\right ) x}\) | \(496\) |
Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \sqrt [3]{x+x^3} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{2} \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \frac {1}{12} \, \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \]
-1/6*sqrt(3)*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^2 + 5 07) - 1274*sqrt(3)*(x^3 + x)^(2/3))/(2205*x^2 + 2197)) + 1/2*(x^3 + x)^(1/ 3)*x - 1/12*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1)
\[ \int \sqrt [3]{x+x^3} \, dx=\int \sqrt [3]{x^{3} + x}\, dx \]
\[ \int \sqrt [3]{x+x^3} \, dx=\int { {\left (x^{3} + x\right )}^{\frac {1}{3}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \sqrt [3]{x+x^3} \, dx=\frac {1}{2} \, x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{12} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{6} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
1/2*x^2*(1/x^2 + 1)^(1/3) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^ (1/3) + 1)) + 1/12*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 1/6*lo g(abs((1/x^2 + 1)^(1/3) - 1))
Time = 5.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.29 \[ \int \sqrt [3]{x+x^3} \, dx=\frac {3\,x\,{\left (x^3+x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -x^2\right )}{4\,{\left (x^2+1\right )}^{1/3}} \]