Integrand size = 13, antiderivative size = 100 \[ \int x^2 \sqrt [3]{x+x^3} \, dx=\frac {1}{12} \sqrt [3]{x+x^3} \left (x+3 x^3\right )+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{6 \sqrt {3}}+\frac {1}{18} \log \left (-x+\sqrt [3]{x+x^3}\right )-\frac {1}{36} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \] Output:
1/12*(x^3+x)^(1/3)*(3*x^3+x)+1/18*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))*3^ (1/2)+1/18*ln(-x+(x^3+x)^(1/3))-1/36*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))
Time = 0.69 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.51 \[ \int x^2 \sqrt [3]{x+x^3} \, dx=\frac {\sqrt [3]{x+x^3} \left (3 x^{4/3} \sqrt [3]{1+x^2}+9 x^{10/3} \sqrt [3]{1+x^2}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )+2 \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )-\log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{36 \sqrt [3]{x} \sqrt [3]{1+x^2}} \] Input:
Integrate[x^2*(x + x^3)^(1/3),x]
Output:
((x + x^3)^(1/3)*(3*x^(4/3)*(1 + x^2)^(1/3) + 9*x^(10/3)*(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)]))/(36*x^(1/3)*(1 + x^2)^(1/3))
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1927, 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.
\(\displaystyle \int x^2 \sqrt [3]{x^3+x} \, dx\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {1}{6} \int \frac {x^3}{\left (x^3+x\right )^{2/3}}dx+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2}{3} \int \frac {x}{\left (x^3+x\right )^{2/3}}dx\right )+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {\sqrt [3]{x}}{\left (x^2+1\right )^{2/3}}dx}{3 \left (x^3+x\right )^{2/3}}\right )+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {2 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x}{\left (x^2+1\right )^{2/3}}d\sqrt [3]{x}}{\left (x^3+x\right )^{2/3}}\right )+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x^{2/3}}{(x+1)^{2/3}}dx^{2/3}}{\left (x^3+x\right )^{2/3}}\right )+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} x \sqrt [3]{x^3+x}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^{2/3}-\sqrt [3]{x+1}\right )\right )}{\left (x^3+x\right )^{2/3}}\right )+\frac {1}{4} \sqrt [3]{x^3+x} x^3\) |
Input:
Int[x^2*(x + x^3)^(1/3),x]
Output:
(x^3*(x + x^3)^(1/3))/4 + ((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
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 = 2.62 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.17
method | result | size |
meijerg | \(\frac {3 x^{\frac {10}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x^{2}\right )}{10}\) | \(17\) |
pseudoelliptic | \(-\frac {x^{2} \left (-9 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}} x^{3}+2 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-3 x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+\ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}}{x^{2}}\right )-2 \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x}{x}\right )\right )}{36 {\left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}\right )}^{2} {\left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x \right )}^{2}}\) | \(150\) |
trager | \(\frac {x \left (3 x^{2}+1\right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{12}-\frac {\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 )}{18}-\frac {\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 )}{6}+\frac {\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 )}{6}\) | \(437\) |
risch | \(\text {Expression too large to display}\) | \(735\) |
Input:
int(x^2*(x^3+x)^(1/3),x,method=_RETURNVERBOSE)
Output:
3/10*x^(10/3)*hypergeom([-1/3,5/3],[8/3],-x^2)
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int x^2 \sqrt [3]{x+x^3} \, dx=\frac {1}{18} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{12} \, {\left (3 \, x^{3} + x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}} + \frac {1}{36} \, \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \] Input:
integrate(x^2*(x^3+x)^(1/3),x, algorithm="fricas")
Output:
1/18*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)) + 1/12*(3*x^3 + x)* (x^3 + x)^(1/3) + 1/36*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1)
\[ \int x^2 \sqrt [3]{x+x^3} \, dx=\int x^{2} \sqrt [3]{x \left (x^{2} + 1\right )}\, dx \] Input:
integrate(x**2*(x**3+x)**(1/3),x)
Output:
Integral(x**2*(x*(x**2 + 1))**(1/3), x)
\[ \int x^2 \sqrt [3]{x+x^3} \, dx=\int { {\left (x^{3} + x\right )}^{\frac {1}{3}} x^{2} \,d x } \] Input:
integrate(x^2*(x^3+x)^(1/3),x, algorithm="maxima")
Output:
integrate((x^3 + x)^(1/3)*x^2, x)
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.77 \[ \int x^2 \sqrt [3]{x+x^3} \, dx=\frac {1}{12} \, {\left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )} x^{4} - \frac {1}{18} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{36} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{18} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:
integrate(x^2*(x^3+x)^(1/3),x, algorithm="giac")
Output:
1/12*((1/x^2 + 1)^(4/3) + 2*(1/x^2 + 1)^(1/3))*x^4 - 1/18*sqrt(3)*arctan(1 /3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) - 1/36*log((1/x^2 + 1)^(2/3) + (1/x^ 2 + 1)^(1/3) + 1) + 1/18*log(abs((1/x^2 + 1)^(1/3) - 1))
Timed out. \[ \int x^2 \sqrt [3]{x+x^3} \, dx=\int x^2\,{\left (x^3+x\right )}^{1/3} \,d x \] Input:
int(x^2*(x + x^3)^(1/3),x)
Output:
int(x^2*(x + x^3)^(1/3), x)
\[ \int x^2 \sqrt [3]{x+x^3} \, dx=\frac {x^{\frac {10}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{4}+\frac {x^{\frac {4}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{12}-\frac {\left (\int \frac {x^{\frac {1}{3}}}{\left (x^{2}+1\right )^{\frac {2}{3}}}d x \right )}{9} \] Input:
int(x^2*(x^3+x)^(1/3),x)
Output:
(9*x**(1/3)*(x**2 + 1)**(1/3)*x**3 + 3*x**(1/3)*(x**2 + 1)**(1/3)*x - 4*in t((x**(1/3)*(x**2 + 1)**(1/3))/(x**2 + 1),x))/36