\(\int \sqrt [3]{x^2+x^3} \, dx\) [1571]

Optimal result

Integrand size = 11, antiderivative size = 108 \[ \int \sqrt [3]{x^2+x^3} \, dx=\frac {1}{6} (1+3 x) \sqrt [3]{x^2+x^3}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (-x+\sqrt [3]{x^2+x^3}\right )-\frac {1}{18} \log \left (x^2+x \sqrt [3]{x^2+x^3}+\left (x^2+x^3\right )^{2/3}\right ) \]

[Out]

1/6*(1+3*x)*(x^3+x^2)^(1/3)+1/9*3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3+x^2)^(1/3)))+1/9*ln(-x+(x^3+x^2)^(1/3))-1/1
8*ln(x^2+x*(x^3+x^2)^(1/3)+(x^3+x^2)^(2/3))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2029, 2049, 2057, 61} \[ \int \sqrt [3]{x^2+x^3} \, dx=\frac {(x+1)^{2/3} x^{4/3} \arctan \left (\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x+1}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} \left (x^3+x^2\right )^{2/3}}+\frac {1}{2} \sqrt [3]{x^3+x^2} x+\frac {1}{6} \sqrt [3]{x^3+x^2}+\frac {(x+1)^{2/3} x^{4/3} \log (x+1)}{18 \left (x^3+x^2\right )^{2/3}}+\frac {(x+1)^{2/3} x^{4/3} \log \left (\frac {\sqrt [3]{x}}{\sqrt [3]{x+1}}-1\right )}{6 \left (x^3+x^2\right )^{2/3}} \]

[In]

Int[(x^2 + x^3)^(1/3),x]

[Out]

(x^2 + x^3)^(1/3)/6 + (x*(x^2 + x^3)^(1/3))/2 + (x^(4/3)*(1 + x)^(2/3)*ArcTan[1/Sqrt[3] + (2*x^(1/3))/(Sqrt[3]
*(1 + x)^(1/3))])/(3*Sqrt[3]*(x^2 + x^3)^(2/3)) + (x^(4/3)*(1 + x)^(2/3)*Log[1 + x])/(18*(x^2 + x^3)^(2/3)) +
(x^(4/3)*(1 + x)^(2/3)*Log[-1 + x^(1/3)/(1 + x)^(1/3)])/(6*(x^2 + x^3)^(2/3))

Rule 61

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt
[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*
((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rule 2029

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Dist[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]

Rule 2049

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] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[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] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x \sqrt [3]{x^2+x^3}+\frac {1}{6} \int \frac {x^2}{\left (x^2+x^3\right )^{2/3}} \, dx \\ & = \frac {1}{6} \sqrt [3]{x^2+x^3}+\frac {1}{2} x \sqrt [3]{x^2+x^3}-\frac {1}{9} \int \frac {x}{\left (x^2+x^3\right )^{2/3}} \, dx \\ & = \frac {1}{6} \sqrt [3]{x^2+x^3}+\frac {1}{2} x \sqrt [3]{x^2+x^3}-\frac {\left (x^{4/3} (1+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} (1+x)^{2/3}} \, dx}{9 \left (x^2+x^3\right )^{2/3}} \\ & = \frac {1}{6} \sqrt [3]{x^2+x^3}+\frac {1}{2} x \sqrt [3]{x^2+x^3}+\frac {x^{4/3} (1+x)^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{1+x}}\right )}{3 \sqrt {3} \left (x^2+x^3\right )^{2/3}}+\frac {x^{4/3} (1+x)^{2/3} \log (1+x)}{18 \left (x^2+x^3\right )^{2/3}}+\frac {x^{4/3} (1+x)^{2/3} \log \left (-1+\frac {\sqrt [3]{x}}{\sqrt [3]{1+x}}\right )}{6 \left (x^2+x^3\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.29 \[ \int \sqrt [3]{x^2+x^3} \, dx=\frac {x^{4/3} (1+x)^{2/3} \left (3 x^{2/3} \sqrt [3]{1+x}+9 x^{5/3} \sqrt [3]{1+x}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2 \sqrt [3]{1+x}}\right )+2 \log \left (-\sqrt [3]{x}+\sqrt [3]{1+x}\right )-\log \left (x^{2/3}+\sqrt [3]{x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{18 \left (x^2 (1+x)\right )^{2/3}} \]

[In]

Integrate[(x^2 + x^3)^(1/3),x]

[Out]

(x^(4/3)*(1 + x)^(2/3)*(3*x^(2/3)*(1 + x)^(1/3) + 9*x^(5/3)*(1 + x)^(1/3) + 2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))
/(x^(1/3) + 2*(1 + x)^(1/3))] + 2*Log[-x^(1/3) + (1 + x)^(1/3)] - Log[x^(2/3) + x^(1/3)*(1 + x)^(1/3) + (1 + x
)^(2/3)]))/(18*(x^2*(1 + x))^(2/3))

Maple [C] (verified)

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

Time = 0.56 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.14

method	result	size
meijerg	\(\frac {3 x^{\frac {5}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x \right )}{5}\)	\(15\)
pseudoelliptic	\(-\frac {x^{4} \left (2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-9 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} x +\ln \left (\frac {\left (x^{2} \left (1+x \right )\right )^{\frac {2}{3}}+\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )-2 \ln \left (\frac {\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}-x}{x}\right )-3 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}\right )}{18 {\left (\left (x^{2} \left (1+x \right )\right )^{\frac {2}{3}}+\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} x +x^{2}\right )}^{2} {\left (\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}-x \right )}^{2}}\)	\(147\)
trager	\(\left (\frac {1}{6}+\frac {x}{2}\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+\frac {\ln \left (-\frac {9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +30 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +25 x^{2}+10 x}{x}\right )}{9}-\frac {\ln \left (\frac {45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+27 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x -57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-48 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -4 x^{2}-3 x}{x}\right )}{9}-\frac {\ln \left (\frac {45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+27 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x -57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-48 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -4 x^{2}-3 x}{x}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{3}\)	\(473\)
risch	\(\text {Expression too large to display}\)	\(659\)

[In]

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

[Out]

3/5*x^(5/3)*hypergeom([-1/3,5/3],[8/3],-x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \sqrt [3]{x^2+x^3} \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{6} \, {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}} {\left (3 \, x + 1\right )} + \frac {1}{9} \, \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} + x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

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

[Out]

-1/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + x^2)^(1/3))/x) + 1/6*(x^3 + x^2)^(1/3)*(3*x + 1) + 1/9*l
og(-(x - (x^3 + x^2)^(1/3))/x) - 1/18*log((x^2 + (x^3 + x^2)^(1/3)*x + (x^3 + x^2)^(2/3))/x^2)

Sympy [F]

\[ \int \sqrt [3]{x^2+x^3} \, dx=\int \sqrt [3]{x^{3} + x^{2}}\, dx \]

[In]

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

[Out]

Integral((x**3 + x**2)**(1/3), x)

Maxima [F]

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

[In]

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

[Out]

integrate((x^3 + x^2)^(1/3), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int \sqrt [3]{x^2+x^3} \, dx=\frac {1}{6} \, {\left ({\left (\frac {1}{x} + 1\right )}^{\frac {4}{3}} + 2 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )} x^{2} - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{18} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

1/6*((1/x + 1)^(4/3) + 2*(1/x + 1)^(1/3))*x^2 - 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x + 1)^(1/3) + 1)) - 1/18
*log((1/x + 1)^(2/3) + (1/x + 1)^(1/3) + 1) + 1/9*log(abs((1/x + 1)^(1/3) - 1))

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.23 \[ \int \sqrt [3]{x^2+x^3} \, dx=\frac {3\,x\,{\left (x^3+x^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {5}{3};\ \frac {8}{3};\ -x\right )}{5\,{\left (x+1\right )}^{1/3}} \]

[In]

int((x^2 + x^3)^(1/3),x)

[Out]

(3*x*(x^2 + x^3)^(1/3)*hypergeom([-1/3, 5/3], 8/3, -x))/(5*(x + 1)^(1/3))