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

Optimal. Leaf size=108 \[ \frac {1}{6} \sqrt [3]{x^3+x^2} (3 x+1)+\frac {1}{9} \log \left (\sqrt [3]{x^3+x^2}-x\right )-\frac {1}{18} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 164, normalized size of antiderivative = 1.52, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2004, 2024, 2032, 59} \begin {gather*} \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}}+\frac {(x+1)^{2/3} x^{4/3} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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 59

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

Rule 2004

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 2024

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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*} \int \sqrt [3]{x^2+x^3} \, dx &=\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} \tan ^{-1}\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*}

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.33 \begin {gather*} \frac {3 x \sqrt [3]{x^2 (x+1)} \, _2F_1\left (-\frac {1}{3},\frac {5}{3};\frac {8}{3};-x\right )}{5 \sqrt [3]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + x^3)^(1/3),x]

[Out]

(3*x*(x^2*(1 + x))^(1/3)*Hypergeometric2F1[-1/3, 5/3, 8/3, -x])/(5*(1 + x)^(1/3))

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IntegrateAlgebraic [A]  time = 0.19, size = 108, normalized size = 1.00 \begin {gather*} \frac {1}{6} (1+3 x) \sqrt [3]{x^2+x^3}+\frac {\tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2 + x^3)^(1/3),x]

[Out]

((1 + 3*x)*(x^2 + x^3)^(1/3))/6 + ArcTan[(Sqrt[3]*x)/(x + 2*(x^2 + x^3)^(1/3))]/(3*Sqrt[3]) + Log[-x + (x^2 +
x^3)^(1/3)]/9 - Log[x^2 + x*(x^2 + x^3)^(1/3) + (x^2 + x^3)^(2/3)]/18

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fricas [A]  time = 0.63, size = 100, normalized size = 0.93 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.18, size = 77, normalized size = 0.71 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 1.04, size = 15, normalized size = 0.14

method	result	size
meijerg	\(\frac {3 x^{\frac {5}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x \right )}{5}\)	\(15\)
risch	\(\frac {\left (1+3 x \right ) \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}}{6}+\frac {\left (\frac {\ln \left (-\frac {x^{2} \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )^{2}+48 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {2}{3}}-30 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x -16 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) x^{2}-\RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )^{2}-30 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}-14 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) x +36 \left (x^{3}+2 x^{2}+x \right )^{\frac {2}{3}}-96 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x +64 x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )-96 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}+112 x +48}{1+x}\right )}{9}+\frac {\RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \ln \left (-\frac {2 x^{2} \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )^{2}-24 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {2}{3}}+9 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x +19 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )^{2}+9 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}+28 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right ) x -30 \left (x^{3}+2 x^{2}+x \right )^{\frac {2}{3}}+48 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x -10 x^{2}+9 \RootOf \left (\textit {\_Z}^{2}+2 \textit {\_Z} +4\right )+48 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}-14 x -4}{1+x}\right )}{18}\right ) \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} \left (\left (1+x \right )^{2} x \right )^{\frac {1}{3}}}{x \left (1+x \right )}\)	\(452\)
trager	\(\left (\frac {1}{6}+\frac {x}{2}\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+\frac {\RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {-36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+45 \left (x^{3}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +33 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+51 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +20 x^{2}+15 x}{x}\right )}{3}-\frac {\ln \left (-\frac {36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+45 \left (x^{3}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +57 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-9 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -5 x^{2}-2 x}{x}\right )}{9}-\frac {\ln \left (-\frac {36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+45 \left (x^{3}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +57 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-9 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -5 x^{2}-2 x}{x}\right ) \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{3}\)	\(485\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 + x^2)^(1/3), x)

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mupad [B]  time = 1.13, size = 25, normalized size = 0.23 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{x^{3} + x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**3 + x**2)**(1/3), x)

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