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

Optimal. Leaf size=97 \[ \sqrt [3]{x^3+x^2}-\frac {1}{3} \log \left (\sqrt [3]{x^3+x^2}-x\right )+\frac {1}{6} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.06, antiderivative size = 142, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2021, 2032, 59} \begin {gather*} \sqrt [3]{x^3+x^2}-\frac {(x+1)^{2/3} x^{4/3} \log (x+1)}{6 \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 )}{2 \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 )}{\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,x]

[Out]

(x^2 + x^3)^(1/3) - (x^(4/3)*(1 + x)^(2/3)*ArcTan[1/Sqrt[3] + (2*x^(1/3))/(Sqrt[3]*(1 + x)^(1/3))])/(Sqrt[3]*(
x^2 + x^3)^(2/3)) - (x^(4/3)*(1 + x)^(2/3)*Log[1 + x])/(6*(x^2 + x^3)^(2/3)) - (x^(4/3)*(1 + x)^(2/3)*Log[-1 +
x^(1/3)/(1 + x)^(1/3)])/(2*(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 2021

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] + Dist[(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] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 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 \frac {\sqrt [3]{x^2+x^3}}{x} \, dx &=\sqrt [3]{x^2+x^3}+\frac {1}{3} \int \frac {x}{\left (x^2+x^3\right )^{2/3}} \, dx\\ &=\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}{3 \left (x^2+x^3\right )^{2/3}}\\ &=\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 )}{\sqrt {3} \left (x^2+x^3\right )^{2/3}}-\frac {x^{4/3} (1+x)^{2/3} \log (1+x)}{6 \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 )}{2 \left (x^2+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.36 \begin {gather*} \frac {3 \sqrt [3]{x^2 (x+1)} \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x\right )}{2 \sqrt [3]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + x^3)^(1/3)/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.23, size = 97, normalized size = 1.00 \begin {gather*} \sqrt [3]{x^2+x^3}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (-x+\sqrt [3]{x^2+x^3}\right )+\frac {1}{6} \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,x]

[Out]

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

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fricas [A]  time = 0.47, size = 93, normalized size = 0.96 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}} - \frac {1}{3} \, \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{6} \, \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,x, algorithm="fricas")

[Out]

1/3*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + x^2)^(1/3))/x) + (x^3 + x^2)^(1/3) - 1/3*log(-(x - (x^3 +
x^2)^(1/3))/x) + 1/6*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 = 0.33, size = 64, normalized size = 0.66 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + x {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + \frac {1}{6} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{3} \, \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,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.95, size = 15, normalized size = 0.15

method	result	size
meijerg	\(\frac {3 x^{\frac {2}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x \right )}{2}\)	\(15\)
trager	\(\left (x^{3}+x^{2}\right )^{\frac {1}{3}}-\frac {\ln \left (-\frac {9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x +24 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+9 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-3 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +16 x^{2}+12 x}{x}\right )}{3}+\RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (\frac {45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x -87 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-18 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +20 x^{2}+8 x}{x}\right )\)	\(318\)
risch	\(\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}+\frac {\left (-\frac {\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{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}-36 \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}}+96 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x +\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-14 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x -64 x^{2}+96 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-112 x -48}{1+x}\right )}{3}+\frac {\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{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}-30 \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}}+48 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}} x -2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-28 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x -10 x^{2}+48 \left (x^{3}+2 x^{2}+x \right )^{\frac {1}{3}}-9 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-14 x -4}{1+x}\right )}{6}\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 )}\)	\(443\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 + x^2)^(1/3)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3+x^2\right )}^{1/3}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^3)^(1/3)/x,x)

[Out]

int((x^2 + x^3)^(1/3)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2*(x + 1))**(1/3)/x, x)

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