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

Optimal. Leaf size=101 \[ \frac {\left (x^3+x^2\right )^{2/3}}{x}+\frac {1}{3} \log \left (\sqrt [3]{x^3+x^2}-x\right )-\frac {1}{6} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 143, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2024, 2011, 59} \begin {gather*} \frac {\left (x^3+x^2\right )^{2/3}}{x}+\frac {x^{2/3} \sqrt [3]{x+1} \log (x)}{6 \sqrt [3]{x^3+x^2}}+\frac {x^{2/3} \sqrt [3]{x+1} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{x}}-1\right )}{2 \sqrt [3]{x^3+x^2}}+\frac {x^{2/3} \sqrt [3]{x+1} \tan ^{-1}\left (\frac {2 \sqrt [3]{x+1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{x^3+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(x^2 + x^3)^(1/3),x]

[Out]

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

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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

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]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt [3]{x^2+x^3}} \, dx &=\frac {\left (x^2+x^3\right )^{2/3}}{x}-\frac {1}{3} \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx\\ &=\frac {\left (x^2+x^3\right )^{2/3}}{x}-\frac {\left (x^{2/3} \sqrt [3]{1+x}\right ) \int \frac {1}{x^{2/3} \sqrt [3]{1+x}} \, dx}{3 \sqrt [3]{x^2+x^3}}\\ &=\frac {\left (x^2+x^3\right )^{2/3}}{x}+\frac {x^{2/3} \sqrt [3]{1+x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{x}}\right )}{\sqrt {3} \sqrt [3]{x^2+x^3}}+\frac {x^{2/3} \sqrt [3]{1+x} \log (x)}{6 \sqrt [3]{x^2+x^3}}+\frac {x^{2/3} \sqrt [3]{1+x} \log \left (-1+\frac {\sqrt [3]{1+x}}{\sqrt [3]{x}}\right )}{2 \sqrt [3]{x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.35 \begin {gather*} \frac {3 \left (x^2 (x+1)\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {4}{3};\frac {7}{3};-x\right )}{4 (x+1)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(x^2 + x^3)^(1/3),x]

[Out]

(3*(x^2*(1 + x))^(2/3)*Hypergeometric2F1[1/3, 4/3, 7/3, -x])/(4*(1 + x)^(2/3))

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IntegrateAlgebraic [A]  time = 0.17, size = 101, normalized size = 1.00 \begin {gather*} \frac {\left (x^2+x^3\right )^{2/3}}{x}-\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/(x^2 + x^3)^(1/3),x]

[Out]

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

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fricas [A]  time = 0.45, size = 103, normalized size = 1.02 \begin {gather*} \frac {2 \, \sqrt {3} x \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + 2 \, x \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - x \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 ) + 6 \, {\left (x^{3} + x^{2}\right )}^{\frac {2}{3}}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 64, normalized size = 0.63 \begin {gather*} x {\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + \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 ) - \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/(x^3+x^2)^(1/3),x, algorithm="giac")

[Out]

x*(1/x + 1)^(2/3) + 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x + 1)^(1/3) + 1)) - 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 = 0.53, size = 15, normalized size = 0.15

method	result	size
meijerg	\(\frac {3 x^{\frac {4}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -x \right )}{4}\)	\(15\)
risch	\(\frac {x \left (1+x \right )}{\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}}-x^{\frac {1}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x \right )\)	\(30\)
trager	\(\frac {\left (x^{3}+x^{2}\right )^{\frac {2}{3}}}{x}+\frac {\ln \left (-\frac {36 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-144 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -36 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x +60 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-18 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x +25 x^{2}+10 x}{x}\right )}{3}+2 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \ln \left (\frac {180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x -114 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x -4 x^{2}-3 x}{x}\right )\)	\(323\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*x^(4/3)*hypergeom([1/3,4/3],[7/3],-x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\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/(x^3+x^2)^(1/3),x, algorithm="maxima")

[Out]

integrate(x/(x^3 + x^2)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2 + x^3)^(1/3),x)

[Out]

int(x/(x^2 + x^3)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x**2*(x + 1))**(1/3), x)

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