3.12.9 \(\int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx\)

Optimal. Leaf size=83 \[ -\log \left (\sqrt [3]{x^3+x^2}-x\right )+\frac {1}{2} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 129, normalized size of antiderivative = 1.55, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2011, 59} \begin {gather*} -\frac {x^{2/3} \sqrt [3]{x+1} \log (x)}{2 \sqrt [3]{x^3+x^2}}-\frac {3 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 {\sqrt {3} 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]{x^3+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + x^3)^(-1/3),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx &=\frac {\left (x^{2/3} \sqrt [3]{1+x}\right ) \int \frac {1}{x^{2/3} \sqrt [3]{1+x}} \, dx}{\sqrt [3]{x^2+x^3}}\\ &=-\frac {\sqrt {3} 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]{x^2+x^3}}-\frac {x^{2/3} \sqrt [3]{1+x} \log (x)}{2 \sqrt [3]{x^2+x^3}}-\frac {3 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 = 34, normalized size = 0.41 \begin {gather*} \frac {3 x \sqrt [3]{x+1} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};-x\right )}{\sqrt [3]{x^2 (x+1)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + x^3)^(-1/3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 83, normalized size = 1.00 \begin {gather*} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{x^2+x^3}\right )+\frac {1}{2} \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]

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

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fricas [A]  time = 0.43, size = 84, normalized size = 1.01 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \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(1/(x^3+x^2)^(1/3),x, algorithm="fricas")

[Out]

-sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + x^2)^(1/3))/x) - log(-(x - (x^3 + x^2)^(1/3))/x) + 1/2*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 = 3.03, size = 55, normalized size = 0.66 \begin {gather*} -\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}{2} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \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(1/(x^3+x^2)^(1/3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.48, size = 15, normalized size = 0.18

method	result	size
meijerg	\(3 x^{\frac {1}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x \right )\)	\(15\)
trager	\(\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x +4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +19 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+9 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+9 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +5 x^{2}+2 x}{x}\right )-\ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +11 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+17 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -20 x^{2}-15 x}{x}\right ) \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+\ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +11 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}+17 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -20 x^{2}-15 x}{x}\right )\)	\(427\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\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(1/(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 = 0.95, size = 25, normalized size = 0.30 \begin {gather*} \frac {3\,x\,{\left (x+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x\right )}{{\left (x^3+x^2\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 + x^3)^(1/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**3 + x**2)**(-1/3), x)

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