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

Optimal. Leaf size=111 \[ \frac {\left (x^3+x^2\right )^{2/3} (3 x-4)}{6 x}-\frac {2}{9} \log \left (\sqrt [3]{x^3+x^2}-x\right )+\frac {1}{9} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 164, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2024, 2011, 59} \begin {gather*} -\frac {2 \left (x^3+x^2\right )^{2/3}}{3 x}+\frac {1}{2} \left (x^3+x^2\right )^{2/3}-\frac {x^{2/3} \sqrt [3]{x+1} \log (x)}{9 \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 )}{3 \sqrt [3]{x^3+x^2}}-\frac {2 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 )}{3 \sqrt {3} \sqrt [3]{x^3+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(x^2 + x^3)^(1/3),x]

[Out]

(x^2 + x^3)^(2/3)/2 - (2*(x^2 + x^3)^(2/3))/(3*x) - (2*x^(2/3)*(1 + x)^(1/3)*ArcTan[1/Sqrt[3] + (2*(1 + x)^(1/
3))/(Sqrt[3]*x^(1/3))])/(3*Sqrt[3]*(x^2 + x^3)^(1/3)) - (x^(2/3)*(1 + x)^(1/3)*Log[x])/(9*(x^2 + x^3)^(1/3)) -
(x^(2/3)*(1 + x)^(1/3)*Log[-1 + (1 + x)^(1/3)/x^(1/3)])/(3*(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^2}{\sqrt [3]{x^2+x^3}} \, dx &=\frac {1}{2} \left (x^2+x^3\right )^{2/3}-\frac {2}{3} \int \frac {x}{\sqrt [3]{x^2+x^3}} \, dx\\ &=\frac {1}{2} \left (x^2+x^3\right )^{2/3}-\frac {2 \left (x^2+x^3\right )^{2/3}}{3 x}+\frac {2}{9} \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx\\ &=\frac {1}{2} \left (x^2+x^3\right )^{2/3}-\frac {2 \left (x^2+x^3\right )^{2/3}}{3 x}+\frac {\left (2 x^{2/3} \sqrt [3]{1+x}\right ) \int \frac {1}{x^{2/3} \sqrt [3]{1+x}} \, dx}{9 \sqrt [3]{x^2+x^3}}\\ &=\frac {1}{2} \left (x^2+x^3\right )^{2/3}-\frac {2 \left (x^2+x^3\right )^{2/3}}{3 x}-\frac {2 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 )}{3 \sqrt {3} \sqrt [3]{x^2+x^3}}-\frac {x^{2/3} \sqrt [3]{1+x} \log (x)}{9 \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 )}{3 \sqrt [3]{x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.34 \begin {gather*} \frac {3 x^3 \sqrt [3]{x+1} \, _2F_1\left (\frac {1}{3},\frac {7}{3};\frac {10}{3};-x\right )}{7 \sqrt [3]{x^2 (x+1)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(x^2 + x^3)^(1/3),x]

[Out]

(3*x^3*(1 + x)^(1/3)*Hypergeometric2F1[1/3, 7/3, 10/3, -x])/(7*(x^2*(1 + x))^(1/3))

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IntegrateAlgebraic [A]  time = 0.29, size = 111, normalized size = 1.00 \begin {gather*} \frac {(-4+3 x) \left (x^2+x^3\right )^{2/3}}{6 x}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^3}}\right )}{3 \sqrt {3}}-\frac {2}{9} \log \left (-x+\sqrt [3]{x^2+x^3}\right )+\frac {1}{9} \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^2 + x^3)^(1/3),x]

[Out]

((-4 + 3*x)*(x^2 + x^3)^(2/3))/(6*x) + (2*ArcTan[(Sqrt[3]*x)/(x + 2*(x^2 + x^3)^(1/3))])/(3*Sqrt[3]) - (2*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)]/9

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fricas [A]  time = 0.47, size = 108, normalized size = 0.97 \begin {gather*} -\frac {4 \, \sqrt {3} x \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + 4 \, x \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - 2 \, 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 ) - 3 \, {\left (x^{3} + x^{2}\right )}^{\frac {2}{3}} {\left (3 \, x - 4\right )}}{18 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 79, normalized size = 0.71 \begin {gather*} -\frac {1}{6} \, {\left (4 \, {\left (\frac {1}{x} + 1\right )}^{\frac {5}{3}} - 7 \, {\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}}\right )} x^{2} - \frac {2}{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}{9} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {2}{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^2/(x^3+x^2)^(1/3),x, algorithm="giac")

[Out]

-1/6*(4*(1/x + 1)^(5/3) - 7*(1/x + 1)^(2/3))*x^2 - 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x + 1)^(1/3) + 1)) + 1
/9*log((1/x + 1)^(2/3) + (1/x + 1)^(1/3) + 1) - 2/9*log(abs((1/x + 1)^(1/3) - 1))

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maple [C]  time = 0.60, size = 15, normalized size = 0.14

method	result	size
meijerg	\(\frac {3 x^{\frac {7}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {7}{3}\right ], \left [\frac {10}{3}\right ], -x \right )}{7}\)	\(15\)
risch	\(\frac {\left (-4+3 x \right ) x \left (1+x \right )}{6 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}}+\frac {2 x^{\frac {1}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x \right )}{3}\)	\(36\)
trager	\(\frac {\left (-4+3 x \right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}}{6 x}-\frac {2 \ln \left (\frac {-36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x +36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x +60 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-18 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x -25 x^{2}-10 x}{x}\right )}{9}+\frac {4 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \ln \left (-\frac {-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x +180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x -114 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x +4 x^{2}+3 x}{x}\right )}{3}\)	\(329\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x^(7/3)*hypergeom([1/3,7/3],[10/3],-x)

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

[Out]

integrate(x^2/(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^2}{{\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^2/(x^2 + x^3)^(1/3),x)

[Out]

int(x^2/(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^{2}}{\sqrt [3]{x^{2} \left (x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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