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

Optimal. Leaf size=100 \[ -\frac {3 \sqrt [3]{x^3+x^2}}{x}-\log \left (\sqrt [3]{x^3+x^2}-x\right )+\frac {1}{2} \log \left (x^2+\sqrt [3]{x^3+x^2} x+\left (x^3+x^2\right )^{2/3}\right )-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x^2}+x}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 147, normalized size of antiderivative = 1.47, 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 = {2020, 2032, 59} \begin {gather*} -\frac {3 \sqrt [3]{x^3+x^2}}{x}-\frac {(x+1)^{2/3} x^{4/3} \log (x+1)}{2 \left (x^3+x^2\right )^{2/3}}-\frac {3 (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 {\sqrt {3} (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 )}{\left (x^3+x^2\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + x^3)^(1/3)/x^2,x]

[Out]

(-3*(x^2 + x^3)^(1/3))/x - (Sqrt[3]*x^(4/3)*(1 + x)^(2/3)*ArcTan[1/Sqrt[3] + (2*x^(1/3))/(Sqrt[3]*(1 + x)^(1/3
))])/(x^2 + x^3)^(2/3) - (x^(4/3)*(1 + x)^(2/3)*Log[1 + x])/(2*(x^2 + x^3)^(2/3)) - (3*x^(4/3)*(1 + x)^(2/3)*L
og[-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 2020

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 + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*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^2} \, dx &=-\frac {3 \sqrt [3]{x^2+x^3}}{x}+\int \frac {x}{\left (x^2+x^3\right )^{2/3}} \, dx\\ &=-\frac {3 \sqrt [3]{x^2+x^3}}{x}+\frac {\left (x^{4/3} (1+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} (1+x)^{2/3}} \, dx}{\left (x^2+x^3\right )^{2/3}}\\ &=-\frac {3 \sqrt [3]{x^2+x^3}}{x}-\frac {\sqrt {3} 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 )}{\left (x^2+x^3\right )^{2/3}}-\frac {x^{4/3} (1+x)^{2/3} \log (1+x)}{2 \left (x^2+x^3\right )^{2/3}}-\frac {3 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 = 36, normalized size = 0.36 \begin {gather*} -\frac {3 \sqrt [3]{x^2 (x+1)} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-x\right )}{x \sqrt [3]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + x^3)^(1/3)/x^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 100, normalized size = 1.00 \begin {gather*} -\frac {3 \sqrt [3]{x^3+x^2}}{x}-\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(x^2 + x^3)^(1/3))/x - 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 = 0.39, size = 102, 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 {1}{3}}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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)^(1/3))/x

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giac [A]  time = 0.32, size = 63, normalized size = 0.63 \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 ) - 3 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + \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((x^3+x^2)^(1/3)/x^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.51, size = 448, normalized size = 4.48 \begin {gather*} -\frac {3 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}}{x}+\frac {\left (-\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 )+\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 )}{2}\right ) \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} \left (x \left (1+x \right )^{2}\right )^{\frac {1}{3}}}{x \left (1+x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+x^2)^(1/3)/x^2,x)

[Out]

-3*(x^2*(1+x))^(1/3)/x+(-ln((-RootOf(_Z^2-2*_Z+4)^2*x^2+48*RootOf(_Z^2-2*_Z+4)*(x^3+2*x^2+x)^(2/3)-30*RootOf(_
Z^2-2*_Z+4)*(x^3+2*x^2+x)^(1/3)*x-16*RootOf(_Z^2-2*_Z+4)*x^2-36*(x^3+2*x^2+x)^(2/3)-30*RootOf(_Z^2-2*_Z+4)*(x^
3+2*x^2+x)^(1/3)+96*(x^3+2*x^2+x)^(1/3)*x+RootOf(_Z^2-2*_Z+4)^2-14*RootOf(_Z^2-2*_Z+4)*x-64*x^2+96*(x^3+2*x^2+
x)^(1/3)+2*RootOf(_Z^2-2*_Z+4)-112*x-48)/(1+x))+1/2*RootOf(_Z^2-2*_Z+4)*ln(-(2*RootOf(_Z^2-2*_Z+4)^2*x^2+24*Ro
otOf(_Z^2-2*_Z+4)*(x^3+2*x^2+x)^(2/3)-9*RootOf(_Z^2-2*_Z+4)*(x^3+2*x^2+x)^(1/3)*x-19*RootOf(_Z^2-2*_Z+4)*x^2-3
0*(x^3+2*x^2+x)^(2/3)-9*RootOf(_Z^2-2*_Z+4)*(x^3+2*x^2+x)^(1/3)+48*(x^3+2*x^2+x)^(1/3)*x-2*RootOf(_Z^2-2*_Z+4)
^2-28*RootOf(_Z^2-2*_Z+4)*x-10*x^2+48*(x^3+2*x^2+x)^(1/3)-9*RootOf(_Z^2-2*_Z+4)-14*x-4)/(1+x)))*(x^2*(1+x))^(1
/3)/x*(x*(1+x)^2)^(1/3)/(1+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^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 + x^2)^(1/3)/x^2, x)

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mupad [B]  time = 1.03, size = 27, normalized size = 0.27 \begin {gather*} -\frac {3\,{\left (x^2\,\left (x+1\right )\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ -x\right )}{x\,{\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^2,x)

[Out]

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

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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^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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