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

Optimal. Leaf size=30 \[ \frac {3 \left (9 x^2+6 x+5\right ) \left (x^3-x^2\right )^{2/3}}{40 x^4} \]

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Rubi [B]  time = 0.08, antiderivative size = 61, normalized size of antiderivative = 2.03, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2016, 2014} \begin {gather*} \frac {27 \left (x^3-x^2\right )^{2/3}}{40 x^2}+\frac {9 \left (x^3-x^2\right )^{2/3}}{20 x^3}+\frac {3 \left (x^3-x^2\right )^{2/3}}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-x^2 + x^3)^(1/3)),x]

[Out]

(3*(-x^2 + x^3)^(2/3))/(8*x^4) + (9*(-x^2 + x^3)^(2/3))/(20*x^3) + (27*(-x^2 + x^3)^(2/3))/(40*x^2)

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt [3]{-x^2+x^3}} \, dx &=\frac {3 \left (-x^2+x^3\right )^{2/3}}{8 x^4}+\frac {3}{4} \int \frac {1}{x^2 \sqrt [3]{-x^2+x^3}} \, dx\\ &=\frac {3 \left (-x^2+x^3\right )^{2/3}}{8 x^4}+\frac {9 \left (-x^2+x^3\right )^{2/3}}{20 x^3}+\frac {9}{20} \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx\\ &=\frac {3 \left (-x^2+x^3\right )^{2/3}}{8 x^4}+\frac {9 \left (-x^2+x^3\right )^{2/3}}{20 x^3}+\frac {27 \left (-x^2+x^3\right )^{2/3}}{40 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.93 \begin {gather*} \frac {3 \left ((x-1) x^2\right )^{2/3} \left (9 x^2+6 x+5\right )}{40 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*((-1 + x)*x^2)^(2/3)*(5 + 6*x + 9*x^2))/(40*x^4)

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IntegrateAlgebraic [A]  time = 0.17, size = 30, normalized size = 1.00 \begin {gather*} \frac {3 \left (5+6 x+9 x^2\right ) \left (-x^2+x^3\right )^{2/3}}{40 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5 + 6*x + 9*x^2)*(-x^2 + x^3)^(2/3))/(40*x^4)

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fricas [A]  time = 0.45, size = 26, normalized size = 0.87 \begin {gather*} \frac {3 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}} {\left (9 \, x^{2} + 6 \, x + 5\right )}}{40 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(x^3 - x^2)^(2/3)*(9*x^2 + 6*x + 5)/x^4

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giac [A]  time = 0.73, size = 41, normalized size = 1.37 \begin {gather*} \frac {3}{8} \, {\left (\frac {1}{x} - 1\right )}^{2} {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} - \frac {6}{5} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{3}} + \frac {3}{2} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^3-x^2)^(1/3),x, algorithm="giac")

[Out]

3/8*(1/x - 1)^2*(-1/x + 1)^(2/3) - 6/5*(-1/x + 1)^(5/3) + 3/2*(-1/x + 1)^(2/3)

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maple [A]  time = 0.07, size = 27, normalized size = 0.90

method	result	size
trager	\(\frac {3 \left (9 x^{2}+6 x +5\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}}{40 x^{4}}\)	\(27\)
gosper	\(\frac {3 \left (-1+x \right ) \left (9 x^{2}+6 x +5\right )}{40 x^{2} \left (x^{3}-x^{2}\right )^{\frac {1}{3}}}\)	\(30\)
risch	\(\frac {-\frac {3}{8}-\frac {3}{40} x -\frac {9}{40} x^{2}+\frac {27}{40} x^{3}}{x^{2} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}\)	\(30\)
meijerg	\(-\frac {3 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{3}} \left (\frac {9}{5} x^{2}+\frac {6}{5} x +1\right ) \left (1-x \right )^{\frac {2}{3}}}{8 \mathrm {signum}\left (-1+x \right )^{\frac {1}{3}} x^{\frac {8}{3}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(9*x^2+6*x+5)*(x^3-x^2)^(2/3)/x^4

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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}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^3 - x^2)^(1/3)*x^3), x)

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mupad [B]  time = 0.19, size = 49, normalized size = 1.63 \begin {gather*} \frac {27\,x^2\,{\left (x^3-x^2\right )}^{2/3}+18\,x\,{\left (x^3-x^2\right )}^{2/3}+15\,{\left (x^3-x^2\right )}^{2/3}}{40\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(x^3 - x^2)^(1/3)),x)

[Out]

(27*x^2*(x^3 - x^2)^(2/3) + 18*x*(x^3 - x^2)^(2/3) + 15*(x^3 - x^2)^(2/3))/(40*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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