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

Optimal. Leaf size=23 \[ \frac {3 \left (x^2-1\right ) \sqrt [3]{x^3-x}}{2 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1590} \begin {gather*} \frac {3 \left (x^3-x\right )^{4/3}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-x + x^3)^(4/3))/(2*x^2)

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.78 \begin {gather*} \frac {3 \left (x \left (x^2-1\right )\right )^{4/3}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x*(-1 + x^2))^(4/3))/(2*x^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 18, normalized size = 0.78 \begin {gather*} \frac {3 \left (-x+x^3\right )^{4/3}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-x + x^3)^(4/3))/(2*x^2)

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fricas [A]  time = 0.45, size = 19, normalized size = 0.83 \begin {gather*} \frac {3 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(x^3 - x)^(1/3)*(x^2 - 1)/x

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giac [A]  time = 0.30, size = 26, normalized size = 1.13 \begin {gather*} \frac {3}{2} \, x^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - \frac {3}{2} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 20, normalized size = 0.87

method	result	size
trager	\(\frac {3 \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{2 x}\)	\(20\)
gosper	\(\frac {3 \left (-1+x \right ) \left (1+x \right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{2 x}\)	\(21\)
risch	\(\frac {3 \left (x \left (x^{2}-1\right )\right )^{\frac {1}{3}} \left (x^{4}-2 x^{2}+1\right )}{2 x \left (x^{2}-1\right )}\)	\(32\)
meijerg	\(-\frac {3 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \hypergeom \left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], x^{2}\right )}{2 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{3}} x^{\frac {2}{3}}}+\frac {9 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{2}\right ) x^{\frac {4}{3}}}{4 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{3}}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 19, normalized size = 0.83 \begin {gather*} \frac {3\,{\left (x^3-x\right )}^{1/3}\,\left (x^2-1\right )}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(x^3 - x)^(1/3)*(x^2 - 1))/(2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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