3.312 \(\int (-3 x+2 x^3) (-3 x^2+x^4)^{3/5} \, dx\)

Optimal. Leaf size=17 \[ \frac {5}{16} \left (x^4-3 x^2\right )^{8/5} \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1588} \[ \frac {5}{16} \left (x^4-3 x^2\right )^{8/5} \]

Antiderivative was successfully verified.

[In]

Int[(-3*x + 2*x^3)*(-3*x^2 + x^4)^(3/5),x]

[Out]

(5*(-3*x^2 + x^4)^(8/5))/16

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \left (-3 x+2 x^3\right ) \left (-3 x^2+x^4\right )^{3/5} \, dx &=\frac {5}{16} \left (-3 x^2+x^4\right )^{8/5}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 75, normalized size = 4.41 \[ \frac {5 \left (x^2 \left (x^2-3\right )\right )^{3/5} \left (16 x^4 \, _2F_1\left (-\frac {3}{5},\frac {13}{5};\frac {18}{5};\frac {x^2}{3}\right )-39 x^2 \, _2F_1\left (-\frac {3}{5},\frac {8}{5};\frac {13}{5};\frac {x^2}{3}\right )\right )}{208 \left (1-\frac {x^2}{3}\right )^{3/5}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-3*x + 2*x^3)*(-3*x^2 + x^4)^(3/5),x]

[Out]

(5*(x^2*(-3 + x^2))^(3/5)*(-39*x^2*Hypergeometric2F1[-3/5, 8/5, 13/5, x^2/3] + 16*x^4*Hypergeometric2F1[-3/5,
13/5, 18/5, x^2/3]))/(208*(1 - x^2/3)^(3/5))

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IntegrateAlgebraic [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {5}{16} \left (x^4-3 x^2\right )^{8/5} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-3*x + 2*x^3)*(-3*x^2 + x^4)^(3/5),x]

[Out]

(5*(-3*x^2 + x^4)^(8/5))/16

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fricas [A]  time = 0.60, size = 13, normalized size = 0.76 \[ \frac {5}{16} \, {\left (x^{4} - 3 \, x^{2}\right )}^{\frac {8}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-3*x)*(x^4-3*x^2)^(3/5),x, algorithm="fricas")

[Out]

5/16*(x^4 - 3*x^2)^(8/5)

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giac [A]  time = 0.63, size = 13, normalized size = 0.76 \[ \frac {5}{16} \, {\left (x^{4} - 3 \, x^{2}\right )}^{\frac {8}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-3*x)*(x^4-3*x^2)^(3/5),x, algorithm="giac")

[Out]

5/16*(x^4 - 3*x^2)^(8/5)

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maple [A]  time = 0.31, size = 14, normalized size = 0.82

method	result	size
default	\(\frac {5 \left (x^{4}-3 x^{2}\right )^{\frac {8}{5}}}{16}\)	\(14\)
gosper	\(\frac {5 \left (x^{4}-3 x^{2}\right )^{\frac {3}{5}} x^{2} \left (x^{2}-3\right )}{16}\)	\(22\)
trager	\(\frac {5 \left (x^{4}-3 x^{2}\right )^{\frac {3}{5}} x^{2} \left (x^{2}-3\right )}{16}\)	\(22\)
risch	\(\frac {5 x^{2} \left (x^{2} \left (x^{2}-3\right )\right )^{\frac {3}{5}} \left (x^{2}-3\right )}{16}\)	\(22\)
meijerg	\(\frac {5 \,3^{\frac {3}{5}} \mathrm {signum}\left (-1+\frac {x^{2}}{3}\right )^{\frac {3}{5}} x^{\frac {26}{5}} \hypergeom \left (\left [-\frac {3}{5}, \frac {13}{5}\right ], \left [\frac {18}{5}\right ], \frac {x^{2}}{3}\right )}{13 \left (-\mathrm {signum}\left (-1+\frac {x^{2}}{3}\right )\right )^{\frac {3}{5}}}-\frac {15 \,3^{\frac {3}{5}} \mathrm {signum}\left (-1+\frac {x^{2}}{3}\right )^{\frac {3}{5}} x^{\frac {16}{5}} \hypergeom \left (\left [-\frac {3}{5}, \frac {8}{5}\right ], \left [\frac {13}{5}\right ], \frac {x^{2}}{3}\right )}{16 \left (-\mathrm {signum}\left (-1+\frac {x^{2}}{3}\right )\right )^{\frac {3}{5}}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3-3*x)*(x^4-3*x^2)^(3/5),x,method=_RETURNVERBOSE)

[Out]

5/16*(x^4-3*x^2)^(8/5)

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maxima [A]  time = 0.60, size = 13, normalized size = 0.76 \[ \frac {5}{16} \, {\left (x^{4} - 3 \, x^{2}\right )}^{\frac {8}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-3*x)*(x^4-3*x^2)^(3/5),x, algorithm="maxima")

[Out]

5/16*(x^4 - 3*x^2)^(8/5)

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mupad [B]  time = 0.28, size = 21, normalized size = 1.24 \[ \frac {5\,x^2\,\left (x^2-3\right )\,{\left (x^4-3\,x^2\right )}^{3/5}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x - 2*x^3)*(x^4 - 3*x^2)^(3/5),x)

[Out]

(5*x^2*(x^2 - 3)*(x^4 - 3*x^2)^(3/5))/16

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sympy [B]  time = 1.31, size = 36, normalized size = 2.12 \[ \frac {5 x^{4} \left (x^{4} - 3 x^{2}\right )^{\frac {3}{5}}}{16} - \frac {15 x^{2} \left (x^{4} - 3 x^{2}\right )^{\frac {3}{5}}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3-3*x)*(x**4-3*x**2)**(3/5),x)

[Out]

5*x**4*(x**4 - 3*x**2)**(3/5)/16 - 15*x**2*(x**4 - 3*x**2)**(3/5)/16

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