\(\int ((-3+x) x)^{2/3} (-3+2 x) \, dx\) [139]

Optimal result

Integrand size = 15, antiderivative size = 16 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} (-((3-x) x))^{5/3} \]

[Out]

3/5*(-(3-x)*x)^(5/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1602} \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} (-((3-x) x))^{5/3} \]

[In]

Int[((-3 + x)*x)^(2/3)*(-3 + 2*x),x]

[Out]

(3*(-((3 - x)*x))^(5/3))/5

Rule 1602

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*} \text {integral}& = \frac {3}{5} (-((3-x) x))^{5/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} ((-3+x) x)^{5/3} \]

[In]

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

[Out]

(3*((-3 + x)*x)^(5/3))/5

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {3 \left (\left (-3+x \right ) x \right )^{\frac {5}{3}}}{5}\)	\(10\)
default	\(\frac {3 \left (\left (-3+x \right ) x \right )^{\frac {5}{3}}}{5}\)	\(10\)
gosper	\(\frac {3 \left (-3+x \right ) x \left (\left (-3+x \right ) x \right )^{\frac {2}{3}}}{5}\)	\(14\)
pseudoelliptic	\(\frac {3 \left (-3+x \right ) x \left (\left (-3+x \right ) x \right )^{\frac {2}{3}}}{5}\)	\(14\)
trager	\(\frac {3 \left (-3+x \right ) x \left (x^{2}-3 x \right )^{\frac {2}{3}}}{5}\)	\(16\)
risch	\(\frac {3 \left (-3+x \right )^{2} x^{2}}{5 \left (\left (-3+x \right ) x \right )^{\frac {1}{3}}}\)	\(18\)
meijerg	\(-\frac {9 \,3^{\frac {2}{3}} \operatorname {signum}\left (-3+x \right )^{\frac {2}{3}} x^{\frac {5}{3}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {2}{3},\frac {5}{3};\frac {8}{3};\frac {x}{3}\right )}{5 \left (-\operatorname {signum}\left (-3+x \right )\right )^{\frac {2}{3}}}+\frac {3 \,3^{\frac {2}{3}} \operatorname {signum}\left (-3+x \right )^{\frac {2}{3}} x^{\frac {8}{3}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {2}{3},\frac {8}{3};\frac {11}{3};\frac {x}{3}\right )}{4 \left (-\operatorname {signum}\left (-3+x \right )\right )^{\frac {2}{3}}}\)	\(64\)

[In]

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

[Out]

3/5*((-3+x)*x)^(5/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} \, {\left (x^{2} - 3 \, x\right )}^{\frac {5}{3}} \]

[In]

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

[Out]

3/5*(x^2 - 3*x)^(5/3)

Sympy [A] (verification not implemented)

Time = 2.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3 \left (x \left (x - 3\right )\right )^{\frac {5}{3}}}{5} \]

[In]

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

[Out]

3*(x*(x - 3))**(5/3)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.56 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} \, \left ({\left (x - 3\right )} x\right )^{\frac {5}{3}} \]

[In]

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

[Out]

3/5*((x - 3)*x)^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3}{5} \, {\left (x^{2} - 3 \, x\right )}^{\frac {5}{3}} \]

[In]

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

[Out]

3/5*(x^2 - 3*x)^(5/3)

Mupad [B] (verification not implemented)

Time = 12.35 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int ((-3+x) x)^{2/3} (-3+2 x) \, dx=\frac {3\,x\,{\left (x\,\left (x-3\right )\right )}^{2/3}\,\left (x-3\right )}{5} \]

[In]

int((2*x - 3)*(x*(x - 3))^(2/3),x)

[Out]

(3*x*(x*(x - 3))^(2/3)*(x - 3))/5