\(\int \frac {1}{3} (-2+2 e^{5/3}+2 x) \, dx\) [62]

Optimal result

Integrand size = 16, antiderivative size = 18 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {1}{3} \left (1-e^{5/3}-x\right )^2 \]

[Out]

1/3*(1-exp(5/3)-x)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {1}{3} \left (-x-e^{5/3}+1\right )^2 \]

[In]

Int[(-2 + 2*E^(5/3) + 2*x)/3,x]

[Out]

(1 - E^(5/3) - x)^2/3

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \left (1-e^{5/3}-x\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {2}{3} \left (-x+e^{5/3} x+\frac {x^2}{2}\right ) \]

[In]

Integrate[(-2 + 2*E^(5/3) + 2*x)/3,x]

[Out]

(2*(-x + E^(5/3)*x + x^2/2))/3

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61

method	result	size
gosper	\(\frac {x \left (x +2 \,{\mathrm e}^{\frac {5}{3}}-2\right )}{3}\)	\(11\)
default	\(\frac {2 x \,{\mathrm e}^{\frac {5}{3}}}{3}+\frac {x^{2}}{3}-\frac {2 x}{3}\)	\(15\)
norman	\(\left (\frac {2 \,{\mathrm e}^{\frac {5}{3}}}{3}-\frac {2}{3}\right ) x +\frac {x^{2}}{3}\)	\(15\)
risch	\(\frac {2 x \,{\mathrm e}^{\frac {5}{3}}}{3}+\frac {x^{2}}{3}-\frac {2 x}{3}\)	\(15\)
parallelrisch	\(\left (\frac {2 \,{\mathrm e}^{\frac {5}{3}}}{3}-\frac {2}{3}\right ) x +\frac {x^{2}}{3}\)	\(15\)
parts	\(\frac {2 x \,{\mathrm e}^{\frac {5}{3}}}{3}+\frac {x^{2}}{3}-\frac {2 x}{3}\)	\(15\)

[In]

int(2/3*exp(5/3)+2/3*x-2/3,x,method=_RETURNVERBOSE)

[Out]

1/3*x*(x+2*exp(5/3)-2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {2}{3} \, x e^{\frac {5}{3}} - \frac {2}{3} \, x \]

[In]

integrate(2/3*exp(5/3)+2/3*x-2/3,x, algorithm="fricas")

[Out]

1/3*x^2 + 2/3*x*e^(5/3) - 2/3*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {x^{2}}{3} + x \left (- \frac {2}{3} + \frac {2 e^{\frac {5}{3}}}{3}\right ) \]

[In]

integrate(2/3*exp(5/3)+2/3*x-2/3,x)

[Out]

x**2/3 + x*(-2/3 + 2*exp(5/3)/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {2}{3} \, x e^{\frac {5}{3}} - \frac {2}{3} \, x \]

[In]

integrate(2/3*exp(5/3)+2/3*x-2/3,x, algorithm="maxima")

[Out]

1/3*x^2 + 2/3*x*e^(5/3) - 2/3*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {2}{3} \, x e^{\frac {5}{3}} - \frac {2}{3} \, x \]

[In]

integrate(2/3*exp(5/3)+2/3*x-2/3,x, algorithm="giac")

[Out]

1/3*x^2 + 2/3*x*e^(5/3) - 2/3*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {1}{3} \left (-2+2 e^{5/3}+2 x\right ) \, dx=\frac {x\,\left (x+2\,{\mathrm {e}}^{5/3}-2\right )}{3} \]

[In]

int((2*x)/3 + (2*exp(5/3))/3 - 2/3,x)

[Out]

(x*(x + 2*exp(5/3) - 2))/3