\(\int \frac {1}{25} (-8 x+10 e x-15 x^2) \, dx\) [5988]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=\frac {1}{25} (-4+5 (e-x)) x^2 \]

[Out]

x^2*(1/5*exp(1)-1/5*x-4/25)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6, 12} \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=-\frac {x^3}{5}-\frac {1}{25} (4-5 e) x^2 \]

[In]

Int[(-8*x + 10*E*x - 15*x^2)/25,x]

[Out]

-1/25*((4 - 5*E)*x^2) - x^3/5

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{25} \left ((-8+10 e) x-15 x^2\right ) \, dx \\ & = \frac {1}{25} \int \left ((-8+10 e) x-15 x^2\right ) \, dx \\ & = -\frac {1}{25} (4-5 e) x^2-\frac {x^3}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=\frac {1}{25} (-4+5 e-5 x) x^2 \]

[In]

Integrate[(-8*x + 10*E*x - 15*x^2)/25,x]

[Out]

((-4 + 5*E - 5*x)*x^2)/25

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
gosper	\(\frac {x^{2} \left (-5 x +5 \,{\mathrm e}-4\right )}{25}\)	\(15\)
norman	\(\left (\frac {{\mathrm e}}{5}-\frac {4}{25}\right ) x^{2}-\frac {x^{3}}{5}\)	\(17\)
default	\(-\frac {x^{3}}{5}+\frac {\left (10 \,{\mathrm e}-8\right ) x^{2}}{50}\)	\(18\)
risch	\(\frac {x^{2} {\mathrm e}}{5}-\frac {x^{3}}{5}-\frac {4 x^{2}}{25}\)	\(19\)
parallelrisch	\(\frac {x^{2} {\mathrm e}}{5}-\frac {x^{3}}{5}-\frac {4 x^{2}}{25}\)	\(19\)
parts	\(\frac {x^{2} {\mathrm e}}{5}-\frac {x^{3}}{5}-\frac {4 x^{2}}{25}\)	\(19\)

[In]

int(2/5*x*exp(1)-3/5*x^2-8/25*x,x,method=_RETURNVERBOSE)

[Out]

1/25*x^2*(-5*x+5*exp(1)-4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=-\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \]

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="fricas")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=- \frac {x^{3}}{5} + x^{2} \left (- \frac {4}{25} + \frac {e}{5}\right ) \]

[In]

integrate(2/5*x*exp(1)-3/5*x**2-8/25*x,x)

[Out]

-x**3/5 + x**2*(-4/25 + E/5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=-\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \]

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="maxima")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=-\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \]

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="giac")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{25} \left (-8 x+10 e x-15 x^2\right ) \, dx=-\frac {x^2\,\left (5\,x-5\,\mathrm {e}+4\right )}{25} \]

[In]

int((2*x*exp(1))/5 - (8*x)/25 - (3*x^2)/5,x)

[Out]

-(x^2*(5*x - 5*exp(1) + 4))/25