\(\int \frac {-4+e^3 (-5+4 x-125 x^4)}{e^3} \, dx\) [8216]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=x \left (-5-\frac {4}{e^3}+2 x-25 x^4\right ) \]

[Out]

x*(2*x-5-4/exp(3)-25*x^4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12} \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-25 x^5+2 x^2-\frac {4 x}{e^3}-5 x \]

[In]

Int[(-4 + E^3*(-5 + 4*x - 125*x^4))/E^3,x]

[Out]

-5*x - (4*x)/E^3 + 2*x^2 - 25*x^5

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}& = \frac {\int \left (-4+e^3 \left (-5+4 x-125 x^4\right )\right ) \, dx}{e^3} \\ & = -\frac {4 x}{e^3}+\int \left (-5+4 x-125 x^4\right ) \, dx \\ & = -5 x-\frac {4 x}{e^3}+2 x^2-25 x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-5 x-\frac {4 x}{e^3}+2 x^2-25 x^5 \]

[In]

Integrate[(-4 + E^3*(-5 + 4*x - 125*x^4))/E^3,x]

[Out]

-5*x - (4*x)/E^3 + 2*x^2 - 25*x^5

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18

method result size
risch \(-25 x^{5}+2 x^{2}-5 x -4 \,{\mathrm e}^{-3} x\) \(20\)
norman \(2 x^{2}-25 x^{5}-{\mathrm e}^{-3} \left (5 \,{\mathrm e}^{3}+4\right ) x\) \(25\)
gosper \(-x \left (25 x^{4} {\mathrm e}^{3}-2 x \,{\mathrm e}^{3}+5 \,{\mathrm e}^{3}+4\right ) {\mathrm e}^{-3}\) \(26\)
parallelrisch \({\mathrm e}^{-3} \left ({\mathrm e}^{3} \left (-25 x^{5}+2 x^{2}-5 x \right )-4 x \right )\) \(27\)
default \({\mathrm e}^{-3} \left (-25 x^{5} {\mathrm e}^{3}+2 x^{2} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3}-4 x \right )\) \(29\)

[In]

int(((-125*x^4+4*x-5)*exp(3)-4)/exp(3),x,method=_RETURNVERBOSE)

[Out]

-25*x^5+2*x^2-5*x-4*exp(-3)*x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]

[In]

integrate(((-125*x^4+4*x-5)*exp(3)-4)/exp(3),x, algorithm="fricas")

[Out]

-((25*x^5 - 2*x^2 + 5*x)*e^3 + 4*x)*e^(-3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=- 25 x^{5} + 2 x^{2} + \frac {x \left (- 5 e^{3} - 4\right )}{e^{3}} \]

[In]

integrate(((-125*x**4+4*x-5)*exp(3)-4)/exp(3),x)

[Out]

-25*x**5 + 2*x**2 + x*(-5*exp(3) - 4)*exp(-3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]

[In]

integrate(((-125*x^4+4*x-5)*exp(3)-4)/exp(3),x, algorithm="maxima")

[Out]

-((25*x^5 - 2*x^2 + 5*x)*e^3 + 4*x)*e^(-3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-{\left ({\left (25 \, x^{5} - 2 \, x^{2} + 5 \, x\right )} e^{3} + 4 \, x\right )} e^{\left (-3\right )} \]

[In]

integrate(((-125*x^4+4*x-5)*exp(3)-4)/exp(3),x, algorithm="giac")

[Out]

-((25*x^5 - 2*x^2 + 5*x)*e^3 + 4*x)*e^(-3)

Mupad [B] (verification not implemented)

Time = 12.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {-4+e^3 \left (-5+4 x-125 x^4\right )}{e^3} \, dx=-25\,x^5+2\,x^2+\left (-4\,{\mathrm {e}}^{-3}-5\right )\,x \]

[In]

int(-exp(-3)*(exp(3)*(125*x^4 - 4*x + 5) + 4),x)

[Out]

2*x^2 - 25*x^5 - x*(4*exp(-3) + 5)