\(\int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx\) [6681]

   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 = 26, antiderivative size = 18 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=e^x-\frac {5 (-16+x) (-2+x)}{e^5 x^3} \]

[Out]

exp(x)-5*(-2+x)/x^3/exp(5)*(x-16)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 14, 2225} \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=-\frac {160}{e^5 x^3}+\frac {90}{e^5 x^2}+e^x-\frac {5}{e^5 x} \]

[In]

Int[(480 - 180*x + 5*x^2 + E^(5 + x)*x^4)/(E^5*x^4),x]

[Out]

E^x - 160/(E^5*x^3) + 90/(E^5*x^2) - 5/(E^5*x)

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {480-180 x+5 x^2+e^{5+x} x^4}{x^4} \, dx}{e^5} \\ & = \frac {\int \left (e^{5+x}+\frac {5 \left (96-36 x+x^2\right )}{x^4}\right ) \, dx}{e^5} \\ & = \frac {\int e^{5+x} \, dx}{e^5}+\frac {5 \int \frac {96-36 x+x^2}{x^4} \, dx}{e^5} \\ & = e^x+\frac {5 \int \left (\frac {96}{x^4}-\frac {36}{x^3}+\frac {1}{x^2}\right ) \, dx}{e^5} \\ & = e^x-\frac {160}{e^5 x^3}+\frac {90}{e^5 x^2}-\frac {5}{e^5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {e^{5+x}-\frac {160}{x^3}+\frac {90}{x^2}-\frac {5}{x}}{e^5} \]

[In]

Integrate[(480 - 180*x + 5*x^2 + E^(5 + x)*x^4)/(E^5*x^4),x]

[Out]

(E^(5 + x) - 160/x^3 + 90/x^2 - 5/x)/E^5

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method result size
risch \(\frac {{\mathrm e}^{-5} \left (-5 x^{2}+90 x -160\right )}{x^{3}}+{\mathrm e}^{x}\) \(20\)
parts \(5 \,{\mathrm e}^{-5} \left (-\frac {1}{x}+\frac {18}{x^{2}}-\frac {32}{x^{3}}\right )+{\mathrm e}^{x}\) \(26\)
default \({\mathrm e}^{-5} \left ({\mathrm e}^{5} {\mathrm e}^{x}-\frac {160}{x^{3}}+\frac {90}{x^{2}}-\frac {5}{x}\right )\) \(27\)
parallelrisch \(\frac {{\mathrm e}^{-5} \left (-160+90 x -5 x^{2}+x^{3} {\mathrm e}^{5} {\mathrm e}^{x}\right )}{x^{3}}\) \(27\)
norman \(\frac {{\mathrm e}^{x} x^{3}-5 x^{2} {\mathrm e}^{-5}+90 \,{\mathrm e}^{-5} x -160 \,{\mathrm e}^{-5}}{x^{3}}\) \(34\)

[In]

int((x^4*exp(5)*exp(x)+5*x^2-180*x+480)/x^4/exp(5),x,method=_RETURNVERBOSE)

[Out]

exp(-5)*(-5*x^2+90*x-160)/x^3+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {{\left (x^{3} e^{\left (x + 5\right )} - 5 \, x^{2} + 90 \, x - 160\right )} e^{\left (-5\right )}}{x^{3}} \]

[In]

integrate((x^4*exp(5)*exp(x)+5*x^2-180*x+480)/x^4/exp(5),x, algorithm="fricas")

[Out]

(x^3*e^(x + 5) - 5*x^2 + 90*x - 160)*e^(-5)/x^3

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=e^{x} + \frac {- 5 x^{2} + 90 x - 160}{x^{3} e^{5}} \]

[In]

integrate((x**4*exp(5)*exp(x)+5*x**2-180*x+480)/x**4/exp(5),x)

[Out]

exp(x) + (-5*x**2 + 90*x - 160)*exp(-5)/x**3

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=-{\left (\frac {5}{x} - \frac {90}{x^{2}} + \frac {160}{x^{3}} - e^{\left (x + 5\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate((x^4*exp(5)*exp(x)+5*x^2-180*x+480)/x^4/exp(5),x, algorithm="maxima")

[Out]

-(5/x - 90/x^2 + 160/x^3 - e^(x + 5))*e^(-5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx=\frac {{\left (x^{3} e^{\left (x + 5\right )} - 5 \, x^{2} + 90 \, x - 160\right )} e^{\left (-5\right )}}{x^{3}} \]

[In]

integrate((x^4*exp(5)*exp(x)+5*x^2-180*x+480)/x^4/exp(5),x, algorithm="giac")

[Out]

(x^3*e^(x + 5) - 5*x^2 + 90*x - 160)*e^(-5)/x^3

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {480-180 x+5 x^2+e^{5+x} x^4}{e^5 x^4} \, dx={\mathrm {e}}^x-\frac {5\,{\mathrm {e}}^{-5}\,x^2-90\,{\mathrm {e}}^{-5}\,x+160\,{\mathrm {e}}^{-5}}{x^3} \]

[In]

int((exp(-5)*(5*x^2 - 180*x + x^4*exp(5)*exp(x) + 480))/x^4,x)

[Out]

exp(x) - (160*exp(-5) - 90*x*exp(-5) + 5*x^2*exp(-5))/x^3