\(\int \frac {e^{-x} (-e^{4+x}+x^2+2 x^3-x^4)}{5 x^2} \, dx\) [6981]

Optimal result

Integrand size = 33, antiderivative size = 23 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {1}{5} e^{-x} \left (-1+\frac {e^{4+x}}{x}+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 6874, 2225, 2207} \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {1}{5} e^{-x} x^2-\frac {e^{-x}}{5}+\frac {e^4}{5 x} \]

[In]

Int[(-E^(4 + x) + x^2 + 2*x^3 - x^4)/(5*E^x*x^2),x]

[Out]

-1/5*1/E^x + E^4/(5*x) + x^2/(5*E^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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {e^4}{5 x}-\frac {1}{5} e^{-x} \left (1-x^2\right ) \]

[In]

Integrate[(-E^(4 + x) + x^2 + 2*x^3 - x^4)/(5*E^x*x^2),x]

[Out]

E^4/(5*x) - (1 - x^2)/(5*E^x)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {{\mathrm e}^{4}}{5 x}+\frac {\left (x^{2}-1\right ) {\mathrm e}^{-x}}{5}\)	\(20\)
parallelrisch	\(\frac {\left (x^{3}+{\mathrm e}^{4} {\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}}{5 x}\)	\(22\)
default	\(-\frac {{\mathrm e}^{-x}}{5}+\frac {x^{2} {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4}}{5 x}\)	\(24\)
norman	\(\frac {\left (-\frac {x}{5}+\frac {x^{3}}{5}+\frac {{\mathrm e}^{4} {\mathrm e}^{x}}{5}\right ) {\mathrm e}^{-x}}{x}\)	\(24\)
parts	\(-\frac {{\mathrm e}^{-x}}{5}+\frac {x^{2} {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4}}{5 x}\)	\(24\)

[In]

int(1/5*(-exp(4)*exp(x)-x^4+2*x^3+x^2)/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {{\left ({\left (x^{3} - x\right )} e^{4} + e^{\left (x + 8\right )}\right )} e^{\left (-x - 4\right )}}{5 \, x} \]

[In]

integrate(1/5*(-exp(4)*exp(x)-x^4+2*x^3+x^2)/exp(x)/x^2,x, algorithm="fricas")

[Out]

1/5*((x^3 - x)*e^4 + e^(x + 8))*e^(-x - 4)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {\left (x^{2} - 1\right ) e^{- x}}{5} + \frac {e^{4}}{5 x} \]

[In]

integrate(1/5*(-exp(4)*exp(x)-x**4+2*x**3+x**2)/exp(x)/x**2,x)

[Out]

(x**2 - 1)*exp(-x)/5 + exp(4)/(5*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {1}{5} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} - \frac {2}{5} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {e^{4}}{5 \, x} - \frac {1}{5} \, e^{\left (-x\right )} \]

[In]

integrate(1/5*(-exp(4)*exp(x)-x^4+2*x^3+x^2)/exp(x)/x^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {{\left (x + 4\right )}^{3} e^{\left (-x\right )} - 12 \, {\left (x + 4\right )}^{2} e^{\left (-x\right )} + 47 \, {\left (x + 4\right )} e^{\left (-x\right )} + e^{4} - 60 \, e^{\left (-x\right )}}{5 \, x} \]

[In]

integrate(1/5*(-exp(4)*exp(x)-x^4+2*x^3+x^2)/exp(x)/x^2,x, algorithm="giac")

[Out]

1/5*((x + 4)^3*e^(-x) - 12*(x + 4)^2*e^(-x) + 47*(x + 4)*e^(-x) + e^4 - 60*e^(-x))/x

Mupad [B] (verification not implemented)

Time = 11.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-e^{4+x}+x^2+2 x^3-x^4\right )}{5 x^2} \, dx=\frac {x^2\,{\mathrm {e}}^{-x}}{5}-\frac {{\mathrm {e}}^{-x}}{5}+\frac {{\mathrm {e}}^4}{5\,x} \]

[In]

int(-(exp(-x)*((exp(4)*exp(x))/5 - x^2/5 - (2*x^3)/5 + x^4/5))/x^2,x)

[Out]

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