\(\int \frac {1+e^{8+x} (5-x-x^2)}{e^8} \, dx\) [7191]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=\frac {3+x}{e^8}-e^x \left (-4-x+x^2\right ) \]

[Out]

1/exp(1)^8*(3+x)-(x^2-x-4)*exp(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=-e^x x^2+e^x x+\frac {x}{e^8}+4 e^x \]

[In]

Int[(1 + E^(8 + x)*(5 - x - x^2))/E^8,x]

[Out]

4*E^x + x/E^8 + E^x*x - E^x*x^2

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 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (1+e^{8+x} \left (5-x-x^2\right )\right ) \, dx}{e^8} \\ & = \frac {x}{e^8}+\frac {\int e^{8+x} \left (5-x-x^2\right ) \, dx}{e^8} \\ & = \frac {x}{e^8}+\frac {\int \left (5 e^{8+x}-e^{8+x} x-e^{8+x} x^2\right ) \, dx}{e^8} \\ & = \frac {x}{e^8}-\frac {\int e^{8+x} x \, dx}{e^8}-\frac {\int e^{8+x} x^2 \, dx}{e^8}+\frac {5 \int e^{8+x} \, dx}{e^8} \\ & = 5 e^x+\frac {x}{e^8}-e^x x-e^x x^2+\frac {\int e^{8+x} \, dx}{e^8}+\frac {2 \int e^{8+x} x \, dx}{e^8} \\ & = 6 e^x+\frac {x}{e^8}+e^x x-e^x x^2-\frac {2 \int e^{8+x} \, dx}{e^8} \\ & = 4 e^x+\frac {x}{e^8}+e^x x-e^x x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=\frac {x}{e^8}-e^x \left (-4-x+x^2\right ) \]

[In]

Integrate[(1 + E^(8 + x)*(5 - x - x^2))/E^8,x]

[Out]

x/E^8 - E^x*(-4 - x + x^2)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(-{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+x \,{\mathrm e}^{-8}\)	\(21\)
parts	\(-{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+x \,{\mathrm e}^{-8}\)	\(23\)
default	\({\mathrm e}^{-8} \left (x +{\mathrm e}^{8} \left ({\mathrm e}^{x} x +4 \,{\mathrm e}^{x}-{\mathrm e}^{x} x^{2}\right )\right )\)	\(29\)
parallelrisch	\({\mathrm e}^{-8} \left (x +{\mathrm e}^{8} \left ({\mathrm e}^{x} x +4 \,{\mathrm e}^{x}-{\mathrm e}^{x} x^{2}\right )\right )\)	\(29\)
norman	\(\left ({\mathrm e}^{-1} x +{\mathrm e}^{7} x \,{\mathrm e}^{x}+4 \,{\mathrm e}^{7} {\mathrm e}^{x}-x^{2} {\mathrm e}^{7} {\mathrm e}^{x}\right ) {\mathrm e}^{-7}\)	\(40\)

[In]

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

[Out]

-exp(x)*x^2+exp(x)*x+4*exp(x)+x*exp(-8)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=-{\left ({\left (x^{2} - x - 4\right )} e^{\left (x + 8\right )} - x\right )} e^{\left (-8\right )} \]

[In]

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

[Out]

-((x^2 - x - 4)*e^(x + 8) - x)*e^(-8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=\frac {x}{e^{8}} + \left (- x^{2} + x + 4\right ) e^{x} \]

[In]

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

[Out]

x*exp(-8) + (-x**2 + x + 4)*exp(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=-{\left ({\left (x^{2} e^{8} - 2 \, x e^{8} + 2 \, e^{8}\right )} e^{x} + {\left (x e^{8} - e^{8}\right )} e^{x} - x - 5 \, e^{\left (x + 8\right )}\right )} e^{\left (-8\right )} \]

[In]

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

[Out]

-((x^2*e^8 - 2*x*e^8 + 2*e^8)*e^x + (x*e^8 - e^8)*e^x - x - 5*e^(x + 8))*e^(-8)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=-{\left ({\left (x^{2} - x - 4\right )} e^{\left (x + 8\right )} - x\right )} e^{\left (-8\right )} \]

[In]

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

[Out]

-((x^2 - x - 4)*e^(x + 8) - x)*e^(-8)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1+e^{8+x} \left (5-x-x^2\right )}{e^8} \, dx=4\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x+x\,{\mathrm {e}}^{-8}+x\,{\mathrm {e}}^x \]

[In]

int(-exp(-8)*(exp(8)*exp(x)*(x + x^2 - 5) - 1),x)

[Out]

4*exp(x) - x^2*exp(x) + x*exp(-8) + x*exp(x)