\(\int e^{-12+x} (e^{12}-e^{12-x}+8 x+4 x^2) \, dx\) [5521]

Optimal result

Integrand size = 27, antiderivative size = 24 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=-\frac {e^5}{5}+e^x-x+4 e^{-12+x} x^2 \]

[Out]

4*x^2/exp(12-x)+exp(x)-x-1/5*exp(5)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6820, 2225, 2227, 2207} \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=4 e^{x-12} x^2-x+e^x \]

[In]

Int[E^(-12 + x)*(E^12 - E^(12 - x) + 8*x + 4*x^2),x]

[Out]

E^x - x + 4*E^(-12 + x)*x^2

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]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+e^x+4 e^{-12+x} x (2+x)\right ) \, dx \\ & = -x+4 \int e^{-12+x} x (2+x) \, dx+\int e^x \, dx \\ & = e^x-x+4 \int \left (2 e^{-12+x} x+e^{-12+x} x^2\right ) \, dx \\ & = e^x-x+4 \int e^{-12+x} x^2 \, dx+8 \int e^{-12+x} x \, dx \\ & = e^x-x+8 e^{-12+x} x+4 e^{-12+x} x^2-8 \int e^{-12+x} \, dx-8 \int e^{-12+x} x \, dx \\ & = -8 e^{-12+x}+e^x-x+4 e^{-12+x} x^2+8 \int e^{-12+x} \, dx \\ & = e^x-x+4 e^{-12+x} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=e^x-x+4 e^{-12+x} x^2 \]

[In]

Integrate[E^(-12 + x)*(E^12 - E^(12 - x) + 8*x + 4*x^2),x]

[Out]

E^x - x + 4*E^(-12 + x)*x^2

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
risch	\(4 \,{\mathrm e}^{x -12} x^{2}+{\mathrm e}^{x}-x\)	\(16\)
norman	\(\left ({\mathrm e}^{2 x}-{\mathrm e}^{x} x +4 \,{\mathrm e}^{-12} x^{2} {\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}\)	\(29\)
parallelrisch	\(\left (4 x^{2}-{\mathrm e}^{12-x} x +{\mathrm e}^{12-x} {\mathrm e}^{x}\right ) {\mathrm e}^{x -12}\)	\(34\)
default	\(-x +8 \,{\mathrm e}^{-12} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+4 \,{\mathrm e}^{-12} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )+{\mathrm e}^{x}\)	\(44\)
parts	\(-x +4 \left (12-x \right )^{2} {\mathrm e}^{x -12}-96 \left (12-x \right ) {\mathrm e}^{x -12}+576 \,{\mathrm e}^{x -12}+{\mathrm e}^{x}\)	\(49\)
meijerg	\(-{\mathrm e}^{x -x \,{\mathrm e}^{-12}+12} \left (1-{\mathrm e}^{x \,{\mathrm e}^{-12}}\right )+\frac {{\mathrm e}^{x -x \,{\mathrm e}^{-12}+12} \left (1-{\mathrm e}^{x \,{\mathrm e}^{-12} \left (-{\mathrm e}^{12}+1\right )}\right )}{-{\mathrm e}^{12}+1}-4 \,{\mathrm e}^{24+x -x \,{\mathrm e}^{-12}} \left (2-\frac {\left (3 x^{2} {\mathrm e}^{-24}-6 x \,{\mathrm e}^{-12}+6\right ) {\mathrm e}^{x \,{\mathrm e}^{-12}}}{3}\right )+8 \,{\mathrm e}^{x -x \,{\mathrm e}^{-12}+12} \left (1-\frac {\left (2-2 x \,{\mathrm e}^{-12}\right ) {\mathrm e}^{x \,{\mathrm e}^{-12}}}{2}\right )\)	\(116\)

[In]

int((exp(12-x)*exp(x)-exp(12-x)+4*x^2+8*x)/exp(12-x),x,method=_RETURNVERBOSE)

[Out]

4*exp(x-12)*x^2+exp(x)-x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx={\left (4 \, x^{2} + e^{12}\right )} e^{\left (x - 12\right )} - x \]

[In]

integrate((exp(12-x)*exp(x)-exp(12-x)+4*x^2+8*x)/exp(12-x),x, algorithm="fricas")

[Out]

(4*x^2 + e^12)*e^(x - 12) - x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=- x + \frac {\left (4 x^{2} + e^{12}\right ) e^{x}}{e^{12}} \]

[In]

integrate((exp(12-x)*exp(x)-exp(12-x)+4*x**2+8*x)/exp(12-x),x)

[Out]

-x + (4*x**2 + exp(12))*exp(-12)*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=4 \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 12\right )} + 8 \, {\left (x - 1\right )} e^{\left (x - 12\right )} - x + e^{x} \]

[In]

integrate((exp(12-x)*exp(x)-exp(12-x)+4*x^2+8*x)/exp(12-x),x, algorithm="maxima")

[Out]

4*(x^2 - 2*x + 2)*e^(x - 12) + 8*(x - 1)*e^(x - 12) - x + e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=4 \, x^{2} e^{\left (x - 12\right )} - x + e^{x} \]

[In]

integrate((exp(12-x)*exp(x)-exp(12-x)+4*x^2+8*x)/exp(12-x),x, algorithm="giac")

[Out]

4*x^2*e^(x - 12) - x + e^x

Mupad [F(-1)]

Timed out. \[ \int e^{-12+x} \left (e^{12}-e^{12-x}+8 x+4 x^2\right ) \, dx=\text {Hanged} \]

[In]

int(exp(x - 12)*(8*x - exp(12 - x) + 4*x^2 + exp(12 - x)*exp(x)),x)

[Out]

\text{Hanged}