\(\int e^{1+x} (1+2 x+x^2+x^3) \, dx\) [6806]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx=1+e^{1+x} (1-x) \left (-5+x-x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 12, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2227, 2225, 2207} \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx=e^{x+1} x^3-2 e^{x+1} x^2+6 e^{x+1} x-5 e^{x+1} \]

[In]

Int[E^(1 + x)*(1 + 2*x + x^2 + x^3),x]

[Out]

-5*E^(1 + x) + 6*E^(1 + x)*x - 2*E^(1 + x)*x^2 + E^(1 + x)*x^3

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}& = \int \left (e^{1+x}+2 e^{1+x} x+e^{1+x} x^2+e^{1+x} x^3\right ) \, dx \\ & = 2 \int e^{1+x} x \, dx+\int e^{1+x} \, dx+\int e^{1+x} x^2 \, dx+\int e^{1+x} x^3 \, dx \\ & = e^{1+x}+2 e^{1+x} x+e^{1+x} x^2+e^{1+x} x^3-2 \int e^{1+x} \, dx-2 \int e^{1+x} x \, dx-3 \int e^{1+x} x^2 \, dx \\ & = -e^{1+x}-2 e^{1+x} x^2+e^{1+x} x^3+2 \int e^{1+x} \, dx+6 \int e^{1+x} x \, dx \\ & = e^{1+x}+6 e^{1+x} x-2 e^{1+x} x^2+e^{1+x} x^3-6 \int e^{1+x} \, dx \\ & = -5 e^{1+x}+6 e^{1+x} x-2 e^{1+x} x^2+e^{1+x} x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx=e^{1+x} \left (-5+6 x-2 x^2+x^3\right ) \]

[In]

Integrate[E^(1 + x)*(1 + 2*x + x^2 + x^3),x]

[Out]

E^(1 + x)*(-5 + 6*x - 2*x^2 + x^3)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
gosper	\(\left (x^{3}-2 x^{2}+6 x -5\right ) {\mathrm e}^{1+x}\)	\(19\)
risch	\(\left (x^{3}-2 x^{2}+6 x -5\right ) {\mathrm e}^{1+x}\)	\(19\)
norman	\(x^{3} {\mathrm e}^{1+x}+6 x \,{\mathrm e}^{1+x}-2 x^{2} {\mathrm e}^{1+x}-5 \,{\mathrm e}^{1+x}\)	\(32\)
parallelrisch	\(x^{3} {\mathrm e}^{1+x}+6 x \,{\mathrm e}^{1+x}-2 x^{2} {\mathrm e}^{1+x}-5 \,{\mathrm e}^{1+x}\)	\(32\)
derivativedivides	\({\mathrm e}^{1+x} \left (1+x \right )^{3}-5 \,{\mathrm e}^{1+x} \left (1+x \right )^{2}+13 \,{\mathrm e}^{1+x} \left (1+x \right )-14 \,{\mathrm e}^{1+x}\)	\(38\)
default	\({\mathrm e}^{1+x} \left (1+x \right )^{3}-5 \,{\mathrm e}^{1+x} \left (1+x \right )^{2}+13 \,{\mathrm e}^{1+x} \left (1+x \right )-14 \,{\mathrm e}^{1+x}\)	\(38\)
parts	\(x^{3} {\mathrm e}^{1+x}+x^{2} {\mathrm e}^{1+x}+2 x \,{\mathrm e}^{1+x}-12 \,{\mathrm e}^{1+x}+10 \,{\mathrm e}^{1+x} \left (1+x \right )-3 \,{\mathrm e}^{1+x} \left (1+x \right )^{2}\)	\(51\)
meijerg	\({\mathrm e} \left (6-\frac {\left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}}{4}\right )-{\mathrm e} \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )+2 \,{\mathrm e} \left (1-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}\right )-\left (1-{\mathrm e}^{x}\right ) {\mathrm e}\)	\(71\)

[In]

int((x^3+x^2+2*x+1)*exp(1+x),x,method=_RETURNVERBOSE)

[Out]

(x^3-2*x^2+6*x-5)*exp(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx={\left (x^{3} - 2 \, x^{2} + 6 \, x - 5\right )} e^{\left (x + 1\right )} \]

[In]

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

[Out]

(x^3 - 2*x^2 + 6*x - 5)*e^(x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx=\left (x^{3} - 2 x^{2} + 6 x - 5\right ) e^{x + 1} \]

[In]

integrate((x**3+x**2+2*x+1)*exp(1+x),x)

[Out]

(x**3 - 2*x**2 + 6*x - 5)*exp(x + 1)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (18) = 36\).

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.00 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx={\left (x^{3} e - 3 \, x^{2} e + 6 \, x e - 6 \, e\right )} e^{x} + {\left (x^{2} e - 2 \, x e + 2 \, e\right )} e^{x} + 2 \, {\left (x e - e\right )} e^{x} + e^{\left (x + 1\right )} \]

[In]

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

[Out]

(x^3*e - 3*x^2*e + 6*x*e - 6*e)*e^x + (x^2*e - 2*x*e + 2*e)*e^x + 2*(x*e - e)*e^x + e^(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx={\left (x^{3} - 2 \, x^{2} + 6 \, x - 5\right )} e^{\left (x + 1\right )} \]

[In]

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

[Out]

(x^3 - 2*x^2 + 6*x - 5)*e^(x + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{1+x} \left (1+2 x+x^2+x^3\right ) \, dx={\mathrm {e}}^{x+1}\,\left (x^3-2\,x^2+6\,x-5\right ) \]

[In]

int(exp(x + 1)*(2*x + x^2 + x^3 + 1),x)

[Out]

exp(x + 1)*(6*x - 2*x^2 + x^3 - 5)