\(\int \frac {1}{2} e^{-x} (e^x (30+e)-30 x+15 x^2) \, dx\) [371]

Optimal result

Integrand size = 25, antiderivative size = 27 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=\frac {1}{2} \left (e x+5 \left (6 x-3 \left (1+e^{-x} x^2\right )\right )\right ) \]

[Out]

1/2*x*exp(1)+15*x-15/2*x^2/exp(x)-15/2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 6820, 2227, 2207, 2225} \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=\frac {1}{2} (30+e) x-\frac {15}{2} e^{-x} x^2 \]

[In]

Int[(E^x*(30 + E) - 30*x + 15*x^2)/(2*E^x),x]

[Out]

((30 + E)*x)/2 - (15*x^2)/(2*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 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}& = \frac {1}{2} \int e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx \\ & = \frac {1}{2} \int \left (30+e+15 e^{-x} (-2+x) x\right ) \, dx \\ & = \frac {1}{2} (30+e) x+\frac {15}{2} \int e^{-x} (-2+x) x \, dx \\ & = \frac {1}{2} (30+e) x+\frac {15}{2} \int \left (-2 e^{-x} x+e^{-x} x^2\right ) \, dx \\ & = \frac {1}{2} (30+e) x+\frac {15}{2} \int e^{-x} x^2 \, dx-15 \int e^{-x} x \, dx \\ & = 15 e^{-x} x+\frac {1}{2} (30+e) x-\frac {15}{2} e^{-x} x^2-15 \int e^{-x} \, dx+15 \int e^{-x} x \, dx \\ & = 15 e^{-x}+\frac {1}{2} (30+e) x-\frac {15}{2} e^{-x} x^2+15 \int e^{-x} \, dx \\ & = \frac {1}{2} (30+e) x-\frac {15}{2} e^{-x} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=\frac {1}{2} \left (30 x+e x-15 e^{-x} x^2\right ) \]

[In]

Integrate[(E^x*(30 + E) - 30*x + 15*x^2)/(2*E^x),x]

[Out]

(30*x + E*x - (15*x^2)/E^x)/2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {x \,{\mathrm e}}{2}+15 x -\frac {15 x^{2} {\mathrm e}^{-x}}{2}\)	\(19\)
risch	\(\frac {x \,{\mathrm e}}{2}+15 x -\frac {15 x^{2} {\mathrm e}^{-x}}{2}\)	\(19\)
parts	\(\frac {x \,{\mathrm e}}{2}+15 x -\frac {15 x^{2} {\mathrm e}^{-x}}{2}\)	\(19\)
norman	\(\left (\left (\frac {{\mathrm e}}{2}+15\right ) x \,{\mathrm e}^{x}-\frac {15 x^{2}}{2}\right ) {\mathrm e}^{-x}\)	\(22\)
parallelrisch	\(\frac {\left (x \,{\mathrm e} \,{\mathrm e}^{x}-15 x^{2}+30 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{2}\)	\(24\)

[In]

int(1/2*((exp(1)+30)*exp(x)+15*x^2-30*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

15*x-15/2*x^2/exp(x)+1/2*x*exp(1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=-\frac {1}{2} \, {\left (15 \, x^{2} - {\left (x e + 30 \, x\right )} e^{x}\right )} e^{\left (-x\right )} \]

[In]

integrate(1/2*((exp(1)+30)*exp(x)+15*x^2-30*x)/exp(x),x, algorithm="fricas")

[Out]

-1/2*(15*x^2 - (x*e + 30*x)*e^x)*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=- \frac {15 x^{2} e^{- x}}{2} + x \left (\frac {e}{2} + 15\right ) \]

[In]

integrate(1/2*((exp(1)+30)*exp(x)+15*x**2-30*x)/exp(x),x)

[Out]

-15*x**2*exp(-x)/2 + x*(E/2 + 15)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=\frac {1}{2} \, x e - \frac {15}{2} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + 15 \, {\left (x + 1\right )} e^{\left (-x\right )} + 15 \, x \]

[In]

integrate(1/2*((exp(1)+30)*exp(x)+15*x^2-30*x)/exp(x),x, algorithm="maxima")

[Out]

1/2*x*e - 15/2*(x^2 + 2*x + 2)*e^(-x) + 15*(x + 1)*e^(-x) + 15*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=-\frac {15}{2} \, x^{2} e^{\left (-x\right )} + \frac {1}{2} \, x e + 15 \, x \]

[In]

integrate(1/2*((exp(1)+30)*exp(x)+15*x^2-30*x)/exp(x),x, algorithm="giac")

[Out]

-15/2*x^2*e^(-x) + 1/2*x*e + 15*x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {1}{2} e^{-x} \left (e^x (30+e)-30 x+15 x^2\right ) \, dx=\frac {x\,\left (\mathrm {e}-15\,x\,{\mathrm {e}}^{-x}+30\right )}{2} \]

[In]

int(exp(-x)*((exp(x)*(exp(1) + 30))/2 - 15*x + (15*x^2)/2),x)

[Out]

(x*(exp(1) - 15*x*exp(-x) + 30))/2