\(\int (162+e^x (-2-2 x)-4 x) \, dx\) [8036]

Optimal result

Integrand size = 14, antiderivative size = 11 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-24-2 x \left (-81+e^x+x\right ) \]

[Out]

-24-2*(x-81+exp(x))*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 2.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2207, 2225} \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2 x^2+162 x+2 e^x-2 e^x (x+1) \]

[In]

Int[162 + E^x*(-2 - 2*x) - 4*x,x]

[Out]

2*E^x + 162*x - 2*x^2 - 2*E^x*(1 + 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]

Rubi steps \begin{align*} \text {integral}& = 162 x-2 x^2+\int e^x (-2-2 x) \, dx \\ & = 162 x-2 x^2-2 e^x (1+x)+2 \int e^x \, dx \\ & = 2 e^x+162 x-2 x^2-2 e^x (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2 \left (-81 x+e^x x+x^2\right ) \]

[In]

Integrate[162 + E^x*(-2 - 2*x) - 4*x,x]

[Out]

-2*(-81*x + E^x*x + x^2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36

method	result	size
default	\(162 x -2 \,{\mathrm e}^{x} x -2 x^{2}\)	\(15\)
norman	\(162 x -2 \,{\mathrm e}^{x} x -2 x^{2}\)	\(15\)
risch	\(162 x -2 \,{\mathrm e}^{x} x -2 x^{2}\)	\(15\)
parallelrisch	\(162 x -2 \,{\mathrm e}^{x} x -2 x^{2}\)	\(15\)
parts	\(162 x -2 \,{\mathrm e}^{x} x -2 x^{2}\)	\(15\)

[In]

int((-2-2*x)*exp(x)-4*x+162,x,method=_RETURNVERBOSE)

[Out]

162*x-2*exp(x)*x-2*x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2 \, x^{2} - 2 \, x e^{x} + 162 \, x \]

[In]

integrate((-2-2*x)*exp(x)-4*x+162,x, algorithm="fricas")

[Out]

-2*x^2 - 2*x*e^x + 162*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=- 2 x^{2} - 2 x e^{x} + 162 x \]

[In]

integrate((-2-2*x)*exp(x)-4*x+162,x)

[Out]

-2*x**2 - 2*x*exp(x) + 162*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2 \, x^{2} - 2 \, x e^{x} + 162 \, x \]

[In]

integrate((-2-2*x)*exp(x)-4*x+162,x, algorithm="maxima")

[Out]

-2*x^2 - 2*x*e^x + 162*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2 \, x^{2} - 2 \, x e^{x} + 162 \, x \]

[In]

integrate((-2-2*x)*exp(x)-4*x+162,x, algorithm="giac")

[Out]

-2*x^2 - 2*x*e^x + 162*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \left (162+e^x (-2-2 x)-4 x\right ) \, dx=-2\,x\,\left (x+{\mathrm {e}}^x-81\right ) \]

[In]

int(162 - exp(x)*(2*x + 2) - 4*x,x)

[Out]

-2*x*(x + exp(x) - 81)