\(\int e^{-x} (180 x+e^x (5+10 x)) \, dx\) [5528]

Optimal result

Integrand size = 19, antiderivative size = 18 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=\frac {19}{5}+5 (1+x) \left (-36 e^{-x}+x\right ) \]

[Out]

5*(1+x)*(x-36/exp(x))+19/5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {6820, 14, 2207, 2225} \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 x^2-180 e^{-x} x+5 x-180 e^{-x} \]

[In]

Int[(180*x + E^x*(5 + 10*x))/E^x,x]

[Out]

-180/E^x + 5*x - (180*x)/E^x + 5*x^2

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 (5+\left (10+180 e^{-x}\right ) x\right ) \, dx \\ & = 5 x+\int \left (10+180 e^{-x}\right ) x \, dx \\ & = 5 x+\int \left (10 x+180 e^{-x} x\right ) \, dx \\ & = 5 x+5 x^2+180 \int e^{-x} x \, dx \\ & = 5 x-180 e^{-x} x+5 x^2+180 \int e^{-x} \, dx \\ & = -180 e^{-x}+5 x-180 e^{-x} x+5 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 e^{-x} (1+x) \left (-36+e^x x\right ) \]

[In]

Integrate[(180*x + E^x*(5 + 10*x))/E^x,x]

[Out]

(5*(1 + x)*(-36 + E^x*x))/E^x

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
risch	\(5 x^{2}+5 x +\left (-180 x -180\right ) {\mathrm e}^{-x}\)	\(20\)
default	\(-180 x \,{\mathrm e}^{-x}+5 x^{2}-180 \,{\mathrm e}^{-x}+5 x\)	\(23\)
norman	\(\left (-180-180 x +5 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}\)	\(23\)
parallelrisch	\(\left (-180-180 x +5 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}\)	\(23\)
parts	\(-180 x \,{\mathrm e}^{-x}+5 x^{2}-180 \,{\mathrm e}^{-x}+5 x\)	\(23\)

[In]

int(((10*x+5)*exp(x)+180*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

5*x^2+5*x+(-180*x-180)*exp(-x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 \, {\left ({\left (x^{2} + x\right )} e^{x} - 36 \, x - 36\right )} e^{\left (-x\right )} \]

[In]

integrate(((10*x+5)*exp(x)+180*x)/exp(x),x, algorithm="fricas")

[Out]

5*((x^2 + x)*e^x - 36*x - 36)*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 x^{2} + 5 x + \left (- 180 x - 180\right ) e^{- x} \]

[In]

integrate(((10*x+5)*exp(x)+180*x)/exp(x),x)

[Out]

5*x**2 + 5*x + (-180*x - 180)*exp(-x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 \, x^{2} - 180 \, {\left (x + 1\right )} e^{\left (-x\right )} + 5 \, x \]

[In]

integrate(((10*x+5)*exp(x)+180*x)/exp(x),x, algorithm="maxima")

[Out]

5*x^2 - 180*(x + 1)*e^(-x) + 5*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5 \, x^{2} - 180 \, {\left (x + 1\right )} e^{\left (-x\right )} + 5 \, x \]

[In]

integrate(((10*x+5)*exp(x)+180*x)/exp(x),x, algorithm="giac")

[Out]

5*x^2 - 180*(x + 1)*e^(-x) + 5*x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int e^{-x} \left (180 x+e^x (5+10 x)\right ) \, dx=5\,\left (x-36\,{\mathrm {e}}^{-x}\right )\,\left (x+1\right ) \]

[In]

int(exp(-x)*(180*x + exp(x)*(10*x + 5)),x)

[Out]

5*(x - 36*exp(-x))*(x + 1)