\(\int \frac {e^{-x} (36+36 x-19 x^2+10 e^x x^2+x^3)}{5 x^2} \, dx\) [3289]

Optimal result

Integrand size = 33, antiderivative size = 29 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=5+\frac {1}{5} e^{-x} \left (6+e^x-\frac {(-6+x)^2}{x}\right )+2 x \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {12, 6874, 2230, 2225, 2208, 2209, 2207} \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=-\frac {1}{5} e^{-x} x+2 x+\frac {18 e^{-x}}{5}-\frac {36 e^{-x}}{5 x} \]

[In]

Int[(36 + 36*x - 19*x^2 + 10*E^x*x^2 + x^3)/(5*E^x*x^2),x]

[Out]

18/(5*E^x) - 36/(5*E^x*x) + 2*x - x/(5*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 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !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 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (10+\frac {e^{-x} \left (36+36 x-19 x^2+x^3\right )}{x^2}\right ) \, dx \\ & = 2 x+\frac {1}{5} \int \frac {e^{-x} \left (36+36 x-19 x^2+x^3\right )}{x^2} \, dx \\ & = 2 x+\frac {1}{5} \int \left (-19 e^{-x}+\frac {36 e^{-x}}{x^2}+\frac {36 e^{-x}}{x}+e^{-x} x\right ) \, dx \\ & = 2 x+\frac {1}{5} \int e^{-x} x \, dx-\frac {19}{5} \int e^{-x} \, dx+\frac {36}{5} \int \frac {e^{-x}}{x^2} \, dx+\frac {36}{5} \int \frac {e^{-x}}{x} \, dx \\ & = \frac {19 e^{-x}}{5}-\frac {36 e^{-x}}{5 x}+2 x-\frac {e^{-x} x}{5}+\frac {36 \text {Ei}(-x)}{5}+\frac {1}{5} \int e^{-x} \, dx-\frac {36}{5} \int \frac {e^{-x}}{x} \, dx \\ & = \frac {18 e^{-x}}{5}-\frac {36 e^{-x}}{5 x}+2 x-\frac {e^{-x} x}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=\frac {1}{5} \left (e^{-x} \left (18-\frac {36}{x}-x\right )+10 x\right ) \]

[In]

Integrate[(36 + 36*x - 19*x^2 + 10*E^x*x^2 + x^3)/(5*E^x*x^2),x]

[Out]

((18 - 36/x - x)/E^x + 10*x)/5

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76

method	result	size
risch	\(2 x -\frac {\left (x^{2}-18 x +36\right ) {\mathrm e}^{-x}}{5 x}\)	\(22\)
norman	\(\frac {\left (-\frac {36}{5}+\frac {18 x}{5}-\frac {x^{2}}{5}+2 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}}{x}\)	\(26\)
default	\(2 x -\frac {x \,{\mathrm e}^{-x}}{5}+\frac {18 \,{\mathrm e}^{-x}}{5}-\frac {36 \,{\mathrm e}^{-x}}{5 x}\)	\(27\)
parallelrisch	\(\frac {\left (-36+10 \,{\mathrm e}^{x} x^{2}-x^{2}+18 x \right ) {\mathrm e}^{-x}}{5 x}\)	\(27\)
parts	\(2 x -\frac {x \,{\mathrm e}^{-x}}{5}+\frac {18 \,{\mathrm e}^{-x}}{5}-\frac {36 \,{\mathrm e}^{-x}}{5 x}\)	\(27\)

[In]

int(1/5*(10*exp(x)*x^2+x^3-19*x^2+36*x+36)/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

2*x-1/5*(x^2-18*x+36)/x*exp(-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=\frac {{\left (10 \, x^{2} e^{x} - x^{2} + 18 \, x - 36\right )} e^{\left (-x\right )}}{5 \, x} \]

[In]

integrate(1/5*(10*exp(x)*x^2+x^3-19*x^2+36*x+36)/exp(x)/x^2,x, algorithm="fricas")

[Out]

1/5*(10*x^2*e^x - x^2 + 18*x - 36)*e^(-x)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=2 x + \frac {\left (- x^{2} + 18 x - 36\right ) e^{- x}}{5 x} \]

[In]

integrate(1/5*(10*exp(x)*x**2+x**3-19*x**2+36*x+36)/exp(x)/x**2,x)

[Out]

2*x + (-x**2 + 18*x - 36)*exp(-x)/(5*x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=-\frac {1}{5} \, {\left (x + 1\right )} e^{\left (-x\right )} + 2 \, x + \frac {36}{5} \, {\rm Ei}\left (-x\right ) + \frac {19}{5} \, e^{\left (-x\right )} - \frac {36}{5} \, \Gamma \left (-1, x\right ) \]

[In]

integrate(1/5*(10*exp(x)*x^2+x^3-19*x^2+36*x+36)/exp(x)/x^2,x, algorithm="maxima")

[Out]

-1/5*(x + 1)*e^(-x) + 2*x + 36/5*Ei(-x) + 19/5*e^(-x) - 36/5*gamma(-1, x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=-\frac {x^{2} e^{\left (-x\right )} - 10 \, x^{2} - 18 \, x e^{\left (-x\right )} + 36 \, e^{\left (-x\right )}}{5 \, x} \]

[In]

integrate(1/5*(10*exp(x)*x^2+x^3-19*x^2+36*x+36)/exp(x)/x^2,x, algorithm="giac")

[Out]

-1/5*(x^2*e^(-x) - 10*x^2 - 18*x*e^(-x) + 36*e^(-x))/x

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x} \left (36+36 x-19 x^2+10 e^x x^2+x^3\right )}{5 x^2} \, dx=2\,x+\frac {18\,{\mathrm {e}}^{-x}}{5}-\frac {x\,{\mathrm {e}}^{-x}}{5}-\frac {36\,{\mathrm {e}}^{-x}}{5\,x} \]

[In]

int((exp(-x)*((36*x)/5 + 2*x^2*exp(x) - (19*x^2)/5 + x^3/5 + 36/5))/x^2,x)

[Out]

2*x + (18*exp(-x))/5 - (x*exp(-x))/5 - (36*exp(-x))/(5*x)