3.33.89 $\int \frac{e^{-x}(36+36x-19x^2+10e^x{x}^2+x^3)}{5x^2}dx$

Optimal. Leaf size=29 $$5+\frac{1}{5}{e}^{-x}(6+e^x-\frac{(-6+x{)}^2}{x})+2x$$

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Rubi [A]  time = 0.30, antiderivative size = 35, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 7, integrand size = 33, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.212, Rules used = {12, 6742, 2199, 2194, 2177, 2178, 2176} $$\begin{array}{c}-\frac{1}{5}{e}^{-x}x+2x+\frac{18e^{-x}}{5}-\frac{36e^{-x}}{5x}\end{array}$$

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

Rule 2194

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 2199

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] &&  !$UseGamma === True

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{5}\int \frac{e^{-x}(36+36x-19x^2+10e^x{x}^2+x^3)}{x^2}dx\\ & =\frac{1}{5}\int (10+\frac{e^{-x}(36+36x-19x^2+x^3)}{x^2})dx\\ & =2x+\frac{1}{5}\int \frac{e^{-x}(36+36x-19x^2+x^3)}{x^2}dx\\ & =2x+\frac{1}{5}\int (-19e^{-x}+\frac{36e^{-x}}{x^2}+\frac{36e^{-x}}{x}+e^{-x}x)dx\\ & =2x+\frac{1}{5}\int e^{-x}xdx-\frac{19}{5}\int e^{-x}dx+\frac{36}{5}\int \frac{e^{-x}}{x^2}dx+\frac{36}{5}\int \frac{e^{-x}}{x}dx\\ & =\frac{19e^{-x}}{5}-\frac{36e^{-x}}{5x}+2x-\frac{e^{-x}x}{5}+\frac{36\text{Ei}(-x)}{5}+\frac{1}{5}\int e^{-x}dx-\frac{36}{5}\int \frac{e^{-x}}{x}dx\\ & =\frac{18e^{-x}}{5}-\frac{36e^{-x}}{5x}+2x-\frac{e^{-x}x}{5}\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 24, normalized size = 0.83 $$\begin{array}{c}\frac{1}{5}(e^{-x}(18-\frac{36}{x}-x)+10x)\end{array}$$

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.55, size = 26, normalized size = 0.90 $$\begin{array}{c}\frac{(10x^2{e}^x-x^2+18x-36)e^{(-x)}}{5x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 32, normalized size = 1.10 $$\begin{array}{c}-\frac{x^2{e}^{(-x)}-10x^2-18x{e}^{(-x)}+36e^{(-x)}}{5x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 22, normalized size = 0.76

method	result	size
risch	$2x-\frac{(x^2-18x+36){\mathrm{e}}^{-x}}{5x}$	$22$
norman	$\frac{(-\frac{36}{5}+\frac{18x}{5}-\frac{x^2}{5}+2{\mathrm{e}}^x{x}^2){\mathrm{e}}^{-x}}{x}$	$26$
default	$2x-\frac{x{\mathrm{e}}^{-x}}{5}+\frac{18{\mathrm{e}}^{-x}}{5}-\frac{36{\mathrm{e}}^{-x}}{5x}$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [C]  time = 0.50, size = 30, normalized size = 1.03 $$\begin{array}{c}-\frac{1}{5}(x+1)e^{(-x)}+2x+\frac{36}{5}\mathrm{E}\mathrm{i}(-x)+\frac{19}{5}{e}^{(-x)}-\frac{36}{5}\mathrm{\Gamma }(-1,x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.09, size = 26, normalized size = 0.90 $$\begin{array}{c}2x+\frac{18{\mathrm{e}}^{-x}}{5}-\frac{x{\mathrm{e}}^{-x}}{5}-\frac{36{\mathrm{e}}^{-x}}{5x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.59 $$\begin{array}{c}2x+\frac{(-x^2+18x-36)e^{-x}}{5x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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