3.69.72 $\int \frac{1}{10}{e}^{-x}(-10+4x-x^2+(-4x+2x^2)\log (3))dx$

Optimal. Leaf size=28 $$\frac{1}{5}(3+e^{-x}(4+x(-1+\frac{x}{2}-x\log (3))))$$

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Rubi [A]  time = 0.10, antiderivative size = 46, normalized size of antiderivative = 1.64, number of steps used = 16, number of rules used = 4, integrand size = 31, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.129, Rules used = {12, 2196, 2194, 2176} $$\begin{array}{c}\frac{1}{10}{e}^{-x}{x}^2-\frac{1}{5}{e}^{-x}{x}^2\log (3)-\frac{e^{-x}x}{5}+\frac{4e^{-x}}{5}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-10 + 4*x - x^2 + (-4*x + 2*x^2)*Log[3])/(10*E^x),x]

[Out]

4/(5*E^x) - x/(5*E^x) + x^2/(10*E^x) - (x^2*Log[3])/(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 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 2196

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] &&  !$UseGamma === True

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{10}\int e^{-x}(-10+4x-x^2+(-4x+2x^2)\log (3))dx\\ & =\frac{1}{10}\int (-10e^{-x}+4e^{-x}x-e^{-x}{x}^2+2e^{-x}(-2+x)x\log (3))dx\\ & =-(\frac{1}{10}\int e^{-x}{x}^{2}dx)+\frac{2}{5}\int e^{-x}xdx+\frac{1}{5}\log (3)\int e^{-x}(-2+x)xdx-\int e^{-x}dx\\ & =e^{-x}-\frac{2e^{-x}x}{5}+\frac{1}{10}{e}^{-x}{x}^2-\frac{1}{5}\int e^{-x}xdx+\frac{2}{5}\int e^{-x}dx+\frac{1}{5}\log (3)\int (-2e^{-x}x+e^{-x}{x}^2)dx\\ & =\frac{3e^{-x}}{5}-\frac{e^{-x}x}{5}+\frac{1}{10}{e}^{-x}{x}^2-\frac{1}{5}\int e^{-x}dx+\frac{1}{5}\log (3)\int e^{-x}{x}^{2}dx-\frac{1}{5}(2\log (3))\int e^{-x}xdx\\ & =\frac{4e^{-x}}{5}-\frac{e^{-x}x}{5}+\frac{1}{10}{e}^{-x}{x}^2+\frac{2}{5}{e}^{-x}x\log (3)-\frac{1}{5}{e}^{-x}{x}^2\log (3)-\frac{1}{5}(2\log (3))\int e^{-x}dx+\frac{1}{5}(2\log (3))\int e^{-x}xdx\\ & =\frac{4e^{-x}}{5}-\frac{e^{-x}x}{5}+\frac{1}{10}{e}^{-x}{x}^2+\frac{2}{5}{e}^{-x}\log (3)-\frac{1}{5}{e}^{-x}{x}^2\log (3)+\frac{1}{5}(2\log (3))\int e^{-x}dx\\ & =\frac{4e^{-x}}{5}-\frac{e^{-x}x}{5}+\frac{1}{10}{e}^{-x}{x}^2-\frac{1}{5}{e}^{-x}{x}^2\log (3)\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.82 $$\begin{array}{c}\frac{1}{10}{e}^{-x}(8-2x-x^2(-1+\log (9)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 4*x - x^2 + (-4*x + 2*x^2)*Log[3])/(10*E^x),x]

[Out]

(8 - 2*x - x^2*(-1 + Log[9]))/(10*E^x)

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fricas [A]  time = 0.61, size = 23, normalized size = 0.82 $$\begin{array}{c}-\frac{1}{10}(2x^2\log (3)-x^2+2x-8)e^{(-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((2*x^2-4*x)*log(3)-x^2+4*x-10)/exp(x),x, algorithm="fricas")

[Out]

-1/10*(2*x^2*log(3) - x^2 + 2*x - 8)*e^(-x)

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giac [A]  time = 0.13, size = 23, normalized size = 0.82 $$\begin{array}{c}-\frac{1}{10}(2x^2\log (3)-x^2+2x-8)e^{(-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((2*x^2-4*x)*log(3)-x^2+4*x-10)/exp(x),x, algorithm="giac")

[Out]

-1/10*(2*x^2*log(3) - x^2 + 2*x - 8)*e^(-x)

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maple [A]  time = 0.03, size = 21, normalized size = 0.75

method	result	size
norman	$(\frac{4}{5}+(-\frac{\ln (3)}{5}+\frac{1}{10})x^2-\frac{x}{5}){\mathrm{e}}^{-x}$	$21$
risch	$\frac{(-2x^2\ln (3)+x^2-2x+8){\mathrm{e}}^{-x}}{10}$	$22$
gosper	$-\frac{(2x^2\ln (3)-x^2+2x-8){\mathrm{e}}^{-x}}{10}$	$24$
default	$\frac{4{\mathrm{e}}^{-x}}{5}-\frac{x{\mathrm{e}}^{-x}}{5}+\frac{x^2{\mathrm{e}}^{-x}}{10}-\frac{{\mathrm{e}}^{-x}{x}^2\ln (3)}{5}$	$35$
meijerg	$-1+{\mathrm{e}}^{-x}+\frac{(2\ln (3)-1)(2-\frac{(3x^2+6x+6){\mathrm{e}}^{-x}}{3})}{10}+\frac{(-4\ln (3)+4)(1-\frac{(2x+2){\mathrm{e}}^{-x}}{2})}{10}$	$54$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((2*x^2-4*x)*ln(3)-x^2+4*x-10)/exp(x),x,method=_RETURNVERBOSE)

[Out]

(4/5+(-1/5*ln(3)+1/10)*x^2-1/5*x)/exp(x)

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maxima [B]  time = 0.37, size = 55, normalized size = 1.96 $$\begin{array}{c}-\frac{1}{5}(x^2+2x+2)e^{(-x)}\log (3)+\frac{2}{5}(x+1)e^{(-x)}\log (3)+\frac{1}{10}(x^2+2x+2)e^{(-x)}-\frac{2}{5}(x+1)e^{(-x)}+e^{(-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((2*x^2-4*x)*log(3)-x^2+4*x-10)/exp(x),x, algorithm="maxima")

[Out]

-1/5*(x^2 + 2*x + 2)*e^(-x)*log(3) + 2/5*(x + 1)*e^(-x)*log(3) + 1/10*(x^2 + 2*x + 2)*e^(-x) - 2/5*(x + 1)*e^(
-x) + e^(-x)

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mupad [B]  time = 0.05, size = 22, normalized size = 0.79 $$\begin{array}{c}-\frac{{\mathrm{e}}^{-x}(2x+x^2\ln (9)-x^2-8)}{10}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*((log(3)*(4*x - 2*x^2))/10 - (2*x)/5 + x^2/10 + 1),x)

[Out]

-(exp(-x)*(2*x + x^2*log(9) - x^2 - 8))/10

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sympy [A]  time = 0.14, size = 20, normalized size = 0.71 $$\begin{array}{c}\frac{(-2x^2\log (3)+x^2-2x+8)e^{-x}}{10}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((2*x**2-4*x)*ln(3)-x**2+4*x-10)/exp(x),x)

[Out]

(-2*x**2*log(3) + x**2 - 2*x + 8)*exp(-x)/10

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