3.33.66 $\int \frac{1}{4}{e}^{-3+x}(-25-5x+10\log (4))dx$

Optimal. Leaf size=18 $$\frac{5}{4}{e}^{-3+x}(-4-x+2\log (4))$$

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 3, number of rules used = 3, integrand size = 18, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.167, Rules used = {12, 2176, 2194} $$\begin{array}{c}\frac{5e^{x-3}}{4}-\frac{5}{4}{e}^{x-3}(x+5-\log (16))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(-3 + x)*(-25 - 5*x + 10*Log[4]))/4,x]

[Out]

(5*E^(-3 + x))/4 - (5*E^(-3 + x)*(5 + x - Log[16]))/4

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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int e^{-3+x}(-25-5x+10\log (4))dx\\ & =-\frac{5}{4}{e}^{-3+x}(5+x-\log (16))+\frac{5}{4}\int e^{-3+x}dx\\ & =\frac{5e^{-3+x}}{4}-\frac{5}{4}{e}^{-3+x}(5+x-\log (16))\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 16, normalized size = 0.89 $$\begin{array}{c}-\frac{5}{4}{e}^{-3+x}(4+x-2\log (4))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + x)*(-25 - 5*x + 10*Log[4]))/4,x]

[Out]

(-5*E^(-3 + x)*(4 + x - 2*Log[4]))/4

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fricas [A]  time = 0.60, size = 13, normalized size = 0.72 $$\begin{array}{c}-\frac{5}{4}(x-4\log (2)+4)e^{(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(20*log(2)-5*x-25)/exp(3-x),x, algorithm="fricas")

[Out]

-5/4*(x - 4*log(2) + 4)*e^(x - 3)

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giac [A]  time = 0.28, size = 13, normalized size = 0.72 $$\begin{array}{c}-\frac{5}{4}(x-4\log (2)+4)e^{(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(20*log(2)-5*x-25)/exp(3-x),x, algorithm="giac")

[Out]

-5/4*(x - 4*log(2) + 4)*e^(x - 3)

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maple [A]  time = 0.04, size = 16, normalized size = 0.89

method	result	size
risch	$\frac{(20\ln (2)-20-5x){\mathrm{e}}^{x-3}}{4}$	$16$
norman	$(-\frac{5x}{4}-5+5\ln (2)){\mathrm{e}}^{x-3}$	$19$
gosper	$\frac{5(4\ln (2)-4-x){\mathrm{e}}^{x-3}}{4}$	$20$
derivativedivides	$\frac{5{\mathrm{e}}^{x-3}(3-x)}{4}-\frac{35{\mathrm{e}}^{x-3}}{4}+5{\mathrm{e}}^{x-3}\ln (2)$	$39$
default	$\frac{5{\mathrm{e}}^{x-3}(3-x)}{4}-\frac{35{\mathrm{e}}^{x-3}}{4}+5{\mathrm{e}}^{x-3}\ln (2)$	$39$
meijerg	$\frac{25{\mathrm{e}}^{-x{\mathrm{e}}^{-3}+x}(1-{\mathrm{e}}^{x{\mathrm{e}}^{-3}})}{4}-5{\mathrm{e}}^{-x{\mathrm{e}}^{-3}+x}\ln (2)(1-{\mathrm{e}}^{x{\mathrm{e}}^{-3}})-\frac{5{\mathrm{e}}^{-x{\mathrm{e}}^{-3}+x+3}(1-\frac{(-2x{\mathrm{e}}^{-3}+2){\mathrm{e}}^{x{\mathrm{e}}^{-3}}}{2})}{4}$	$69$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(20*ln(2)-5*x-25)/exp(3-x),x,method=_RETURNVERBOSE)

[Out]

1/4*(20*ln(2)-20-5*x)*exp(x-3)

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maxima [A]  time = 0.48, size = 24, normalized size = 1.33 $$\begin{array}{c}-\frac{5}{4}(x-1)e^{(x-3)}+5e^{(x-3)}\log (2)-\frac{25}{4}{e}^{(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(20*log(2)-5*x-25)/exp(3-x),x, algorithm="maxima")

[Out]

-5/4*(x - 1)*e^(x - 3) + 5*e^(x - 3)*log(2) - 25/4*e^(x - 3)

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mupad [B]  time = 1.97, size = 15, normalized size = 0.83 $$\begin{array}{c}-{\mathrm{e}}^{-3}{\mathrm{e}}^x(\frac{5x}{4}-5\ln (2)+5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x - 3)*((5*x)/4 - 5*log(2) + 25/4),x)

[Out]

-exp(-3)*exp(x)*((5*x)/4 - 5*log(2) + 5)

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sympy [A]  time = 0.12, size = 15, normalized size = 0.83 $$\begin{array}{c}\frac{(-5x-20+20\log (2))e^{x-3}}{4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(20*ln(2)-5*x-25)/exp(3-x),x)

[Out]

(-5*x - 20 + 20*log(2))*exp(x - 3)/4

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