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

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 {gather*} \frac {5 e^{x-3}}{4}-\frac {5}{4} e^{x-3} (x+5-\log (16)) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-3+x} (-25-5 x+10 \log (4)) \, dx\\ &=-\frac {5}{4} e^{-3+x} (5+x-\log (16))+\frac {5}{4} \int e^{-3+x} \, dx\\ &=\frac {5 e^{-3+x}}{4}-\frac {5}{4} e^{-3+x} (5+x-\log (16))\\ \end {aligned} \end {gather*}

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

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 {gather*} -\frac {5}{4} \, {\left (x - 4 \, \log \relax (2) + 4\right )} e^{\left (x - 3\right )} \end {gather*}

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 {gather*} -\frac {5}{4} \, {\left (x - 4 \, \log \relax (2) + 4\right )} e^{\left (x - 3\right )} \end {gather*}

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	\(\frac {\left (20 \ln \relax (2)-20-5 x \right ) {\mathrm e}^{x -3}}{4}\)	\(16\)
norman	\(\left (-\frac {5 x}{4}-5+5 \ln \relax (2)\right ) {\mathrm e}^{x -3}\)	\(19\)
gosper	\(\frac {5 \left (4 \ln \relax (2)-4-x \right ) {\mathrm e}^{x -3}}{4}\)	\(20\)
derivativedivides	\(\frac {5 \,{\mathrm e}^{x -3} \left (3-x \right )}{4}-\frac {35 \,{\mathrm e}^{x -3}}{4}+5 \,{\mathrm e}^{x -3} \ln \relax (2)\)	\(39\)
default	\(\frac {5 \,{\mathrm e}^{x -3} \left (3-x \right )}{4}-\frac {35 \,{\mathrm e}^{x -3}}{4}+5 \,{\mathrm e}^{x -3} \ln \relax (2)\)	\(39\)
meijerg	\(\frac {25 \,{\mathrm e}^{-x \,{\mathrm e}^{-3}+x} \left (1-{\mathrm e}^{x \,{\mathrm e}^{-3}}\right )}{4}-5 \,{\mathrm e}^{-x \,{\mathrm e}^{-3}+x} \ln \relax (2) \left (1-{\mathrm e}^{x \,{\mathrm e}^{-3}}\right )-\frac {5 \,{\mathrm e}^{-x \,{\mathrm e}^{-3}+x +3} \left (1-\frac {\left (-2 x \,{\mathrm e}^{-3}+2\right ) {\mathrm e}^{x \,{\mathrm e}^{-3}}}{2}\right )}{4}\)	\(69\)

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 {gather*} -\frac {5}{4} \, {\left (x - 1\right )} e^{\left (x - 3\right )} + 5 \, e^{\left (x - 3\right )} \log \relax (2) - \frac {25}{4} \, e^{\left (x - 3\right )} \end {gather*}

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 {gather*} -{\mathrm {e}}^{-3}\,{\mathrm {e}}^x\,\left (\frac {5\,x}{4}-5\,\ln \relax (2)+5\right ) \end {gather*}

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 {gather*} \frac {\left (- 5 x - 20 + 20 \log {\relax (2 )}\right ) e^{x - 3}}{4} \end {gather*}

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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