\(\int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx\) [3266]

Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=\frac {5}{4} e^{-3+x} (-4-x+2 \log (4)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2207, 2225} \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=\frac {5 e^{x-3}}{4}-\frac {5}{4} e^{x-3} (x+5-\log (16)) \]

[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 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 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]

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=-\frac {5}{4} e^{-3+x} (4+x-2 \log (4)) \]

[In]

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

[Out]

(-5*E^(-3 + x)*(4 + x - 2*Log[4]))/4

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {\left (20 \ln \left (2\right )-20-5 x \right ) {\mathrm e}^{-3+x}}{4}\)	\(16\)
norman	\(\left (-\frac {5 x}{4}-5+5 \ln \left (2\right )\right ) {\mathrm e}^{-3+x}\)	\(19\)
gosper	\(\frac {5 \left (4 \ln \left (2\right )-4-x \right ) {\mathrm e}^{-3+x}}{4}\)	\(20\)
parallelrisch	\(\frac {\left (20 \ln \left (2\right )-20-5 x \right ) {\mathrm e}^{-3+x}}{4}\)	\(20\)
derivativedivides	\(\frac {5 \,{\mathrm e}^{-3+x} \left (-x +3\right )}{4}-\frac {35 \,{\mathrm e}^{-3+x}}{4}+5 \,{\mathrm e}^{-3+x} \ln \left (2\right )\)	\(39\)
default	\(\frac {5 \,{\mathrm e}^{-3+x} \left (-x +3\right )}{4}-\frac {35 \,{\mathrm e}^{-3+x}}{4}+5 \,{\mathrm e}^{-3+x} \ln \left (2\right )\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=-\frac {5}{4} \, {\left (x - 4 \, \log \left (2\right ) + 4\right )} e^{\left (x - 3\right )} \]

[In]

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

[Out]

-5/4*(x - 4*log(2) + 4)*e^(x - 3)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=\frac {\left (- 5 x - 20 + 20 \log {\left (2 \right )}\right ) e^{x - 3}}{4} \]

[In]

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

[Out]

(-5*x - 20 + 20*log(2))*exp(x - 3)/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=-\frac {5}{4} \, {\left (x - 1\right )} e^{\left (x - 3\right )} + 5 \, e^{\left (x - 3\right )} \log \left (2\right ) - \frac {25}{4} \, e^{\left (x - 3\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=-\frac {5}{4} \, {\left (x - 4 \, \log \left (2\right ) + 4\right )} e^{\left (x - 3\right )} \]

[In]

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

[Out]

-5/4*(x - 4*log(2) + 4)*e^(x - 3)

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{4} e^{-3+x} (-25-5 x+10 \log (4)) \, dx=-{\mathrm {e}}^{-3}\,{\mathrm {e}}^x\,\left (\frac {5\,x}{4}-5\,\ln \left (2\right )+5\right ) \]

[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)