\(\int \frac {e^{-x} (e^x (-24+8 e^4)+10 x-5 x^2)}{8 \log (5)} \, dx\) [7539]

Optimal result

Integrand size = 33, antiderivative size = 33 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {-1+2 x+e^4 x+5 \left (-x+\frac {1}{8} e^{-x} x^2\right )}{\log (5)} \]

[Out]

(5/8*x^2/exp(x)-3*x+x*exp(4)-1)/ln(5)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6820, 2227, 2207, 2225} \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {5 e^{-x} x^2}{8 \log (5)}-\frac {\left (3-e^4\right ) x}{\log (5)} \]

[In]

Int[(E^x*(-24 + 8*E^4) + 10*x - 5*x^2)/(8*E^x*Log[5]),x]

[Out]

-(((3 - E^4)*x)/Log[5]) + (5*x^2)/(8*E^x*Log[5])

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]

Rule 2227

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] &&  !TrueQ[$UseGamma]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right ) \, dx}{8 \log (5)} \\ & = \frac {\int \left (-24+8 e^4-5 e^{-x} (-2+x) x\right ) \, dx}{8 \log (5)} \\ & = -\frac {\left (3-e^4\right ) x}{\log (5)}-\frac {5 \int e^{-x} (-2+x) x \, dx}{8 \log (5)} \\ & = -\frac {\left (3-e^4\right ) x}{\log (5)}-\frac {5 \int \left (-2 e^{-x} x+e^{-x} x^2\right ) \, dx}{8 \log (5)} \\ & = -\frac {\left (3-e^4\right ) x}{\log (5)}-\frac {5 \int e^{-x} x^2 \, dx}{8 \log (5)}+\frac {5 \int e^{-x} x \, dx}{4 \log (5)} \\ & = -\frac {5 e^{-x} x}{4 \log (5)}-\frac {\left (3-e^4\right ) x}{\log (5)}+\frac {5 e^{-x} x^2}{8 \log (5)}+\frac {5 \int e^{-x} \, dx}{4 \log (5)}-\frac {5 \int e^{-x} x \, dx}{4 \log (5)} \\ & = -\frac {5 e^{-x}}{4 \log (5)}-\frac {\left (3-e^4\right ) x}{\log (5)}+\frac {5 e^{-x} x^2}{8 \log (5)}-\frac {5 \int e^{-x} \, dx}{4 \log (5)} \\ & = -\frac {\left (3-e^4\right ) x}{\log (5)}+\frac {5 e^{-x} x^2}{8 \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {x \left (-24+8 e^4+5 e^{-x} x\right )}{8 \log (5)} \]

[In]

Integrate[(E^x*(-24 + 8*E^4) + 10*x - 5*x^2)/(8*E^x*Log[5]),x]

[Out]

(x*(-24 + 8*E^4 + (5*x)/E^x))/(8*Log[5])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {-24 x +5 x^{2} {\mathrm e}^{-x}+8 x \,{\mathrm e}^{4}}{8 \ln \left (5\right )}\)	\(25\)
parts	\(\frac {\left ({\mathrm e}^{4}-3\right ) x}{\ln \left (5\right )}+\frac {5 x^{2} {\mathrm e}^{-x}}{8 \ln \left (5\right )}\)	\(25\)
norman	\(\left (\frac {\left ({\mathrm e}^{4}-3\right ) x \,{\mathrm e}^{x}}{\ln \left (5\right )}+\frac {5 x^{2}}{8 \ln \left (5\right )}\right ) {\mathrm e}^{-x}\)	\(28\)
parallelrisch	\(\frac {\left (8 x \,{\mathrm e}^{4} {\mathrm e}^{x}+5 x^{2}-24 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{8 \ln \left (5\right )}\)	\(29\)
risch	\(\frac {x \,{\mathrm e}^{4}}{\ln \left (5\right )}-\frac {3 x}{\ln \left (5\right )}+\frac {5 x^{2} {\mathrm e}^{-x}}{8 \ln \left (5\right )}\)	\(30\)

[In]

int(1/8*((8*exp(4)-24)*exp(x)-5*x^2+10*x)/exp(x)/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/8/ln(5)*(-24*x+5*x^2/exp(x)+8*x*exp(4))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {{\left (5 \, x^{2} + 8 \, {\left (x e^{4} - 3 \, x\right )} e^{x}\right )} e^{\left (-x\right )}}{8 \, \log \left (5\right )} \]

[In]

integrate(1/8*((8*exp(4)-24)*exp(x)-5*x^2+10*x)/exp(x)/log(5),x, algorithm="fricas")

[Out]

1/8*(5*x^2 + 8*(x*e^4 - 3*x)*e^x)*e^(-x)/log(5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {5 x^{2} e^{- x}}{8 \log {\left (5 \right )}} + \frac {x \left (-3 + e^{4}\right )}{\log {\left (5 \right )}} \]

[In]

integrate(1/8*((8*exp(4)-24)*exp(x)-5*x**2+10*x)/exp(x)/ln(5),x)

[Out]

5*x**2*exp(-x)/(8*log(5)) + x*(-3 + exp(4))/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {8 \, x e^{4} + 5 \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} - 10 \, {\left (x + 1\right )} e^{\left (-x\right )} - 24 \, x}{8 \, \log \left (5\right )} \]

[In]

integrate(1/8*((8*exp(4)-24)*exp(x)-5*x^2+10*x)/exp(x)/log(5),x, algorithm="maxima")

[Out]

1/8*(8*x*e^4 + 5*(x^2 + 2*x + 2)*e^(-x) - 10*(x + 1)*e^(-x) - 24*x)/log(5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {5 \, x^{2} e^{\left (-x\right )} + 8 \, x e^{4} - 24 \, x}{8 \, \log \left (5\right )} \]

[In]

integrate(1/8*((8*exp(4)-24)*exp(x)-5*x^2+10*x)/exp(x)/log(5),x, algorithm="giac")

[Out]

1/8*(5*x^2*e^(-x) + 8*x*e^4 - 24*x)/log(5)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-x} \left (e^x \left (-24+8 e^4\right )+10 x-5 x^2\right )}{8 \log (5)} \, dx=\frac {x\,\left (8\,{\mathrm {e}}^4+5\,x\,{\mathrm {e}}^{-x}-24\right )}{8\,\ln \left (5\right )} \]

[In]

int((exp(-x)*((5*x)/4 + (exp(x)*(8*exp(4) - 24))/8 - (5*x^2)/8))/log(5),x)

[Out]

(x*(8*exp(4) + 5*x*exp(-x) - 24))/(8*log(5))