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

Optimal. Leaf size=33 \[ \frac {-1+2 x+e^4 x+5 \left (-x+\frac {1}{8} e^{-x} x^2\right )}{\log (5)} \]

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Rubi [A]  time = 0.10, antiderivative size = 31, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6688, 2196, 2176, 2194} \begin {gather*} \frac {5 e^{-x} x^2}{8 \log (5)}-\frac {\left (3-e^4\right ) x}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[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 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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 0.73 \begin {gather*} \frac {x \left (-24+8 e^4+5 e^{-x} x\right )}{8 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 4.39, size = 28, normalized size = 0.85 \begin {gather*} \frac {{\left (5 \, x^{2} + 8 \, {\left (x e^{4} - 3 \, x\right )} e^{x}\right )} e^{\left (-x\right )}}{8 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 24, normalized size = 0.73 \begin {gather*} \frac {5 \, x^{2} e^{\left (-x\right )} + 8 \, x e^{4} - 24 \, x}{8 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 25, normalized size = 0.76

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.38, size = 38, normalized size = 1.15 \begin {gather*} \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 \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.08, size = 20, normalized size = 0.61 \begin {gather*} \frac {x\,\left (8\,{\mathrm {e}}^4+5\,x\,{\mathrm {e}}^{-x}-24\right )}{8\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.13, size = 22, normalized size = 0.67 \begin {gather*} \frac {5 x^{2} e^{- x}}{8 \log {\relax (5 )}} + \frac {x \left (-3 + e^{4}\right )}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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