3.49.98 \(\int \frac {1}{2} e^{-x} (50+(-5+5 x-5 \log (4)) \log (5)) \, dx\)

Optimal. Leaf size=21 \[ \frac {5}{2} e^{-x} (-10-(x-\log (4)) \log (5)) \]

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 2187, 2176, 2194} \begin {gather*} -\frac {5}{2} e^{-x} (x \log (5)+10-(1+\log (4)) \log (5))-\frac {5}{2} e^{-x} \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 + (-5 + 5*x - 5*Log[4])*Log[5])/(2*E^x),x]

[Out]

(-5*Log[5])/(2*E^x) - (5*(10 + x*Log[5] - (1 + Log[4])*Log[5]))/(2*E^x)

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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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}{2} \int e^{-x} (50+(-5+5 x-5 \log (4)) \log (5)) \, dx\\ &=\frac {1}{2} \int e^{-x} (5 x \log (5)+5 (10-(1+\log (4)) \log (5))) \, dx\\ &=-\frac {5}{2} e^{-x} (10+x \log (5)-(1+\log (4)) \log (5))+\frac {1}{2} (5 \log (5)) \int e^{-x} \, dx\\ &=-\frac {5}{2} e^{-x} \log (5)-\frac {5}{2} e^{-x} (10+x \log (5)-(1+\log (4)) \log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{2} e^{-x} (-50-5 x \log (5)+5 \log (4) \log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 + (-5 + 5*x - 5*Log[4])*Log[5])/(2*E^x),x]

[Out]

(-50 - 5*x*Log[5] + 5*Log[4]*Log[5])/(2*E^x)

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fricas [A]  time = 0.60, size = 17, normalized size = 0.81 \begin {gather*} -\frac {5}{2} \, {\left ({\left (x - 2 \, \log \relax (2)\right )} \log \relax (5) + 10\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*log(2)+5*x-5)*log(5)+50)/exp(x),x, algorithm="fricas")

[Out]

-5/2*((x - 2*log(2))*log(5) + 10)*e^(-x)

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giac [A]  time = 0.16, size = 18, normalized size = 0.86 \begin {gather*} -\frac {5}{2} \, {\left (x \log \relax (5) - 2 \, \log \relax (5) \log \relax (2) + 10\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*log(2)+5*x-5)*log(5)+50)/exp(x),x, algorithm="giac")

[Out]

-5/2*(x*log(5) - 2*log(5)*log(2) + 10)*e^(-x)

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maple [A]  time = 0.03, size = 19, normalized size = 0.90

method	result	size
norman	\(\left (-\frac {5 x \ln \relax (5)}{2}-25+5 \ln \relax (2) \ln \relax (5)\right ) {\mathrm e}^{-x}\)	\(19\)
gosper	\(\frac {5 \left (2 \ln \relax (2) \ln \relax (5)-x \ln \relax (5)-10\right ) {\mathrm e}^{-x}}{2}\)	\(20\)
risch	\(\frac {\left (10 \ln \relax (2) \ln \relax (5)-5 x \ln \relax (5)-50\right ) {\mathrm e}^{-x}}{2}\)	\(20\)
default	\(-25 \,{\mathrm e}^{-x}-\frac {5 \,{\mathrm e}^{-x} x \ln \relax (5)}{2}+5 \,{\mathrm e}^{-x} \ln \relax (2) \ln \relax (5)\)	\(27\)
meijerg	\(25-25 \,{\mathrm e}^{-x}-5 \ln \relax (2) \ln \relax (5) \left (1-{\mathrm e}^{-x}\right )+\frac {5 \ln \relax (5) \left (1-\frac {\left (2 x +2\right ) {\mathrm e}^{-x}}{2}\right )}{2}-\frac {5 \ln \relax (5) \left (1-{\mathrm e}^{-x}\right )}{2}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-10*ln(2)+5*x-5)*ln(5)+50)/exp(x),x,method=_RETURNVERBOSE)

[Out]

(-5/2*x*ln(5)-25+5*ln(2)*ln(5))/exp(x)

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maxima [B]  time = 0.36, size = 36, normalized size = 1.71 \begin {gather*} -\frac {5}{2} \, {\left (x + 1\right )} e^{\left (-x\right )} \log \relax (5) + 5 \, e^{\left (-x\right )} \log \relax (5) \log \relax (2) + \frac {5}{2} \, e^{\left (-x\right )} \log \relax (5) - 25 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*log(2)+5*x-5)*log(5)+50)/exp(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 18, normalized size = 0.86 \begin {gather*} -\frac {5\,{\mathrm {e}}^{-x}\,\left (x\,\ln \relax (5)-2\,\ln \relax (2)\,\ln \relax (5)+10\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*((log(5)*(10*log(2) - 5*x + 5))/2 - 25),x)

[Out]

-(5*exp(-x)*(x*log(5) - 2*log(2)*log(5) + 10))/2

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sympy [A]  time = 0.15, size = 20, normalized size = 0.95 \begin {gather*} \frac {\left (- 5 x \log {\relax (5 )} - 50 + 10 \log {\relax (2 )} \log {\relax (5 )}\right ) e^{- x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*ln(2)+5*x-5)*ln(5)+50)/exp(x),x)

[Out]

(-5*x*log(5) - 50 + 10*log(2)*log(5))*exp(-x)/2

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