3.9.57 \(\int \frac {e^{-x} (1+(1-x) \log (5)-\log (-e^{12+4 e^3+x}))}{\log (5)} \, dx\)

Optimal. Leaf size=27 \[ e^{-x} \left (x+\frac {\log \left (-e^{4 \left (3+e^3\right )+x}\right )}{\log (5)}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6742, 2194, 2176, 2554} \begin {gather*} -e^{-x} (1-x)+e^{-x}+\frac {e^{-x} \log \left (-e^{x+4 \left (3+e^3\right )}\right )}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + (1 - x)*Log[5] - Log[-E^(12 + 4*E^3 + x)])/(E^x*Log[5]),x]

[Out]

E^(-x) - (1 - x)/E^x + Log[-E^(4*(3 + E^3) + x)]/(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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{-x} \left (1+(1-x) \log (5)-\log \left (-e^{12+4 e^3+x}\right )\right ) \, dx}{\log (5)}\\ &=\frac {\int \left (e^{-x}-e^{-x} (-1+x) \log (5)-e^{-x} \log \left (-e^{12+4 e^3+x}\right )\right ) \, dx}{\log (5)}\\ &=\frac {\int e^{-x} \, dx}{\log (5)}-\frac {\int e^{-x} \log \left (-e^{12+4 e^3+x}\right ) \, dx}{\log (5)}-\int e^{-x} (-1+x) \, dx\\ &=-e^{-x} (1-x)-\frac {e^{-x}}{\log (5)}+\frac {e^{-x} \log \left (-e^{4 \left (3+e^3\right )+x}\right )}{\log (5)}-\frac {\int e^{-x} \, dx}{\log (5)}-\int e^{-x} \, dx\\ &=e^{-x}-e^{-x} (1-x)+\frac {e^{-x} \log \left (-e^{4 \left (3+e^3\right )+x}\right )}{\log (5)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 28, normalized size = 1.04 \begin {gather*} \frac {e^{-x} \left (x \log (5)+\log \left (-e^{12+4 e^3+x}\right )\right )}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + (1 - x)*Log[5] - Log[-E^(12 + 4*E^3 + x)])/(E^x*Log[5]),x]

[Out]

(x*Log[5] + Log[-E^(12 + 4*E^3 + x)])/(E^x*Log[5])

________________________________________________________________________________________

fricas [C]  time = 0.66, size = 23, normalized size = 0.85 \begin {gather*} \frac {{\left (i \, \pi + x \log \relax (5) + x + 4 \, e^{3} + 12\right )} e^{\left (-x\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-exp(4*exp(3)+x+12))+(-x+1)*log(5)+1)/exp(x)/log(5),x, algorithm="fricas")

[Out]

(I*pi + x*log(5) + x + 4*e^3 + 12)*e^(-x)/log(5)

________________________________________________________________________________________

giac [A]  time = 0.58, size = 34, normalized size = 1.26 \begin {gather*} \frac {x e^{\left (-x\right )} \log \relax (5) + x e^{\left (-x\right )} + 12 \, e^{\left (-x\right )} + 4 \, e^{\left (-x + 3\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-exp(4*exp(3)+x+12))+(-x+1)*log(5)+1)/exp(x)/log(5),x, algorithm="giac")

[Out]

(x*e^(-x)*log(5) + x*e^(-x) + 12*e^(-x) + 4*e^(-x + 3))/log(5)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 25, normalized size = 0.93

method	result	size
norman	\(\left (x +\frac {\ln \left (-{\mathrm e}^{4 \,{\mathrm e}^{3}+12} {\mathrm e}^{x}\right )}{\ln \relax (5)}\right ) {\mathrm e}^{-x}\)	\(25\)
default	\(\frac {12 \,{\mathrm e}^{-x}+x \,{\mathrm e}^{-x}+{\mathrm e}^{-x} \left (\ln \left (-{\mathrm e}^{4 \,{\mathrm e}^{3}+x +12}\right )-4 \,{\mathrm e}^{3}-x -12\right )+4 \,{\mathrm e}^{-x} {\mathrm e}^{3}+{\mathrm e}^{-x} x \ln \relax (5)}{\ln \relax (5)}\)	\(60\)
risch	\(\frac {{\mathrm e}^{-x} \ln \left ({\mathrm e}^{x}\right )}{\ln \relax (5)}+\frac {\left (-2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}+2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}+2 i \pi +8 \,{\mathrm e}^{3}+24+2 x \ln \relax (5)\right ) {\mathrm e}^{-x}}{2 \ln \relax (5)}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(-exp(4*exp(3)+x+12))+(1-x)*ln(5)+1)/exp(x)/ln(5),x,method=_RETURNVERBOSE)

[Out]

(x+1/ln(5)*ln(-exp(4*exp(3)+12)*exp(x)))/exp(x)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 40, normalized size = 1.48 \begin {gather*} \frac {{\left (x + 1\right )} e^{\left (-x\right )} \log \relax (5) - e^{\left (-x\right )} \log \relax (5) + e^{\left (-x\right )} \log \left (-e^{\left (x + 4 \, e^{3} + 12\right )}\right )}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-exp(4*exp(3)+x+12))+(-x+1)*log(5)+1)/exp(x)/log(5),x, algorithm="maxima")

[Out]

((x + 1)*e^(-x)*log(5) - e^(-x)*log(5) + e^(-x)*log(-e^(x + 4*e^3 + 12)))/log(5)

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 24, normalized size = 0.89 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,\left (4\,{\mathrm {e}}^3+\ln \left (-{\mathrm {e}}^x\right )+x\,\ln \relax (5)+12\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(log(-exp(x + 4*exp(3) + 12)) + log(5)*(x - 1) - 1))/log(5),x)

[Out]

(exp(-x)*(4*exp(3) + log(-exp(x)) + x*log(5) + 12))/log(5)

________________________________________________________________________________________

sympy [C]  time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} - \frac {\left (- x \log {\relax (5 )} - x - 4 e^{3} - 12 - i \pi \right ) e^{- x}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(-exp(4*exp(3)+x+12))+(-x+1)*ln(5)+1)/exp(x)/ln(5),x)

[Out]

-(-x*log(5) - x - 4*exp(3) - 12 - I*pi)*exp(-x)/log(5)

________________________________________________________________________________________