3.58.66 \(\int e^{10-11 x} (-11-11 \log (5)) \, dx\)

Optimal. Leaf size=29 \[ e^{-x+\frac {10 x}{-1+\frac {x}{x-x^2}}} (1+\log (5)) \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.41, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2194} \begin {gather*} e^{10-11 x} (1+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(10 - 11*x)*(-11 - 11*Log[5]),x]

[Out]

E^(10 - 11*x)*(1 + 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 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} &=-\left ((11 (1+\log (5))) \int e^{10-11 x} \, dx\right )\\ &=e^{10-11 x} (1+\log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.41 \begin {gather*} e^{10-11 x} (1+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(10 - 11*x)*(-11 - 11*Log[5]),x]

[Out]

E^(10 - 11*x)*(1 + Log[5])

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fricas [A]  time = 0.82, size = 11, normalized size = 0.38 \begin {gather*} {\left (\log \relax (5) + 1\right )} e^{\left (-11 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11*log(5)-11)/exp(11*x-10),x, algorithm="fricas")

[Out]

(log(5) + 1)*e^(-11*x + 10)

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giac [A]  time = 0.13, size = 11, normalized size = 0.38 \begin {gather*} {\left (\log \relax (5) + 1\right )} e^{\left (-11 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11*log(5)-11)/exp(11*x-10),x, algorithm="giac")

[Out]

(log(5) + 1)*e^(-11*x + 10)

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maple [A]  time = 0.05, size = 14, normalized size = 0.48

method	result	size
gosper	\(\left (\ln \relax (5)+1\right ) {\mathrm e}^{-11 x +10}\)	\(14\)
norman	\(\left (\ln \relax (5)+1\right ) {\mathrm e}^{-11 x +10}\)	\(14\)
derivativedivides	\(-\left (-\ln \relax (5)-1\right ) {\mathrm e}^{-11 x +10}\)	\(17\)
default	\(-\frac {\left (-11 \ln \relax (5)-11\right ) {\mathrm e}^{-11 x +10}}{11}\)	\(17\)
risch	\({\mathrm e}^{-11 x +10} \ln \relax (5)+{\mathrm e}^{-11 x +10}\)	\(17\)
meijerg	\(-\ln \relax (5) {\mathrm e}^{-11 x +11 x \,{\mathrm e}^{10}} \left (1-{\mathrm e}^{-11 x \,{\mathrm e}^{10}}\right )-{\mathrm e}^{-11 x +11 x \,{\mathrm e}^{10}} \left (1-{\mathrm e}^{-11 x \,{\mathrm e}^{10}}\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-11*ln(5)-11)/exp(11*x-10),x,method=_RETURNVERBOSE)

[Out]

(ln(5)+1)/exp(11*x-10)

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maxima [A]  time = 0.35, size = 11, normalized size = 0.38 \begin {gather*} {\left (\log \relax (5) + 1\right )} e^{\left (-11 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11*log(5)-11)/exp(11*x-10),x, algorithm="maxima")

[Out]

(log(5) + 1)*e^(-11*x + 10)

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mupad [B]  time = 0.06, size = 11, normalized size = 0.38 \begin {gather*} {\mathrm {e}}^{10-11\,x}\,\left (\ln \relax (5)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(10 - 11*x)*(11*log(5) + 11),x)

[Out]

exp(10 - 11*x)*(log(5) + 1)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.34 \begin {gather*} \left (1 + \log {\relax (5 )}\right ) e^{10 - 11 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11*ln(5)-11)/exp(11*x-10),x)

[Out]

(1 + log(5))*exp(10 - 11*x)

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