3.102.20 \(\int \frac {e^{11+x} (1+2 \log (3))}{\log (3)} \, dx\)

Optimal. Leaf size=12 \[ e^{11+x} \left (2+\frac {1}{\log (3)}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2194} \begin {gather*} \frac {e^{x+11} (1+\log (9))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(11 + x)*(1 + 2*Log[3]))/Log[3],x]

[Out]

(E^(11 + x)*(1 + Log[9]))/Log[3]

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} &=\frac {(1+\log (9)) \int e^{11+x} \, dx}{\log (3)}\\ &=\frac {e^{11+x} (1+\log (9))}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.17 \begin {gather*} \frac {e^{11+x} (1+\log (9))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(11 + x)*(1 + 2*Log[3]))/Log[3],x]

[Out]

(E^(11 + x)*(1 + Log[9]))/Log[3]

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fricas [A]  time = 0.58, size = 16, normalized size = 1.33 \begin {gather*} e^{\left (x + \log \left (\frac {2 \, \log \relax (3) + 1}{\log \relax (3)}\right ) + 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((2*log(3)+1)/log(3))+11+x),x, algorithm="fricas")

[Out]

e^(x + log((2*log(3) + 1)/log(3)) + 11)

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giac [A]  time = 0.26, size = 16, normalized size = 1.33 \begin {gather*} e^{\left (x + \log \left (\frac {2 \, \log \relax (3) + 1}{\log \relax (3)}\right ) + 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((2*log(3)+1)/log(3))+11+x),x, algorithm="giac")

[Out]

e^(x + log((2*log(3) + 1)/log(3)) + 11)

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maple [A]  time = 0.05, size = 17, normalized size = 1.42

method	result	size
gosper	\({\mathrm e}^{\ln \left (\frac {2 \ln \relax (3)+1}{\ln \relax (3)}\right )+11+x}\)	\(17\)
derivativedivides	\({\mathrm e}^{\ln \left (\frac {2 \ln \relax (3)+1}{\ln \relax (3)}\right )+11+x}\)	\(17\)
default	\({\mathrm e}^{\ln \left (\frac {2 \ln \relax (3)+1}{\ln \relax (3)}\right )+11+x}\)	\(17\)
norman	\({\mathrm e}^{\ln \left (\frac {2 \ln \relax (3)+1}{\ln \relax (3)}\right )+11+x}\)	\(17\)
risch	\(2 \,{\mathrm e}^{11+x}+\frac {{\mathrm e}^{11+x}}{\ln \relax (3)}\)	\(17\)
meijerg	\(-{\mathrm e}^{\ln \left (\frac {2 \ln \relax (3)+1}{\ln \relax (3)}\right )+11} \left (1-{\mathrm e}^{x}\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(ln((2*ln(3)+1)/ln(3))+11+x),x,method=_RETURNVERBOSE)

[Out]

exp(ln((2*ln(3)+1)/ln(3))+11+x)

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maxima [A]  time = 0.34, size = 15, normalized size = 1.25 \begin {gather*} \frac {{\left (2 \, \log \relax (3) + 1\right )} e^{\left (x + 11\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((2*log(3)+1)/log(3))+11+x),x, algorithm="maxima")

[Out]

(2*log(3) + 1)*e^(x + 11)/log(3)

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mupad [B]  time = 0.04, size = 15, normalized size = 1.25 \begin {gather*} \frac {{\mathrm {e}}^{11}\,{\mathrm {e}}^x\,\left (2\,\ln \relax (3)+1\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x + log((2*log(3) + 1)/log(3)) + 11),x)

[Out]

(exp(11)*exp(x)*(2*log(3) + 1))/log(3)

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sympy [A]  time = 0.09, size = 14, normalized size = 1.17 \begin {gather*} \frac {\left (1 + 2 \log {\relax (3 )}\right ) e^{x + 11}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(ln((2*ln(3)+1)/ln(3))+11+x),x)

[Out]

(1 + 2*log(3))*exp(x + 11)/log(3)

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