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

Optimal. Leaf size=19 \[ \log \left (6-\frac {x+\frac {1}{2} (5+x+\log (5))}{e^3}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.68, 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, 31} \begin {gather*} \log \left (3 x-12 e^3+5+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[3/(5 - 12*E^3 + 3*x + Log[5]),x]

[Out]

Log[5 - 12*E^3 + 3*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.68 \begin {gather*} \log \left (5-12 e^3+3 x+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[5 - 12*E^3 + 3*x + Log[5]]

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fricas [A]  time = 0.64, size = 12, normalized size = 0.63 \begin {gather*} \log \left (3 \, x - 12 \, e^{3} + \log \relax (5) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x - 12*e^3 + log(5) + 5)

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giac [A]  time = 0.13, size = 13, normalized size = 0.68 \begin {gather*} \log \left ({\left | 3 \, x - 12 \, e^{3} + \log \relax (5) + 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(3*x - 12*e^3 + log(5) + 5))

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maple [A]  time = 0.26, size = 13, normalized size = 0.68

method	result	size
default	\(\ln \left (\ln \relax (5)-12 \,{\mathrm e}^{3}+3 x +5\right )\)	\(13\)
norman	\(\ln \left (\ln \relax (5)-12 \,{\mathrm e}^{3}+3 x +5\right )\)	\(13\)
risch	\(\ln \left (\ln \relax (5)-12 \,{\mathrm e}^{3}+3 x +5\right )\)	\(13\)
meijerg	\(\frac {3 \left (\frac {\ln \relax (5)}{3}-4 \,{\mathrm e}^{3}+\frac {5}{3}\right ) \ln \left (1+\frac {3 x}{\ln \relax (5)-12 \,{\mathrm e}^{3}+5}\right )}{\ln \relax (5)-12 \,{\mathrm e}^{3}+5}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(5)-12*exp(3)+3*x+5)

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maxima [A]  time = 0.37, size = 12, normalized size = 0.63 \begin {gather*} \log \left (3 \, x - 12 \, e^{3} + \log \relax (5) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x - 12*e^3 + log(5) + 5)

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mupad [B]  time = 0.09, size = 12, normalized size = 0.63 \begin {gather*} \ln \left (3\,x-12\,{\mathrm {e}}^3+\ln \relax (5)+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x - 12*exp(3) + log(5) + 5)

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sympy [A]  time = 0.06, size = 14, normalized size = 0.74 \begin {gather*} \log {\left (3 x - 12 e^{3} + \log {\relax (5 )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x - 12*exp(3) + log(5) + 5)

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