3.47.10 \(\int \frac {10}{e^4+2 x-\log (\frac {4}{3})} \, dx\)

Optimal. Leaf size=20 \[ 5 \log \left (\frac {1}{2} \left (-e^4-2 x+\log \left (\frac {4}{3}\right )\right )\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 31} \begin {gather*} 5 \log \left (2 x+e^4-\log \left (\frac {4}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[10/(E^4 + 2*x - Log[4/3]),x]

[Out]

5*Log[E^4 + 2*x - Log[4/3]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 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} &=10 \int \frac {1}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx\\ &=5 \log \left (e^4+2 x-\log \left (\frac {4}{3}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.80 \begin {gather*} 5 \log \left (e^4+2 x-\log \left (\frac {4}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[10/(E^4 + 2*x - Log[4/3]),x]

[Out]

5*Log[E^4 + 2*x - Log[4/3]]

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fricas [A]  time = 0.51, size = 11, normalized size = 0.55 \begin {gather*} 5 \, \log \left (2 \, x + e^{4} + \log \left (\frac {3}{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(log(3/4)+2*x+exp(4)),x, algorithm="fricas")

[Out]

5*log(2*x + e^4 + log(3/4))

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giac [A]  time = 0.18, size = 12, normalized size = 0.60 \begin {gather*} 5 \, \log \left ({\left | 2 \, x + e^{4} + \log \left (\frac {3}{4}\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(log(3/4)+2*x+exp(4)),x, algorithm="giac")

[Out]

5*log(abs(2*x + e^4 + log(3/4)))

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maple [A]  time = 0.09, size = 12, normalized size = 0.60

method	result	size
default	\(5 \ln \left (\ln \left (\frac {3}{4}\right )+2 x +{\mathrm e}^{4}\right )\)	\(12\)
norman	\(5 \ln \left (\ln \left (\frac {3}{4}\right )+2 x +{\mathrm e}^{4}\right )\)	\(12\)
risch	\(5 \ln \left (\ln \relax (3)-2 \ln \relax (2)+2 x +{\mathrm e}^{4}\right )\)	\(16\)
meijerg	\(\frac {10 \left (\frac {\ln \left (\frac {3}{4}\right )}{2}+\frac {{\mathrm e}^{4}}{2}\right ) \ln \left (1+\frac {2 x}{\ln \left (\frac {3}{4}\right )+{\mathrm e}^{4}}\right )}{\ln \left (\frac {3}{4}\right )+{\mathrm e}^{4}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10/(ln(3/4)+2*x+exp(4)),x,method=_RETURNVERBOSE)

[Out]

5*ln(ln(3/4)+2*x+exp(4))

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maxima [A]  time = 0.37, size = 11, normalized size = 0.55 \begin {gather*} 5 \, \log \left (2 \, x + e^{4} + \log \left (\frac {3}{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(log(3/4)+2*x+exp(4)),x, algorithm="maxima")

[Out]

5*log(2*x + e^4 + log(3/4))

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mupad [B]  time = 0.22, size = 11, normalized size = 0.55 \begin {gather*} 5\,\ln \left (2\,x+{\mathrm {e}}^4+\ln \left (\frac {3}{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10/(2*x + exp(4) + log(3/4)),x)

[Out]

5*log(2*x + exp(4) + log(3/4))

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sympy [A]  time = 0.07, size = 17, normalized size = 0.85 \begin {gather*} 5 \log {\left (2 x - 2 \log {\relax (2 )} + \log {\relax (3 )} + e^{4} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(ln(3/4)+2*x+exp(4)),x)

[Out]

5*log(2*x - 2*log(2) + log(3) + exp(4))

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