\(\int \frac {10}{e^4+2 x-\log (\frac {4}{3})} \, dx\) [4610]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \log \left (\frac {1}{2} \left (-e^4-2 x+\log \left (\frac {4}{3}\right )\right )\right ) \]

[Out]

5*ln(-x-1/2*ln(3/4)-1/2*exp(4))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 31} \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \log \left (2 x+e^4-\log \left (\frac {4}{3}\right )\right ) \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \log \left (e^4+2 x-\log \left (\frac {4}{3}\right )\right ) \]

[In]

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

[Out]

5*Log[E^4 + 2*x - Log[4/3]]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60

method	result	size
default	\(5 \ln \left (\ln \left (\frac {3}{4}\right )+2 x +{\mathrm e}^{4}\right )\)	\(12\)
norman	\(5 \ln \left (\ln \left (\frac {3}{4}\right )+2 x +{\mathrm e}^{4}\right )\)	\(12\)
parallelrisch	\(5 \ln \left (\frac {\ln \left (\frac {3}{4}\right )}{2}+x +\frac {{\mathrm e}^{4}}{2}\right )\)	\(14\)
risch	\(5 \ln \left (\ln \left (3\right )-2 \ln \left (2\right )+2 x +{\mathrm e}^{4}\right )\)	\(16\)
meijerg	\(\frac {10 \left (-\ln \left (2\right )+\frac {\ln \left (3\right )}{2}+\frac {{\mathrm e}^{4}}{2}\right ) \ln \left (1+\frac {2 x}{-2 \ln \left (2\right )+\ln \left (3\right )+{\mathrm e}^{4}}\right )}{\ln \left (\frac {3}{4}\right )+{\mathrm e}^{4}}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \, \log \left (2 \, x + e^{4} + \log \left (\frac {3}{4}\right )\right ) \]

[In]

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

[Out]

5*log(2*x + e^4 + log(3/4))

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \log {\left (2 x - 2 \log {\left (2 \right )} + \log {\left (3 \right )} + e^{4} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \, \log \left (2 \, x + e^{4} + \log \left (\frac {3}{4}\right )\right ) \]

[In]

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

[Out]

5*log(2*x + e^4 + log(3/4))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5 \, \log \left ({\left | 2 \, x + e^{4} + \log \left (\frac {3}{4}\right ) \right |}\right ) \]

[In]

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

[Out]

5*log(abs(2*x + e^4 + log(3/4)))

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {10}{e^4+2 x-\log \left (\frac {4}{3}\right )} \, dx=5\,\ln \left (2\,x+{\mathrm {e}}^4+\ln \left (\frac {3}{4}\right )\right ) \]

[In]

int(10/(2*x + exp(4) + log(3/4)),x)

[Out]

5*log(2*x + exp(4) + log(3/4))