\(\int \frac {e^x}{4+6 e^x} \, dx\) [1]

Optimal result

Integrand size = 13, antiderivative size = 12 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \log \left (2+3 e^x\right ) \]

[Out]

1/6*ln(2+3*exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2278, 31} \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \log \left (3 e^x+2\right ) \]

[In]

Int[E^x/(4 + 6*E^x),x]

[Out]

Log[2 + 3*E^x]/6

Rule 31

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

Rule 2278

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{4+6 x} \, dx,x,e^x\right ) \\ & = \frac {1}{6} \log \left (2+3 e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \log \left (4+6 e^x\right ) \]

[In]

Integrate[E^x/(4 + 6*E^x),x]

[Out]

Log[4 + 6*E^x]/6

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\ln \left (\frac {2}{3}+{\mathrm e}^{x}\right )}{6}\)	\(8\)
parallelrisch	\(\frac {\ln \left (\frac {2}{3}+{\mathrm e}^{x}\right )}{6}\)	\(8\)
derivativedivides	\(\frac {\ln \left (2+3 \,{\mathrm e}^{x}\right )}{6}\)	\(10\)
default	\(\frac {\ln \left (2+3 \,{\mathrm e}^{x}\right )}{6}\)	\(10\)
norman	\(\frac {\ln \left (4+6 \,{\mathrm e}^{x}\right )}{6}\)	\(10\)

[In]

int(exp(x)/(4+6*exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/6*ln(2/3+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \, \log \left (3 \, e^{x} + 2\right ) \]

[In]

integrate(exp(x)/(4+6*exp(x)),x, algorithm="fricas")

[Out]

1/6*log(3*e^x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {\log {\left (e^{x} + \frac {2}{3} \right )}}{6} \]

[In]

integrate(exp(x)/(4+6*exp(x)),x)

[Out]

log(exp(x) + 2/3)/6

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \, \log \left (3 \, e^{x} + 2\right ) \]

[In]

integrate(exp(x)/(4+6*exp(x)),x, algorithm="maxima")

[Out]

1/6*log(3*e^x + 2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {1}{6} \, \log \left (3 \, e^{x} + 2\right ) \]

[In]

integrate(exp(x)/(4+6*exp(x)),x, algorithm="giac")

[Out]

1/6*log(3*e^x + 2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{4+6 e^x} \, dx=\frac {\ln \left (3\,{\mathrm {e}}^x+2\right )}{6} \]

[In]

int(exp(x)/(6*exp(x) + 4),x)

[Out]

log(3*exp(x) + 2)/6