\(\int \frac {F^x}{a+b F^x} \, dx\) [8]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left (a+b F^x\right )}{b \log (F)} \]

[Out]

ln(a+b*F^x)/b/ln(F)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, 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} \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left (a+b F^x\right )}{b \log (F)} \]

[In]

Int[F^x/(a + b*F^x),x]

[Out]

Log[a + b*F^x]/(b*Log[F])

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}& = \frac {\text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,F^x\right )}{\log (F)} \\ & = \frac {\log \left (a+b F^x\right )}{b \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left (a+b F^x\right )}{b \log (F)} \]

[In]

Integrate[F^x/(a + b*F^x),x]

[Out]

Log[a + b*F^x]/(b*Log[F])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \,F^{x}\right )}{b \ln \left (F \right )}\)	\(17\)
default	\(\frac {\ln \left (a +b \,F^{x}\right )}{b \ln \left (F \right )}\)	\(17\)
parallelrisch	\(\frac {\ln \left (a +b \,F^{x}\right )}{b \ln \left (F \right )}\)	\(17\)
norman	\(\frac {\ln \left (a +b \,{\mathrm e}^{x \ln \left (F \right )}\right )}{b \ln \left (F \right )}\)	\(19\)
risch	\(\frac {\ln \left (F^{x}+\frac {a}{b}\right )}{b \ln \left (F \right )}\)	\(19\)

[In]

int(F^x/(a+b*F^x),x,method=_RETURNVERBOSE)

[Out]

ln(a+b*F^x)/b/ln(F)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left (F^{x} b + a\right )}{b \log \left (F\right )} \]

[In]

integrate(F^x/(a+b*F^x),x, algorithm="fricas")

[Out]

log(F^x*b + a)/(b*log(F))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log {\left (F^{x} + \frac {a}{b} \right )}}{b \log {\left (F \right )}} \]

[In]

integrate(F**x/(a+b*F**x),x)

[Out]

log(F**x + a/b)/(b*log(F))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left (F^{x} b + a\right )}{b \log \left (F\right )} \]

[In]

integrate(F^x/(a+b*F^x),x, algorithm="maxima")

[Out]

log(F^x*b + a)/(b*log(F))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\log \left ({\left | F^{x} b + a \right |}\right )}{b \log \left (F\right )} \]

[In]

integrate(F^x/(a+b*F^x),x, algorithm="giac")

[Out]

log(abs(F^x*b + a))/(b*log(F))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {F^x}{a+b F^x} \, dx=\frac {\ln \left (a+F^x\,b\right )}{b\,\ln \left (F\right )} \]

[In]

int(F^x/(a + F^x*b),x)

[Out]

log(a + F^x*b)/(b*log(F))