3.9 \(\int \frac {F^{d x}}{a+b F^{c+d x}} \, dx\)

Optimal. Leaf size=28 \[ \frac {F^{-c} \log \left (a+b F^{c+d x}\right )}{b d \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2247, 2246, 31} \[ \frac {F^{-c} \log \left (a+b F^{c+d x}\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(d*x)/(a + b*F^(c + d*x)),x]

[Out]

Log[a + b*F^(c + d*x)]/(b*d*F^c*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 2246

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]

Rule 2247

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {F^{-c} \log \left (a+b F^{c+d x}\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(d*x)/(a + b*F^(c + d*x)),x]

[Out]

Log[a + b*F^(c + d*x)]/(b*d*F^c*Log[F])

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fricas [A]  time = 0.41, size = 28, normalized size = 1.00 \[ \frac {\log \left (F^{d x + c} b + a\right )}{F^{c} b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(F^(d*x + c)*b + a)/(F^c*b*d*log(F))

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giac [A]  time = 0.32, size = 30, normalized size = 1.07 \[ \frac {\log \left ({\left | F^{d x} F^{c} b + a \right |}\right )}{F^{c} b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(F^(d*x)*F^c*b + a))/(F^c*b*d*log(F))

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maple [A]  time = 0.03, size = 33, normalized size = 1.18 \[ \frac {F^{-c} \ln \left (b \,{\mathrm e}^{c \ln \relax (F )} {\mathrm e}^{d x \ln \relax (F )}+a \right )}{b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x)/(b*F^(d*x+c)+a),x)

[Out]

1/(F^c)/b/ln(F)/d*ln(a+b*exp(c*ln(F))*exp(d*ln(F)*x))

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maxima [A]  time = 0.43, size = 28, normalized size = 1.00 \[ \frac {\log \left (F^{d x + c} b + a\right )}{F^{c} b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(F^(d*x + c)*b + a)/(F^c*b*d*log(F))

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mupad [B]  time = 3.58, size = 28, normalized size = 1.00 \[ \frac {\ln \left (a+F^{c+d\,x}\,b\right )}{F^c\,b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x)/(a + F^(c + d*x)*b),x)

[Out]

log(a + F^(c + d*x)*b)/(F^c*b*d*log(F))

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sympy [A]  time = 0.39, size = 24, normalized size = 0.86 \[ \frac {e^{- c \log {\relax (F )}} \log {\left (F^{c + d x} + \frac {a}{b} \right )}}{b d \log {\relax (F )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-c*log(F))*log(F**(c + d*x) + a/b)/(b*d*log(F))

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