3.34 \(\int \frac {f^{a+2 b x}}{c+d f^{e+2 b x}} \, dx\)

Optimal. Leaf size=34 \[ \frac {f^{a-e} \log \left (d f^{2 b x+e}+c\right )}{2 b d \log (f)} \]

[Out]

1/2*f^(a-e)*ln(c+d*f^(2*b*x+e))/b/d/ln(f)

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Rubi [A]  time = 0.08, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2247, 2246, 31} \[ \frac {f^{a-e} \log \left (d f^{2 b x+e}+c\right )}{2 b d \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + 2*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(f^(a - e)*Log[c + d*f^(e + 2*b*x)])/(2*b*d*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^{a+2 b x}}{c+d f^{e+2 b x}} \, dx &=f^{a-e} \int \frac {f^{e+2 b x}}{c+d f^{e+2 b x}} \, dx\\ &=\frac {f^{a-e} \operatorname {Subst}\left (\int \frac {1}{c+d x} \, dx,x,f^{e+2 b x}\right )}{2 b \log (f)}\\ &=\frac {f^{a-e} \log \left (c+d f^{e+2 b x}\right )}{2 b d \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 1.00 \[ \frac {f^{a-e} \log \left (d f^{2 b x+e}+c\right )}{2 b d \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + 2*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(f^(a - e)*Log[c + d*f^(e + 2*b*x)])/(2*b*d*Log[f])

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fricas [A]  time = 0.42, size = 32, normalized size = 0.94 \[ \frac {f^{a - e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(2*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="fricas")

[Out]

1/2*f^(a - e)*log(d*f^(2*b*x + e) + c)/(b*d*log(f))

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giac [A]  time = 0.33, size = 37, normalized size = 1.09 \[ \frac {f^{a} \log \left ({\left | d f^{2 \, b x} f^{e} + c \right |}\right )}{2 \, b d f^{e} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(2*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="giac")

[Out]

1/2*f^a*log(abs(d*f^(2*b*x)*f^e + c))/(b*d*f^e*log(f))

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maple [A]  time = 0.03, size = 47, normalized size = 1.38 \[ \frac {f^{a} f^{-e} \ln \left (d \,{\mathrm e}^{\left (2 b x +a \right ) \ln \relax (f )} {\mathrm e}^{-a \ln \relax (f )+e \ln \relax (f )}+c \right )}{2 b d \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(2*b*x+a)/(c+d*f^(2*b*x+e)),x)

[Out]

1/2/(f^e)/d/ln(f)/b*f^a*ln(c+d*exp(-ln(f)*a+ln(f)*e)*exp((2*b*x+a)*ln(f)))

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maxima [A]  time = 0.43, size = 32, normalized size = 0.94 \[ \frac {f^{a - e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(2*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="maxima")

[Out]

1/2*f^(a - e)*log(d*f^(2*b*x + e) + c)/(b*d*log(f))

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mupad [B]  time = 3.63, size = 37, normalized size = 1.09 \[ \frac {f^{a-e}\,\ln \left (d\,f^{a+e+2\,b\,x}+c\,f^a\right )}{2\,b\,d\,\ln \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + 2*b*x)/(c + d*f^(e + 2*b*x)),x)

[Out]

(f^(a - e)*log(d*f^(a + e + 2*b*x) + c*f^a))/(2*b*d*log(f))

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sympy [A]  time = 0.77, size = 42, normalized size = 1.24 \[ \frac {e^{\left (a - e\right ) \log {\relax (f )}} \log {\left (\frac {c e^{a \log {\relax (f )}} e^{- e \log {\relax (f )}}}{d} + f^{a + 2 b x} \right )}}{2 b d \log {\relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(2*b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

exp((a - e)*log(f))*log(c*exp(a*log(f))*exp(-e*log(f))/d + f**(a + 2*b*x))/(2*b*d*log(f))

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