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]

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

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Rubi [A]  time = 0.0811105, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]]

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 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*} \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*}

Mathematica [A]  time = 0.017154, size = 34, normalized size = 1. \[ \frac{f^{a-e} \log \left (d f^{2 b x+e}+c\right )}{2 b d \log (f)} \]

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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Maple [A]  time = 0.01, size = 47, normalized size = 1.4 \begin{align*}{\frac{{f}^{a}\ln \left ( c+d{{\rm e}^{-\ln \left ( f \right ) a+\ln \left ( f \right ) e}}{{\rm e}^{ \left ( 2\,bx+a \right ) \ln \left ( f \right ) }} \right ) }{2\,{f}^{e}d\ln \left ( f \right ) b}} \end{align*}

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 = 1.1564, size = 43, normalized size = 1.26 \begin{align*} \frac{f^{a - e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d \log \left (f\right )} \end{align*}

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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Fricas [A]  time = 1.78473, size = 72, normalized size = 2.12 \begin{align*} \frac{f^{a - e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d \log \left (f\right )} \end{align*}

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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Sympy [A]  time = 0.914736, size = 42, normalized size = 1.24 \begin{align*} \frac{e^{\left (a - e\right ) \log{\left (f \right )}} \log{\left (\frac{c e^{a \log{\left (f \right )}} e^{- e \log{\left (f \right )}}}{d} + f^{a + 2 b x} \right )}}{2 b d \log{\left (f \right )}} \end{align*}

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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Giac [A]  time = 1.47007, size = 50, normalized size = 1.47 \begin{align*} \frac{f^{a} \log \left ({\left | d f^{2 \, b x} f^{e} + c \right |}\right )}{2 \, b d f^{e} \log \left (f\right )} \end{align*}

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))