3.36 \(\int \frac{f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx\)

Optimal. Leaf size=61 \[ \frac{f^{a+2 b x-e}}{2 b d \log (f)}-\frac{c f^{a-2 e} \log \left (d f^{2 b x+e}+c\right )}{2 b d^2 \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.060494, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2248, 43} \[ \frac{f^{a+2 b x-e}}{2 b d \log (f)}-\frac{c f^{a-2 e} \log \left (d f^{2 b x+e}+c\right )}{2 b d^2 \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + 4*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

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

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx &=\frac{f^{a-2 e} \operatorname{Subst}\left (\int \frac{x}{c+d x} \, dx,x,f^{e+2 b x}\right )}{2 b \log (f)}\\ &=\frac{f^{a-2 e} \operatorname{Subst}\left (\int \left (\frac{1}{d}-\frac{c}{d (c+d x)}\right ) \, dx,x,f^{e+2 b x}\right )}{2 b \log (f)}\\ &=\frac{f^{a-e+2 b x}}{2 b d \log (f)}-\frac{c f^{a-2 e} \log \left (c+d f^{e+2 b x}\right )}{2 b d^2 \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.04202, size = 48, normalized size = 0.79 \[ \frac{f^{a-2 e} \left (d f^{2 b x+e}-c \log \left (d f^{2 b x+e}+c\right )\right )}{2 b d^2 \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + 4*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

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

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Maple [A]  time = 0.016, size = 76, normalized size = 1.3 \begin{align*}{\frac{{{\rm e}^{ \left ( 2\,bx+e \right ) \ln \left ( f \right ) }}}{2\, \left ({f}^{e} \right ) ^{2}\ln \left ( f \right ) bd} \left ({f}^{{\frac{a}{2}}} \right ) ^{2}}-{\frac{c\ln \left ( c+d{{\rm e}^{ \left ( 2\,bx+e \right ) \ln \left ( f \right ) }} \right ) }{2\,{d}^{2}b\ln \left ( f \right ) \left ({f}^{e} \right ) ^{2}} \left ({f}^{{\frac{a}{2}}} \right ) ^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(4*b*x+a)/(c+d*f^(2*b*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.13063, size = 112, normalized size = 1.84 \begin{align*} -\frac{c f^{a - 2 \, e} \log \left (d \sqrt{f^{4 \, b x + a}} f^{-\frac{1}{2} \, a + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} + \frac{{\left (d \sqrt{f^{4 \, b x + a}} f^{-\frac{1}{2} \, a + e} + c\right )} f^{a - 2 \, e}}{2 \, b d^{2} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(4*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="maxima")

[Out]

-1/2*c*f^(a - 2*e)*log(d*sqrt(f^(4*b*x + a))*f^(-1/2*a + e) + c)/(b*d^2*log(f)) + 1/2*(d*sqrt(f^(4*b*x + a))*f
^(-1/2*a + e) + c)*f^(a - 2*e)/(b*d^2*log(f))

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Fricas [A]  time = 1.62445, size = 123, normalized size = 2.02 \begin{align*} \frac{d f^{2 \, b x + e} f^{a - 2 \, e} - c f^{a - 2 \, e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(4*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.67611, size = 218, normalized size = 3.57 \begin{align*} \begin{cases} \frac{e^{\frac{a \log{\left (f \right )}}{2}} e^{- e \log{\left (f \right )}} \sqrt{e^{\left (a + 4 b x\right ) \log{\left (f \right )}}}}{2 b d \log{\left (f \right )}} & \text{for}\: 2 b d e^{e \log{\left (f \right )}} \log{\left (f \right )} \neq 0 \\\frac{x \left (c^{2} e^{\frac{3 a \log{\left (f \right )}}{2}} + 2 c d e^{a \log{\left (f \right )}} e^{e \log{\left (f \right )}} + d^{2} e^{\frac{a \log{\left (f \right )}}{2}} e^{2 e \log{\left (f \right )}}\right )}{c^{2} d e^{a \log{\left (f \right )}} e^{e \log{\left (f \right )}} + 2 c d^{2} e^{\frac{a \log{\left (f \right )}}{2}} e^{2 e \log{\left (f \right )}} + d^{3} e^{3 e \log{\left (f \right )}}} & \text{otherwise} \end{cases} - \frac{c e^{\left (a - 2 e\right ) \log{\left (f \right )}} \log{\left (\frac{c e^{\frac{a \log{\left (f \right )}}{2}} e^{- e \log{\left (f \right )}}}{d} + \sqrt{e^{\left (a + 4 b x\right ) \log{\left (f \right )}}} \right )}}{2 b d^{2} \log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(4*b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

Piecewise((exp(a*log(f)/2)*exp(-e*log(f))*sqrt(exp((a + 4*b*x)*log(f)))/(2*b*d*log(f)), Ne(2*b*d*exp(e*log(f))
*log(f), 0)), (x*(c**2*exp(3*a*log(f)/2) + 2*c*d*exp(a*log(f))*exp(e*log(f)) + d**2*exp(a*log(f)/2)*exp(2*e*lo
g(f)))/(c**2*d*exp(a*log(f))*exp(e*log(f)) + 2*c*d**2*exp(a*log(f)/2)*exp(2*e*log(f)) + d**3*exp(3*e*log(f))),
True)) - c*exp((a - 2*e)*log(f))*log(c*exp(a*log(f)/2)*exp(-e*log(f))/d + sqrt(exp((a + 4*b*x)*log(f))))/(2*b
*d**2*log(f))

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Giac [A]  time = 1.20974, size = 89, normalized size = 1.46 \begin{align*} \frac{1}{2} \, f^{a}{\left (\frac{f^{2 \, b x}}{b d f^{e} \log \left (f\right )} - \frac{c \log \left ({\left | d f^{2 \, b x} f^{e} + c \right |}\right )}{b d^{2} f^{2 \, e} \log \left (f\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(4*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="giac")

[Out]

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