\(\int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx\) [36]

Optimal result

Integrand size = 23, antiderivative size = 61 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-e+2 b x}}{2 b d \log (f)}-\frac {c f^{a-2 e} \log \left (c+d f^{e+2 b x}\right )}{2 b d^2 \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2280, 45} \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a+2 b x-e}}{2 b d \log (f)}-\frac {c f^{a-2 e} \log \left (d f^{2 b x+e}+c\right )}{2 b d^2 \log (f)} \]

[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 45

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])

Rule 2280

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]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a-2 e} \text {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} \text {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] (verified)

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-2 e} \left (d f^{e+2 b x}-2 b c x \log (f)-c \log \left (c+d f^{e+2 b x}\right )+c \log \left (b d^3 f^{e+2 b x} \log (f)\right )\right )}{2 b d^2 \log (f)} \]

[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) - 2*b*c*x*Log[f] - c*Log[c + d*f^(e + 2*b*x)] + c*Log[b*d^3*f^(e + 2*b*x)*Log[f]
]))/(2*b*d^2*Log[f])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25

method	result	size
norman	\(\frac {f^{-2 e} f^{a} {\mathrm e}^{\left (2 b x +e \right ) \ln \left (f \right )}}{2 \ln \left (f \right ) b d}-\frac {c \,f^{-2 e} f^{a} \ln \left (c +d \,{\mathrm e}^{\left (2 b x +e \right ) \ln \left (f \right )}\right )}{2 \ln \left (f \right ) b \,d^{2}}\)	\(76\)
risch	\(\frac {f^{a} f^{-2 e} f^{2 b x +e}}{2 \ln \left (f \right ) b d}+\frac {f^{a} f^{-2 e} c e}{2 b \,d^{2}}-\frac {f^{a} f^{-2 e} c \ln \left (f^{2 b x +e}+\frac {c}{d}\right )}{2 \ln \left (f \right ) b \,d^{2}}\)	\(96\)

[In]

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

[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)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {d f^{2 \, b x + e} f^{a - 2 \, e} - c f^{a - 2 \, e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.51 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {\left (\begin {cases} x & \text {for}\: b = 0 \vee f = 1 \\\frac {e^{2 b x \log {\left (f \right )}}}{2 b \log {\left (f \right )}} & \text {otherwise} \end {cases}\right ) e^{a \log {\left (f \right )}} e^{- e \log {\left (f \right )}}}{d} - \frac {c e^{\left (a - 2 e\right ) \log {\left (f \right )}} \log {\left (\frac {c e^{\frac {a \log {\left (f \right )}}{2}} e^{- e \log {\left (f \right )}}}{d} + \sqrt {e^{\left (a + 4 b x\right ) \log {\left (f \right )}}} \right )}}{2 b d^{2} \log {\left (f \right )}} \]

[In]

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

[Out]

Piecewise((x, Eq(b, 0) | Eq(f, 1)), (exp(2*b*x*log(f))/(2*b*log(f)), True))*exp(a*log(f))*exp(-e*log(f))/d - 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
))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {c f^{a - 2 \, e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} + \frac {{\left (d f^{2 \, b x + e} + c\right )} f^{a - 2 \, e}}{2 \, b d^{2} \log \left (f\right )} \]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a} {\left (\frac {f^{2 \, b x} f^{e}}{d f^{2 \, e} \log \left (f\right )} - \frac {c \log \left ({\left | d f^{2 \, b x} f^{e} + c \right |}\right )}{d^{2} f^{2 \, e} \log \left (f\right )}\right )}}{2 \, b} \]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {f^{a-2\,e}\,\left (\frac {c\,\ln \left (c+d\,f^{e+2\,b\,x}\right )}{2}-\frac {d\,f^{e+2\,b\,x}}{2}\right )}{b\,d^2\,\ln \left (f\right )} \]

[In]

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

[Out]

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