\(\int \frac {e^{2 x}}{(a+b e^x)^4} \, dx\) [21]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=\frac {a}{3 b^2 \left (a+b e^x\right )^3}-\frac {1}{2 b^2 \left (a+b e^x\right )^2} \]

[Out]

1/3*a/b^2/(a+b*exp(x))^3-1/2/b^2/(a+b*exp(x))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2280, 45} \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=\frac {a}{3 b^2 \left (a+b e^x\right )^3}-\frac {1}{2 b^2 \left (a+b e^x\right )^2} \]

[In]

Int[E^(2*x)/(a + b*E^x)^4,x]

[Out]

a/(3*b^2*(a + b*E^x)^3) - 1/(2*b^2*(a + b*E^x)^2)

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}& = \text {Subst}\left (\int \frac {x}{(a+b x)^4} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^4}+\frac {1}{b (a+b x)^3}\right ) \, dx,x,e^x\right ) \\ & = \frac {a}{3 b^2 \left (a+b e^x\right )^3}-\frac {1}{2 b^2 \left (a+b e^x\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=\frac {-a-3 b e^x}{6 b^2 \left (a+b e^x\right )^3} \]

[In]

Integrate[E^(2*x)/(a + b*E^x)^4,x]

[Out]

(-a - 3*b*E^x)/(6*b^2*(a + b*E^x)^3)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {3 b \,{\mathrm e}^{x}+a}{6 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{3}}\)	\(21\)
norman	\(\frac {-\frac {a}{6 b^{2}}-\frac {{\mathrm e}^{x}}{2 b}}{\left (a +b \,{\mathrm e}^{x}\right )^{3}}\)	\(24\)
default	\(\frac {a}{3 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{3}}-\frac {1}{2 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{2}}\)	\(29\)
parallelrisch	\(\frac {b^{4} {\mathrm e}^{2 x} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} b^{3} a}{6 b^{3} a^{2} \left (a +b \,{\mathrm e}^{x}\right )^{3}}\)	\(38\)

[In]

int(exp(2*x)/(a+b*exp(x))^4,x,method=_RETURNVERBOSE)

[Out]

-1/6*(3*b*exp(x)+a)/b^2/(a+b*exp(x))^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=-\frac {3 \, b e^{x} + a}{6 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} \]

[In]

integrate(exp(2*x)/(a+b*exp(x))^4,x, algorithm="fricas")

[Out]

-1/6*(3*b*e^x + a)/(b^5*e^(3*x) + 3*a*b^4*e^(2*x) + 3*a^2*b^3*e^x + a^3*b^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=\frac {- a - 3 b e^{x}}{6 a^{3} b^{2} + 18 a^{2} b^{3} e^{x} + 18 a b^{4} e^{2 x} + 6 b^{5} e^{3 x}} \]

[In]

integrate(exp(2*x)/(a+b*exp(x))**4,x)

[Out]

(-a - 3*b*exp(x))/(6*a**3*b**2 + 18*a**2*b**3*exp(x) + 18*a*b**4*exp(2*x) + 6*b**5*exp(3*x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (28) = 56\).

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.50 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=-\frac {b e^{x}}{2 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} - \frac {a}{6 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} \]

[In]

integrate(exp(2*x)/(a+b*exp(x))^4,x, algorithm="maxima")

[Out]

-1/2*b*e^x/(b^5*e^(3*x) + 3*a*b^4*e^(2*x) + 3*a^2*b^3*e^x + a^3*b^2) - 1/6*a/(b^5*e^(3*x) + 3*a*b^4*e^(2*x) +
3*a^2*b^3*e^x + a^3*b^2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=-\frac {3 \, b e^{x} + a}{6 \, {\left (b e^{x} + a\right )}^{3} b^{2}} \]

[In]

integrate(exp(2*x)/(a+b*exp(x))^4,x, algorithm="giac")

[Out]

-1/6*(3*b*e^x + a)/((b*e^x + a)^3*b^2)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {e^{2 x}}{\left (a+b e^x\right )^4} \, dx=\frac {\frac {{\mathrm {e}}^{2\,x}}{2\,a}+\frac {b\,{\mathrm {e}}^{3\,x}}{6\,a^2}}{a^3+3\,{\mathrm {e}}^x\,a^2\,b+3\,{\mathrm {e}}^{2\,x}\,a\,b^2+{\mathrm {e}}^{3\,x}\,b^3} \]

[In]

int(exp(2*x)/(a + b*exp(x))^4,x)

[Out]

(exp(2*x)/(2*a) + (b*exp(3*x))/(6*a^2))/(b^3*exp(3*x) + a^3 + 3*a^2*b*exp(x) + 3*a*b^2*exp(2*x))