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

Optimal. Leaf size=34 \[ \frac{a}{3 b^2 \left (a+b e^x\right )^3}-\frac{1}{2 b^2 \left (a+b e^x\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0359505, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2248, 43} \[ \frac{a}{3 b^2 \left (a+b e^x\right )^3}-\frac{1}{2 b^2 \left (a+b e^x\right )^2} \]

Antiderivative was successfully verified.

[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 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{e^{2 x}}{\left (a+b e^x\right )^4} \, dx &=\operatorname{Subst}\left (\int \frac{x}{(a+b x)^4} \, dx,x,e^x\right )\\ &=\operatorname{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]  time = 0.0211904, size = 24, normalized size = 0.71 \[ -\frac{a+3 b e^x}{6 b^2 \left (a+b e^x\right )^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.006, size = 29, normalized size = 0.9 \begin{align*}{\frac{a}{3\,{b}^{2} \left ( a+b{{\rm e}^{x}} \right ) ^{3}}}-{\frac{1}{2\,{b}^{2} \left ( a+b{{\rm e}^{x}} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.08893, size = 115, normalized size = 3.38 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.52872, size = 105, normalized size = 3.09 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.146506, size = 51, normalized size = 1.5 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.20175, size = 27, normalized size = 0.79 \begin{align*} -\frac{3 \, b e^{x} + a}{6 \,{\left (b e^{x} + a\right )}^{3} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)