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

Optimal. Leaf size=27 \[ \frac{a}{b^2 \left (a+b e^x\right )}+\frac{\log \left (a+b e^x\right )}{b^2} \]

[Out]

a/(b^2*(a + b*E^x)) + Log[a + b*E^x]/b^2

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Rubi [A]  time = 0.0337294, antiderivative size = 27, 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} \[ \frac{a}{b^2 \left (a+b e^x\right )}+\frac{\log \left (a+b e^x\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(a + b*E^x)^2,x]

[Out]

a/(b^2*(a + b*E^x)) + Log[a + b*E^x]/b^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 )^2} \, dx &=\operatorname{Subst}\left (\int \frac{x}{(a+b x)^2} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{a}{b (a+b x)^2}+\frac{1}{b (a+b x)}\right ) \, dx,x,e^x\right )\\ &=\frac{a}{b^2 \left (a+b e^x\right )}+\frac{\log \left (a+b e^x\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.0240013, size = 24, normalized size = 0.89 \[ \frac{\frac{a}{a+b e^x}+\log \left (a+b e^x\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(a + b*E^x)^2,x]

[Out]

(a/(a + b*E^x) + Log[a + b*E^x])/b^2

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Maple [A]  time = 0.007, size = 26, normalized size = 1. \begin{align*}{\frac{a}{{b}^{2} \left ( a+b{{\rm e}^{x}} \right ) }}+{\frac{\ln \left ( a+b{{\rm e}^{x}} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/b^2/(a+b*exp(x))+ln(a+b*exp(x))/b^2

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Maxima [A]  time = 1.06359, size = 38, normalized size = 1.41 \begin{align*} \frac{a}{b^{3} e^{x} + a b^{2}} + \frac{\log \left (b e^{x} + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(b^3*e^x + a*b^2) + log(b*e^x + a)/b^2

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Fricas [A]  time = 1.45637, size = 70, normalized size = 2.59 \begin{align*} \frac{{\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b^{3} e^{x} + a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*e^x + a)*log(b*e^x + a) + a)/(b^3*e^x + a*b^2)

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Sympy [A]  time = 0.164065, size = 24, normalized size = 0.89 \begin{align*} \frac{a}{a b^{2} + b^{3} e^{x}} + \frac{\log{\left (\frac{a}{b} + e^{x} \right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(a*b**2 + b**3*exp(x)) + log(a/b + exp(x))/b**2

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Giac [A]  time = 1.30259, size = 35, normalized size = 1.3 \begin{align*} \frac{\log \left ({\left | b e^{x} + a \right |}\right )}{b^{2}} + \frac{a}{{\left (b e^{x} + a\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*e^x + a))/b^2 + a/((b*e^x + a)*b^2)