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

Optimal. Leaf size=22 \[ \frac{e^x}{b}-\frac{a \log \left (a+b e^x\right )}{b^2} \]

[Out]

E^x/b - (a*Log[a + b*E^x])/b^2

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Rubi [A]  time = 0.0320372, antiderivative size = 22, 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} \[ \frac{e^x}{b}-\frac{a \log \left (a+b e^x\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(a + b*E^x),x]

[Out]

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

Mathematica [A]  time = 0.0152858, size = 22, normalized size = 1. \[ \frac{e^x}{b}-\frac{a \log \left (a+b e^x\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(a + b*E^x),x]

[Out]

E^x/b - (a*Log[a + b*E^x])/b^2

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Maple [A]  time = 0.005, size = 21, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{x}}}{b}}-{\frac{a\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)),x)

[Out]

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

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Maxima [A]  time = 1.19547, size = 27, normalized size = 1.23 \begin{align*} \frac{e^{x}}{b} - \frac{a \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)),x, algorithm="maxima")

[Out]

e^x/b - a*log(b*e^x + a)/b^2

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Fricas [A]  time = 1.52141, size = 43, normalized size = 1.95 \begin{align*} \frac{b e^{x} - a \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)),x, algorithm="fricas")

[Out]

(b*e^x - a*log(b*e^x + a))/b^2

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Sympy [A]  time = 0.134158, size = 20, normalized size = 0.91 \begin{align*} - \frac{a \log{\left (\frac{a}{b} + e^{x} \right )}}{b^{2}} + \begin{cases} \frac{e^{x}}{b} & \text{for}\: b \neq 0 \\\frac{x}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(a/b + exp(x))/b**2 + Piecewise((exp(x)/b, Ne(b, 0)), (x/b, True))

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Giac [A]  time = 1.3437, size = 28, normalized size = 1.27 \begin{align*} \frac{e^{x}}{b} - \frac{a \log \left ({\left | b e^{x} + a \right |}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x/b - a*log(abs(b*e^x + a))/b^2