\(\int \frac {e^{2 x}}{a+b e^x} \, dx\) [18]

Optimal result

Integrand size = 15, antiderivative size = 22 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {e^x}{b}-\frac {a \log \left (a+b e^x\right )}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, 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} \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {e^x}{b}-\frac {a \log \left (a+b e^x\right )}{b^2} \]

[In]

Int[E^(2*x)/(a + b*E^x),x]

[Out]

E^x/b - (a*Log[a + b*E^x])/b^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} \, dx,x,e^x\right ) \\ & = \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {e^x}{b}-\frac {a \log \left (a+b e^x\right )}{b^2} \]

[In]

Integrate[E^(2*x)/(a + b*E^x),x]

[Out]

E^x/b - (a*Log[a + b*E^x])/b^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {{\mathrm e}^{x}}{b}-\frac {a \ln \left (a +b \,{\mathrm e}^{x}\right )}{b^{2}}\)	\(21\)
norman	\(\frac {{\mathrm e}^{x}}{b}-\frac {a \ln \left (a +b \,{\mathrm e}^{x}\right )}{b^{2}}\)	\(21\)
risch	\(\frac {{\mathrm e}^{x}}{b}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a}{b}\right )}{b^{2}}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {b e^{x} - a \log \left (b e^{x} + a\right )}{b^{2}} \]

[In]

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

[Out]

(b*e^x - a*log(b*e^x + a))/b^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=- \frac {a \log {\left (\frac {a}{b} + e^{x} \right )}}{b^{2}} + \begin {cases} \frac {e^{x}}{b} & \text {for}\: b \neq 0 \\\frac {x}{b} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {e^{x}}{b} - \frac {a \log \left (b e^{x} + a\right )}{b^{2}} \]

[In]

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

[Out]

e^x/b - a*log(b*e^x + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=\frac {e^{x}}{b} - \frac {a \log \left ({\left | b e^{x} + a \right |}\right )}{b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 x}}{a+b e^x} \, dx=-\frac {a\,\ln \left (a+b\,{\mathrm {e}}^x\right )-b\,{\mathrm {e}}^x}{b^2} \]

[In]

int(exp(2*x)/(a + b*exp(x)),x)

[Out]

-(a*log(a + b*exp(x)) - b*exp(x))/b^2