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]

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

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \[ \frac {e^x}{b}-\frac {a \log \left (a+b e^x\right )}{b^2} \]

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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fricas [A]  time = 0.43, size = 19, normalized size = 0.86 \[ \frac {b e^{x} - a \log \left (b e^{x} + a\right )}{b^{2}} \]

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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giac [A]  time = 0.39, size = 21, normalized size = 0.95 \[ \frac {e^{x}}{b} - \frac {a \log \left ({\left | b e^{x} + a \right |}\right )}{b^{2}} \]

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

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maple [A]  time = 0.01, size = 21, normalized size = 0.95 \[ -\frac {a \ln \left (b \,{\mathrm e}^{x}+a \right )}{b^{2}}+\frac {{\mathrm e}^{x}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 20, normalized size = 0.91 \[ \frac {e^{x}}{b} - \frac {a \log \left (b e^{x} + a\right )}{b^{2}} \]

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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mupad [B]  time = 3.57, size = 20, normalized size = 0.91 \[ -\frac {a\,\ln \left (a+b\,{\mathrm {e}}^x\right )-b\,{\mathrm {e}}^x}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(a + b*exp(x)),x)

[Out]

-(a*log(a + b*exp(x)) - b*exp(x))/b^2

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sympy [A]  time = 0.13, size = 20, normalized size = 0.91 \[ - \frac {a \log {\left (\frac {a}{b} + e^{x} \right )}}{b^{2}} + \begin {cases} \frac {e^{x}}{b} & \text {for}\: b \neq 0 \\\frac {x}{b} & \text {otherwise} \end {cases} \]

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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