3.665 \(\int \frac {e^x}{16-e^{2 x}} \, dx\)

Optimal. Leaf size=12 \[ \frac {1}{4} \tanh ^{-1}\left (\frac {e^x}{4}\right ) \]

[Out]

1/4*arctanh(1/4*exp(x))

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Rubi [A]  time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2249, 206} \[ \frac {1}{4} \tanh ^{-1}\left (\frac {e^x}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {align*} \int \frac {e^x}{16-e^{2 x}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,e^x\right )\\ &=\frac {1}{4} \tanh ^{-1}\left (\frac {e^x}{4}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \[ \frac {1}{4} \tanh ^{-1}\left (\frac {e^x}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

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fricas [B]  time = 0.40, size = 15, normalized size = 1.25 \[ \frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left (e^{x} - 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="fricas")

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

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giac [B]  time = 0.19, size = 16, normalized size = 1.33 \[ \frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left ({\left | e^{x} - 4 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="giac")

[Out]

1/8*log(e^x + 4) - 1/8*log(abs(e^x - 4))

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maple [B]  time = 0.04, size = 16, normalized size = 1.33 \[ -\frac {\ln \left ({\mathrm e}^{x}-4\right )}{8}+\frac {\ln \left ({\mathrm e}^{x}+4\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(16-exp(2*x)),x)

[Out]

1/8*ln(exp(x)+4)-1/8*ln(exp(x)-4)

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maxima [B]  time = 1.05, size = 15, normalized size = 1.25 \[ \frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left (e^{x} - 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="maxima")

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

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mupad [B]  time = 3.65, size = 15, normalized size = 1.25 \[ \frac {\ln \left ({\mathrm {e}}^x+4\right )}{8}-\frac {\ln \left ({\mathrm {e}}^x-4\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)/(exp(2*x) - 16),x)

[Out]

log(exp(x) + 4)/8 - log(exp(x) - 4)/8

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sympy [B]  time = 0.11, size = 15, normalized size = 1.25 \[ - \frac {\log {\left (e^{x} - 4 \right )}}{8} + \frac {\log {\left (e^{x} + 4 \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x)

[Out]

-log(exp(x) - 4)/8 + log(exp(x) + 4)/8

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