\(\int \frac {e^x}{16-e^{2 x}} \, dx\) [665]

Optimal result

Integrand size = 15, antiderivative size = 12 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{4} \text {arctanh}\left (\frac {e^x}{4}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2281, 212} \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{4} \text {arctanh}\left (\frac {e^x}{4}\right ) \]

[In]

Int[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

Rule 212

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

Rule 2281

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*} \text {integral}& = \text {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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{4} \text {arctanh}\left (\frac {e^x}{4}\right ) \]

[In]

Integrate[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(15\) vs. \(2(7)=14\).

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {\ln \left ({\mathrm e}^{x}+4\right )}{8}-\frac {\ln \left ({\mathrm e}^{x}-4\right )}{8}\)	\(16\)
norman	\(\frac {\ln \left ({\mathrm e}^{x}+4\right )}{8}-\frac {\ln \left ({\mathrm e}^{x}-4\right )}{8}\)	\(16\)
risch	\(\frac {\ln \left ({\mathrm e}^{x}+4\right )}{8}-\frac {\ln \left ({\mathrm e}^{x}-4\right )}{8}\)	\(16\)

[In]

int(exp(x)/(16-exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left (e^{x} - 4\right ) \]

[In]

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

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=- \frac {\log {\left (e^{x} - 4 \right )}}{8} + \frac {\log {\left (e^{x} + 4 \right )}}{8} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left (e^{x} - 4\right ) \]

[In]

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

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (7) = 14\).

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {1}{8} \, \log \left (e^{x} + 4\right ) - \frac {1}{8} \, \log \left ({\left | e^{x} - 4 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{16-e^{2 x}} \, dx=\frac {\ln \left ({\mathrm {e}}^x+4\right )}{8}-\frac {\ln \left ({\mathrm {e}}^x-4\right )}{8} \]

[In]

int(-exp(x)/(exp(2*x) - 16),x)

[Out]

log(exp(x) + 4)/8 - log(exp(x) - 4)/8