3.704 \(\int \frac {e^{3 x}}{-1+e^{2 x}} \, dx\)

Optimal. Leaf size=10 \[ e^x-\tanh ^{-1}\left (e^x\right ) \]

[Out]

exp(x)-arctanh(exp(x))

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2248, 321, 207} \[ e^x-\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(3*x)/(-1 + E^(2*x)),x]

[Out]

E^x - ArcTanh[E^x]

Rule 207

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^{3 x}}{-1+e^{2 x}} \, dx &=\operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,e^x\right )\\ &=e^x+\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^x\right )\\ &=e^x-\tanh ^{-1}\left (e^x\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ e^x-\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x)/(-1 + E^(2*x)),x]

[Out]

E^x - ArcTanh[E^x]

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fricas [B]  time = 0.41, size = 17, normalized size = 1.70 \[ e^{x} - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(-1+exp(2*x)),x, algorithm="fricas")

[Out]

e^x - 1/2*log(e^x + 1) + 1/2*log(e^x - 1)

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giac [B]  time = 0.21, size = 18, normalized size = 1.80 \[ e^{x} - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(-1+exp(2*x)),x, algorithm="giac")

[Out]

e^x - 1/2*log(e^x + 1) + 1/2*log(abs(e^x - 1))

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maple [B]  time = 0.03, size = 18, normalized size = 1.80 \[ {\mathrm e}^{x}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x)/(exp(2*x)-1),x)

[Out]

exp(x)+1/2*ln(exp(x)-1)-1/2*ln(exp(x)+1)

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maxima [B]  time = 0.95, size = 17, normalized size = 1.70 \[ e^{x} - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(-1+exp(2*x)),x, algorithm="maxima")

[Out]

e^x - 1/2*log(e^x + 1) + 1/2*log(e^x - 1)

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mupad [B]  time = 0.10, size = 17, normalized size = 1.70 \[ \frac {\ln \left ({\mathrm {e}}^x-1\right )}{2}-\frac {\ln \left ({\mathrm {e}}^x+1\right )}{2}+{\mathrm {e}}^x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x)/(exp(2*x) - 1),x)

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2 + exp(x)

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sympy [B]  time = 0.12, size = 19, normalized size = 1.90 \[ e^{x} + \frac {\log {\left (e^{x} - 1 \right )}}{2} - \frac {\log {\left (e^{x} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(-1+exp(2*x)),x)

[Out]

exp(x) + log(exp(x) - 1)/2 - log(exp(x) + 1)/2

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