3.640 \(\int e^{3 x/2} \log (-1+e^x) \, dx\)

Optimal. Leaf size=52 \[ -\frac{4 e^{x/2}}{3}-\frac{4}{9} e^{3 x/2}+\frac{2}{3} e^{3 x/2} \log \left (e^x-1\right )+\frac{4}{3} \tanh ^{-1}\left (e^{x/2}\right ) \]

[Out]

(-4*E^(x/2))/3 - (4*E^((3*x)/2))/9 + (4*ArcTanh[E^(x/2)])/3 + (2*E^((3*x)/2)*Log[-1 + E^x])/3

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Rubi [A]  time = 0.0436347, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2194, 2554, 12, 2248, 302, 207} \[ -\frac{4 e^{x/2}}{3}-\frac{4}{9} e^{3 x/2}+\frac{2}{3} e^{3 x/2} \log \left (e^x-1\right )+\frac{4}{3} \tanh ^{-1}\left (e^{x/2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^((3*x)/2)*Log[-1 + E^x],x]

[Out]

(-4*E^(x/2))/3 - (4*E^((3*x)/2))/9 + (4*ArcTanh[E^(x/2)])/3 + (2*E^((3*x)/2)*Log[-1 + E^x])/3

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

\begin{align*} \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx &=\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\int \frac{2 e^{5 x/2}}{3 \left (-1+e^x\right )} \, dx\\ &=\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac{2}{3} \int \frac{e^{5 x/2}}{-1+e^x} \, dx\\ &=\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac{4}{3} \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,e^{x/2}\right )\\ &=\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac{4}{3} \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,e^{x/2}\right )\\ &=-\frac{4 e^{x/2}}{3}-\frac{4}{9} e^{3 x/2}+\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac{4}{3} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{x/2}\right )\\ &=-\frac{4 e^{x/2}}{3}-\frac{4}{9} e^{3 x/2}+\frac{4}{3} \tanh ^{-1}\left (e^{x/2}\right )+\frac{2}{3} e^{3 x/2} \log \left (-1+e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0333664, size = 42, normalized size = 0.81 \[ \frac{2}{9} \left (e^{x/2} \left (3 e^x \log \left (e^x-1\right )-2 \left (e^x+3\right )\right )+6 \tanh ^{-1}\left (e^{x/2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^((3*x)/2)*Log[-1 + E^x],x]

[Out]

(2*(6*ArcTanh[E^(x/2)] + E^(x/2)*(-2*(3 + E^x) + 3*E^x*Log[-1 + E^x])))/9

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Maple [A]  time = 0.023, size = 43, normalized size = 0.8 \begin{align*}{\frac{2\,\ln \left ( -1+{{\rm e}^{x}} \right ) }{3}{{\rm e}^{{\frac{3\,x}{2}}}}}-{\frac{4}{9}{{\rm e}^{{\frac{3\,x}{2}}}}}-{\frac{4}{3}{{\rm e}^{{\frac{x}{2}}}}}+{\frac{2}{3}\ln \left ({{\rm e}^{{\frac{x}{2}}}}+1 \right ) }-{\frac{2}{3}\ln \left ( -1+{{\rm e}^{{\frac{x}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*exp(3/2*x)*ln(-1+exp(x))-4/9*exp(3/2*x)-4/3*exp(1/2*x)+2/3*ln(exp(1/2*x)+1)-2/3*ln(-1+exp(1/2*x))

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Maxima [A]  time = 0.950337, size = 57, normalized size = 1.1 \begin{align*} \frac{2}{3} \, e^{\left (\frac{3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac{4}{9} \, e^{\left (\frac{3}{2} \, x\right )} - \frac{4}{3} \, e^{\left (\frac{1}{2} \, x\right )} + \frac{2}{3} \, \log \left (e^{\left (\frac{1}{2} \, x\right )} + 1\right ) - \frac{2}{3} \, \log \left (e^{\left (\frac{1}{2} \, x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^(3/2*x)*log(e^x - 1) - 4/9*e^(3/2*x) - 4/3*e^(1/2*x) + 2/3*log(e^(1/2*x) + 1) - 2/3*log(e^(1/2*x) - 1)

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Fricas [A]  time = 2.37824, size = 149, normalized size = 2.87 \begin{align*} \frac{2}{3} \, e^{\left (\frac{3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac{4}{9} \, e^{\left (\frac{3}{2} \, x\right )} - \frac{4}{3} \, e^{\left (\frac{1}{2} \, x\right )} + \frac{2}{3} \, \log \left (e^{\left (\frac{1}{2} \, x\right )} + 1\right ) - \frac{2}{3} \, \log \left (e^{\left (\frac{1}{2} \, x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^(3/2*x)*log(e^x - 1) - 4/9*e^(3/2*x) - 4/3*e^(1/2*x) + 2/3*log(e^(1/2*x) + 1) - 2/3*log(e^(1/2*x) - 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.07113, size = 58, normalized size = 1.12 \begin{align*} \frac{2}{3} \, e^{\left (\frac{3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac{4}{9} \, e^{\left (\frac{3}{2} \, x\right )} - \frac{4}{3} \, e^{\left (\frac{1}{2} \, x\right )} + \frac{2}{3} \, \log \left (e^{\left (\frac{1}{2} \, x\right )} + 1\right ) - \frac{2}{3} \, \log \left ({\left | e^{\left (\frac{1}{2} \, x\right )} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^(3/2*x)*log(e^x - 1) - 4/9*e^(3/2*x) - 4/3*e^(1/2*x) + 2/3*log(e^(1/2*x) + 1) - 2/3*log(abs(e^(1/2*x) -
1))