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

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

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 12

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

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*}

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Mathematica [A]  time = 0.04, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.88, size = 42, normalized size = 0.81 \[ \frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} - 1\right ) \]

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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giac [A]  time = 0.86, size = 43, normalized size = 0.83 \[ \frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left ({\left | e^{\left (\frac {1}{2} \, x\right )} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.03, size = 43, normalized size = 0.83

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{\frac {3 x}{2}} \ln \left (-1+{\mathrm e}^{x}\right )}{3}-\frac {4 \,{\mathrm e}^{\frac {3 x}{2}}}{9}-\frac {4 \,{\mathrm e}^{\frac {x}{2}}}{3}-\frac {2 \ln \left (-1+{\mathrm e}^{\frac {x}{2}}\right )}{3}+\frac {2 \ln \left ({\mathrm e}^{\frac {x}{2}}+1\right )}{3}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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maxima [A]  time = 0.45, size = 42, normalized size = 0.81 \[ \frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} - 1\right ) \]

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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mupad [B]  time = 0.55, size = 31, normalized size = 0.60 \[ \frac {4\,\mathrm {atanh}\left (\sqrt {{\mathrm {e}}^x}\right )}{3}-\frac {4\,{\mathrm {e}}^{\frac {3\,x}{2}}}{9}-\frac {4\,{\mathrm {e}}^{x/2}}{3}+\frac {2\,{\mathrm {e}}^{\frac {3\,x}{2}}\,\ln \left ({\mathrm {e}}^x-1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((3*x)/2)*log(exp(x) - 1),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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