\(\int \frac {e^x (15-5 x)+(2+e^x (-2-x)+x) \log (-1+e^x)}{(3+e^x (-3+x)-x) \log (-1+e^x)} \, dx\) [7685]

Optimal result

Integrand size = 52, antiderivative size = 26 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x+5 \left (e^3-\log \left (4 (3-x) \log \left (-1+e^x\right )\right )\right ) \]

[Out]

5*exp(3)-5*ln(4*ln(-1+exp(x))*(-x+3))-x

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6820, 45, 2320, 2437, 2339, 29} \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x-5 \log (3-x)-5 \log \left (\log \left (e^x-1\right )\right ) \]

[In]

Int[(E^x*(15 - 5*x) + (2 + E^x*(-2 - x) + x)*Log[-1 + E^x])/((3 + E^x*(-3 + x) - x)*Log[-1 + E^x]),x]

[Out]

-x - 5*Log[3 - x] - 5*Log[Log[-1 + E^x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2+x}{3-x}-\frac {5 e^x}{\left (-1+e^x\right ) \log \left (-1+e^x\right )}\right ) \, dx \\ & = -\left (5 \int \frac {e^x}{\left (-1+e^x\right ) \log \left (-1+e^x\right )} \, dx\right )+\int \frac {2+x}{3-x} \, dx \\ & = -\left (5 \text {Subst}\left (\int \frac {1}{(-1+x) \log (-1+x)} \, dx,x,e^x\right )\right )+\int \left (-1-\frac {5}{-3+x}\right ) \, dx \\ & = -x-5 \log (3-x)-5 \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-1+e^x\right ) \\ & = -x-5 \log (3-x)-5 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (-1+e^x\right )\right ) \\ & = -x-5 \log (3-x)-5 \log \left (\log \left (-1+e^x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x-5 \log (3-x)-5 \log \left (\log \left (-1+e^x\right )\right ) \]

[In]

Integrate[(E^x*(15 - 5*x) + (2 + E^x*(-2 - x) + x)*Log[-1 + E^x])/((3 + E^x*(-3 + x) - x)*Log[-1 + E^x]),x]

[Out]

-x - 5*Log[3 - x] - 5*Log[Log[-1 + E^x]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73

method	result	size
norman	\(-x -5 \ln \left (\ln \left ({\mathrm e}^{x}-1\right )\right )-5 \ln \left (-3+x \right )\)	\(19\)
risch	\(-x -5 \ln \left (\ln \left ({\mathrm e}^{x}-1\right )\right )-5 \ln \left (-3+x \right )\)	\(19\)
parallelrisch	\(-x -5 \ln \left (\ln \left ({\mathrm e}^{x}-1\right )\right )-5 \ln \left (-3+x \right )\)	\(19\)

[In]

int((((-2-x)*exp(x)+2+x)*ln(exp(x)-1)+(15-5*x)*exp(x))/((-3+x)*exp(x)+3-x)/ln(exp(x)-1),x,method=_RETURNVERBOS
E)

[Out]

-x-5*ln(ln(exp(x)-1))-5*ln(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \]

[In]

integrate((((-2-x)*exp(x)+2+x)*log(exp(x)-1)+(15-5*x)*exp(x))/((-3+x)*exp(x)+3-x)/log(exp(x)-1),x, algorithm="
fricas")

[Out]

-x - 5*log(x - 3) - 5*log(log(e^x - 1))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=- x - 5 \log {\left (x - 3 \right )} - 5 \log {\left (\log {\left (e^{x} - 1 \right )} \right )} \]

[In]

integrate((((-2-x)*exp(x)+2+x)*ln(exp(x)-1)+(15-5*x)*exp(x))/((-3+x)*exp(x)+3-x)/ln(exp(x)-1),x)

[Out]

-x - 5*log(x - 3) - 5*log(log(exp(x) - 1))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \]

[In]

integrate((((-2-x)*exp(x)+2+x)*log(exp(x)-1)+(15-5*x)*exp(x))/((-3+x)*exp(x)+3-x)/log(exp(x)-1),x, algorithm="
maxima")

[Out]

-x - 5*log(x - 3) - 5*log(log(e^x - 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \]

[In]

integrate((((-2-x)*exp(x)+2+x)*log(exp(x)-1)+(15-5*x)*exp(x))/((-3+x)*exp(x)+3-x)/log(exp(x)-1),x, algorithm="
giac")

[Out]

-x - 5*log(x - 3) - 5*log(log(e^x - 1))

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^x (15-5 x)+\left (2+e^x (-2-x)+x\right ) \log \left (-1+e^x\right )}{\left (3+e^x (-3+x)-x\right ) \log \left (-1+e^x\right )} \, dx=-x-5\,\ln \left (\ln \left ({\mathrm {e}}^x-1\right )\right )-5\,\ln \left (x-3\right ) \]

[In]

int(-(exp(x)*(5*x - 15) - log(exp(x) - 1)*(x - exp(x)*(x + 2) + 2))/(log(exp(x) - 1)*(exp(x)*(x - 3) - x + 3))
,x)

[Out]

- x - 5*log(log(exp(x) - 1)) - 5*log(x - 3)