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

Optimal. Leaf size=26 \[ -x+5 \left (e^3-\log \left (4 (3-x) \log \left (-1+e^x\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 6, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6688, 43, 2282, 2390, 2302, 29} \begin {gather*} -x-5 \log (3-x)-5 \log \left (\log \left (e^x-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2282

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 2302

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 2390

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-1+e^x\right )\\ &=-x-5 \log (3-x)-5 \operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 21, normalized size = 0.81 \begin {gather*} -x-5 \log (3-x)-5 \log \left (\log \left (-1+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 18, normalized size = 0.69 \begin {gather*} -x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.29, size = 18, normalized size = 0.69 \begin {gather*} -x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 19, normalized size = 0.73

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 18, normalized size = 0.69 \begin {gather*} -x - 5 \, \log \left (x - 3\right ) - 5 \, \log \left (\log \left (e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.92, size = 18, normalized size = 0.69 \begin {gather*} -x-5\,\ln \left (\ln \left ({\mathrm {e}}^x-1\right )\right )-5\,\ln \left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.17, size = 19, normalized size = 0.73 \begin {gather*} - x - 5 \log {\left (x - 3 \right )} - 5 \log {\left (\log {\left (e^{x} - 1 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________