3.3.97 \(\int \frac {-4-2 e^{\frac {1}{2} (-x+2 \log (3))}+e^{\frac {1}{2} (-x+2 \log (3))} (1-x) \log (1-x)}{(-12+12 x+e^{\frac {1}{2} (-x+2 \log (3))} (-6+6 x)) \log (1-x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{3} \log \left (\frac {2 \left (2+3 e^{-x/2}\right )}{\log (1-x)}\right ) \]

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Rubi [A]  time = 0.91, antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 11, number of rules used = 9, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {6688, 12, 6742, 2282, 36, 29, 31, 2390, 2302} \begin {gather*} -\frac {x}{6}+\frac {1}{3} \log \left (2 e^{x/2}+3\right )-\frac {1}{3} \log (\log (1-x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 2*E^((-x + 2*Log[3])/2) + E^((-x + 2*Log[3])/2)*(1 - x)*Log[1 - x])/((-12 + 12*x + E^((-x + 2*Log[3]
)/2)*(-6 + 6*x))*Log[1 - x]),x]

[Out]

-1/6*x + Log[3 + 2*E^(x/2)]/3 - Log[Log[1 - 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 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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]]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+4 e^{x/2}+3 (-1+x) \log (1-x)}{6 \left (3+2 e^{x/2}\right ) (1-x) \log (1-x)} \, dx\\ &=\frac {1}{6} \int \frac {6+4 e^{x/2}+3 (-1+x) \log (1-x)}{\left (3+2 e^{x/2}\right ) (1-x) \log (1-x)} \, dx\\ &=\frac {1}{6} \int \left (-\frac {3}{3+2 e^{x/2}}-\frac {2}{(-1+x) \log (1-x)}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{(-1+x) \log (1-x)} \, dx\right )-\frac {1}{2} \int \frac {1}{3+2 e^{x/2}} \, dx\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,1-x\right )\right )-\operatorname {Subst}\left (\int \frac {1}{x (3+2 x)} \, dx,x,e^{x/2}\right )\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{x/2}\right )\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (1-x)\right )+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{3+2 x} \, dx,x,e^{x/2}\right )\\ &=-\frac {x}{6}+\frac {1}{3} \log \left (3+2 e^{x/2}\right )-\frac {1}{3} \log (\log (1-x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{6} \left (-x+2 \log \left (3+2 e^{x/2}\right )-2 \log (\log (1-x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 2*E^((-x + 2*Log[3])/2) + E^((-x + 2*Log[3])/2)*(1 - x)*Log[1 - x])/((-12 + 12*x + E^((-x + 2*
Log[3])/2)*(-6 + 6*x))*Log[1 - x]),x]

[Out]

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

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fricas [A]  time = 0.72, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{3} \, \log \left (e^{\left (-\frac {1}{2} \, x + \log \relax (3)\right )} + 2\right ) - \frac {1}{3} \, \log \left (\log \left (-x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(log(3)-1/2*x)*log(-x+1)-2*exp(log(3)-1/2*x)-4)/((6*x-6)*exp(log(3)-1/2*x)+12*x-12)/log(-
x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 21, normalized size = 0.81 \begin {gather*} \frac {1}{3} \, \log \left (3 \, e^{\left (-\frac {1}{2} \, x\right )} + 2\right ) - \frac {1}{3} \, \log \left (\log \left (-x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(log(3)-1/2*x)*log(-x+1)-2*exp(log(3)-1/2*x)-4)/((6*x-6)*exp(log(3)-1/2*x)+12*x-12)/log(-
x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 23, normalized size = 0.88

method	result	size
norman	\(-\frac {\ln \left (\ln \left (1-x \right )\right )}{3}+\frac {\ln \left ({\mathrm e}^{\ln \relax (3)-\frac {x}{2}}+2\right )}{3}\)	\(23\)
risch	\(-\frac {\ln \relax (3)}{3}+\frac {\ln \left (3 \,{\mathrm e}^{-\frac {x}{2}}+2\right )}{3}-\frac {\ln \left (\ln \left (1-x \right )\right )}{3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*exp(ln(3)-1/2*x)*ln(1-x)-2*exp(ln(3)-1/2*x)-4)/((6*x-6)*exp(ln(3)-1/2*x)+12*x-12)/ln(1-x),x,method=
_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.23, size = 22, normalized size = 0.85 \begin {gather*} -\frac {1}{6} \, x + \frac {1}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + \frac {3}{2}\right ) - \frac {1}{3} \, \log \left (\log \left (-x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(log(3)-1/2*x)*log(-x+1)-2*exp(log(3)-1/2*x)-4)/((6*x-6)*exp(log(3)-1/2*x)+12*x-12)/log(-
x+1),x, algorithm="maxima")

[Out]

-1/6*x + 1/3*log(e^(1/2*x) + 3/2) - 1/3*log(log(-x + 1))

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mupad [B]  time = 0.54, size = 21, normalized size = 0.81 \begin {gather*} \frac {\ln \left (\frac {3}{\sqrt {{\mathrm {e}}^x}}+2\right )}{3}-\frac {\ln \left (\ln \left (1-x\right )\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(log(3) - x/2) + exp(log(3) - x/2)*log(1 - x)*(x - 1) + 4)/(log(1 - x)*(12*x + exp(log(3) - x/2)*(6
*x - 6) - 12)),x)

[Out]

log(3/exp(x)^(1/2) + 2)/3 - log(log(1 - x))/3

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sympy [A]  time = 0.39, size = 20, normalized size = 0.77 \begin {gather*} \frac {\log {\left (\frac {2}{3} + e^{- \frac {x}{2}} \right )}}{3} - \frac {\log {\left (\log {\left (1 - x \right )} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(ln(3)-1/2*x)*ln(-x+1)-2*exp(ln(3)-1/2*x)-4)/((6*x-6)*exp(ln(3)-1/2*x)+12*x-12)/ln(-x+1),
x)

[Out]

log(2/3 + exp(-x/2))/3 - log(log(1 - x))/3

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