3.59.39 \(\int \frac {132 \log (e^{-x} (-22+e^x))}{-22+e^x} \, dx\)

Optimal. Leaf size=14 \[ 3 \log ^2\left (1-22 e^{-x}\right ) \]

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Rubi [C]  time = 0.17, antiderivative size = 69, normalized size of antiderivative = 4.93, number of steps used = 12, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {12, 2282, 2466, 2462, 260, 2416, 2390, 2301, 2394, 2315, 2391} \begin {gather*} -6 \text {Li}_2\left (22 e^{-x}\right )+6 \text {Li}_2\left (1-\frac {e^x}{22}\right )-3 \log ^2\left (e^x-22\right )+6 \log \left (\frac {e^x}{22}\right ) \log \left (e^x-22\right )+6 \log \left (1-22 e^{-x}\right ) \log \left (e^x-22\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(132*Log[(-22 + E^x)/E^x])/(-22 + E^x),x]

[Out]

6*Log[E^x/22]*Log[-22 + E^x] + 6*Log[1 - 22/E^x]*Log[-22 + E^x] - 3*Log[-22 + E^x]^2 - 6*PolyLog[2, 22/E^x] +
6*PolyLog[2, 1 - E^x/22]

Rule 12

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

Rule 260

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

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2394

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2462

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

Rule 2466

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_))^(r_.), x_S
ymbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e,
f, g, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=132 \int \frac {\log \left (e^{-x} \left (-22+e^x\right )\right )}{-22+e^x} \, dx\\ &=132 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {22}{x}\right )}{(-22+x) x} \, dx,x,e^x\right )\\ &=132 \operatorname {Subst}\left (\int \left (\frac {\log \left (1-\frac {22}{x}\right )}{22 (-22+x)}-\frac {\log \left (1-\frac {22}{x}\right )}{22 x}\right ) \, dx,x,e^x\right )\\ &=6 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {22}{x}\right )}{-22+x} \, dx,x,e^x\right )-6 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {22}{x}\right )}{x} \, dx,x,e^x\right )\\ &=6 \log \left (1-22 e^{-x}\right ) \log \left (-22+e^x\right )-6 \text {Li}_2\left (22 e^{-x}\right )-132 \operatorname {Subst}\left (\int \frac {\log (-22+x)}{\left (1-\frac {22}{x}\right ) x^2} \, dx,x,e^x\right )\\ &=6 \log \left (1-22 e^{-x}\right ) \log \left (-22+e^x\right )-6 \text {Li}_2\left (22 e^{-x}\right )-132 \operatorname {Subst}\left (\int \left (\frac {\log (-22+x)}{22 (-22+x)}-\frac {\log (-22+x)}{22 x}\right ) \, dx,x,e^x\right )\\ &=6 \log \left (1-22 e^{-x}\right ) \log \left (-22+e^x\right )-6 \text {Li}_2\left (22 e^{-x}\right )-6 \operatorname {Subst}\left (\int \frac {\log (-22+x)}{-22+x} \, dx,x,e^x\right )+6 \operatorname {Subst}\left (\int \frac {\log (-22+x)}{x} \, dx,x,e^x\right )\\ &=6 \log \left (\frac {e^x}{22}\right ) \log \left (-22+e^x\right )+6 \log \left (1-22 e^{-x}\right ) \log \left (-22+e^x\right )-6 \text {Li}_2\left (22 e^{-x}\right )-6 \operatorname {Subst}\left (\int \frac {\log \left (\frac {x}{22}\right )}{-22+x} \, dx,x,e^x\right )-6 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-22+e^x\right )\\ &=6 \log \left (\frac {e^x}{22}\right ) \log \left (-22+e^x\right )+6 \log \left (1-22 e^{-x}\right ) \log \left (-22+e^x\right )-3 \log ^2\left (-22+e^x\right )-6 \text {Li}_2\left (22 e^{-x}\right )+6 \text {Li}_2\left (1-\frac {e^x}{22}\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.03, size = 122, normalized size = 8.71 \begin {gather*} 132 \left (\frac {1}{22} \log \left (22 e^{-x}\right ) \log \left (-e^{-x} \left (22-e^x\right )\right )+\frac {1}{22} \log \left (\frac {e^x}{22}\right ) \log \left (-22+e^x\right )+\frac {1}{22} \log \left (-e^{-x} \left (22-e^x\right )\right ) \log \left (-22+e^x\right )-\frac {1}{44} \log ^2\left (-22+e^x\right )+\frac {1}{22} \text {Li}_2\left (\frac {1}{22} \left (22-e^x\right )\right )+\frac {1}{22} \text {Li}_2\left (-e^{-x} \left (22-e^x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(132*Log[(-22 + E^x)/E^x])/(-22 + E^x),x]

[Out]

132*((Log[22/E^x]*Log[-((22 - E^x)/E^x)])/22 + (Log[E^x/22]*Log[-22 + E^x])/22 + (Log[-((22 - E^x)/E^x)]*Log[-
22 + E^x])/22 - Log[-22 + E^x]^2/44 + PolyLog[2, (22 - E^x)/22]/22 + PolyLog[2, -((22 - E^x)/E^x)]/22)

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fricas [A]  time = 0.75, size = 14, normalized size = 1.00 \begin {gather*} 3 \, \log \left ({\left (e^{x} - 22\right )} e^{\left (-x\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(132*log((exp(x)-22)/exp(x))/(exp(x)-22),x, algorithm="fricas")

[Out]

3*log((e^x - 22)*e^(-x))^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {132 \, \log \left ({\left (e^{x} - 22\right )} e^{\left (-x\right )}\right )}{e^{x} - 22}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(132*log((exp(x)-22)/exp(x))/(exp(x)-22),x, algorithm="giac")

[Out]

integrate(132*log((e^x - 22)*e^(-x))/(e^x - 22), x)

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maple [A]  time = 0.30, size = 14, normalized size = 1.00

method	result	size
derivativedivides	\(3 \ln \left (1-22 \,{\mathrm e}^{-x}\right )^{2}\)	\(14\)
default	\(3 \ln \left (1-22 \,{\mathrm e}^{-x}\right )^{2}\)	\(14\)
norman	\(3 \ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{2}\)	\(15\)
risch	\(132 \left (\frac {x}{22}-\frac {\ln \left ({\mathrm e}^{x}-22\right )}{22}\right ) \ln \left ({\mathrm e}^{x}\right )+3 \ln \left ({\mathrm e}^{x}-22\right )^{2}-3 x^{2}-3 i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{2}+3 i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )+3 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{3}-3 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )+3 i \pi \ln \left ({\mathrm e}^{x}-22\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{2}-3 i \pi \ln \left ({\mathrm e}^{x}-22\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-3 i \pi \ln \left ({\mathrm e}^{x}-22\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{3}+3 i \pi \ln \left ({\mathrm e}^{x}-22\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}-22\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(132*ln((exp(x)-22)/exp(x))/(exp(x)-22),x,method=_RETURNVERBOSE)

[Out]

3*ln(1-22/exp(x))^2

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maxima [B]  time = 0.36, size = 44, normalized size = 3.14 \begin {gather*} -3 \, x^{2} - 6 \, {\left (x - \log \left (e^{x} - 22\right )\right )} \log \left ({\left (e^{x} - 22\right )} e^{\left (-x\right )}\right ) + 6 \, x \log \left (e^{x} - 22\right ) - 3 \, \log \left (e^{x} - 22\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(132*log((exp(x)-22)/exp(x))/(exp(x)-22),x, algorithm="maxima")

[Out]

-3*x^2 - 6*(x - log(e^x - 22))*log((e^x - 22)*e^(-x)) + 6*x*log(e^x - 22) - 3*log(e^x - 22)^2

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mupad [B]  time = 0.20, size = 13, normalized size = 0.93 \begin {gather*} 3\,{\ln \left (1-22\,{\mathrm {e}}^{-x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((132*log(exp(-x)*(exp(x) - 22)))/(exp(x) - 22),x)

[Out]

3*log(1 - 22*exp(-x))^2

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sympy [A]  time = 0.12, size = 12, normalized size = 0.86 \begin {gather*} 3 \log {\left (\left (e^{x} - 22\right ) e^{- x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(132*ln((exp(x)-22)/exp(x))/(exp(x)-22),x)

[Out]

3*log((exp(x) - 22)*exp(-x))**2

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