3.79.60 \(\int \frac {-4+e^{6 x} (5+30 x)+(4-5 e^{6 x}) \log (\frac {5 e^6}{-4+5 e^{6 x}})}{-4+5 e^{6 x}} \, dx\)

Optimal. Leaf size=21 \[ x-x \log \left (\frac {e^6}{-\frac {4}{5}+e^{6 x}}\right ) \]

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Rubi [C]  time = 0.22, antiderivative size = 78, normalized size of antiderivative = 3.71, number of steps used = 9, number of rules used = 8, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6742, 2184, 2190, 2279, 2391, 2282, 2394, 2315} \begin {gather*} \frac {1}{6} \text {Li}_2\left (\frac {5 e^{6 x}}{4}\right )+\frac {1}{6} \text {Li}_2\left (1-\frac {5 e^{6 x}}{4}\right )-5 x+x \log \left (1-\frac {5 e^{6 x}}{4}\right )-\frac {1}{6} \log \left (\frac {5 e^{6 x}}{4}\right ) \log \left (-\frac {5}{4-5 e^{6 x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^(6*x)*(5 + 30*x) + (4 - 5*E^(6*x))*Log[(5*E^6)/(-4 + 5*E^(6*x))])/(-4 + 5*E^(6*x)),x]

[Out]

-5*x - (Log[(5*E^(6*x))/4]*Log[-5/(4 - 5*E^(6*x))])/6 + x*Log[1 - (5*E^(6*x))/4] + PolyLog[2, (5*E^(6*x))/4]/6
+ PolyLog[2, 1 - (5*E^(6*x))/4]/6

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5+6 x+\frac {24 x}{-4+5 e^{6 x}}-\log \left (\frac {5}{-4+5 e^{6 x}}\right )\right ) \, dx\\ &=-5 x+3 x^2+24 \int \frac {x}{-4+5 e^{6 x}} \, dx-\int \log \left (\frac {5}{-4+5 e^{6 x}}\right ) \, dx\\ &=-5 x-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log \left (\frac {5}{-4+5 x}\right )}{x} \, dx,x,e^{6 x}\right )+30 \int \frac {e^{6 x} x}{-4+5 e^{6 x}} \, dx\\ &=-5 x-\frac {1}{6} \log \left (\frac {5 e^{6 x}}{4}\right ) \log \left (-\frac {5}{4-5 e^{6 x}}\right )+x \log \left (1-\frac {5 e^{6 x}}{4}\right )-\frac {5}{6} \operatorname {Subst}\left (\int \frac {\log \left (\frac {5 x}{4}\right )}{-4+5 x} \, dx,x,e^{6 x}\right )-\int \log \left (1-\frac {5 e^{6 x}}{4}\right ) \, dx\\ &=-5 x-\frac {1}{6} \log \left (\frac {5 e^{6 x}}{4}\right ) \log \left (-\frac {5}{4-5 e^{6 x}}\right )+x \log \left (1-\frac {5 e^{6 x}}{4}\right )+\frac {1}{6} \text {Li}_2\left (1-\frac {5 e^{6 x}}{4}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {5 x}{4}\right )}{x} \, dx,x,e^{6 x}\right )\\ &=-5 x-\frac {1}{6} \log \left (\frac {5 e^{6 x}}{4}\right ) \log \left (-\frac {5}{4-5 e^{6 x}}\right )+x \log \left (1-\frac {5 e^{6 x}}{4}\right )+\frac {1}{6} \text {Li}_2\left (\frac {5 e^{6 x}}{4}\right )+\frac {1}{6} \text {Li}_2\left (1-\frac {5 e^{6 x}}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.10, size = 80, normalized size = 3.81 \begin {gather*} \frac {1}{6} \left (-30 x+18 x^2+6 x \log \left (1-\frac {4 e^{-6 x}}{5}\right )-\log \left (\frac {5 e^{6 x}}{4}\right ) \log \left (\frac {5}{-4+5 e^{6 x}}\right )-\text {Li}_2\left (\frac {4 e^{-6 x}}{5}\right )+\text {Li}_2\left (1-\frac {5 e^{6 x}}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^(6*x)*(5 + 30*x) + (4 - 5*E^(6*x))*Log[(5*E^6)/(-4 + 5*E^(6*x))])/(-4 + 5*E^(6*x)),x]

[Out]

(-30*x + 18*x^2 + 6*x*Log[1 - 4/(5*E^(6*x))] - Log[(5*E^(6*x))/4]*Log[5/(-4 + 5*E^(6*x))] - PolyLog[2, 4/(5*E^
(6*x))] + PolyLog[2, 1 - (5*E^(6*x))/4])/6

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fricas [A]  time = 0.77, size = 20, normalized size = 0.95 \begin {gather*} -x \log \left (\frac {5 \, e^{6}}{5 \, e^{\left (6 \, x\right )} - 4}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(6*x)+4)*log(5*exp(3)^2/(5*exp(6*x)-4))+(30*x+5)*exp(6*x)-4)/(5*exp(6*x)-4),x, algorithm="fr
icas")

[Out]

-x*log(5*e^6/(5*e^(6*x) - 4)) + x

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giac [A]  time = 0.18, size = 20, normalized size = 0.95 \begin {gather*} -x \log \left (\frac {5}{5 \, e^{\left (6 \, x\right )} - 4}\right ) - 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(6*x)+4)*log(5*exp(3)^2/(5*exp(6*x)-4))+(30*x+5)*exp(6*x)-4)/(5*exp(6*x)-4),x, algorithm="gi
ac")

[Out]

-x*log(5/(5*e^(6*x) - 4)) - 5*x

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maple [A]  time = 0.08, size = 14, normalized size = 0.67

method	result	size
risch	\(x \ln \left ({\mathrm e}^{6 x}-\frac {4}{5}\right )-5 x\)	\(14\)
norman	\(x -x \ln \left (\frac {5 \,{\mathrm e}^{6}}{5 \,{\mathrm e}^{6 x}-4}\right )\)	\(23\)
default	\(\frac {\ln \left ({\mathrm e}^{6 x}\right )}{6}+\frac {\left (6 x -\ln \left (\frac {5 \,{\mathrm e}^{6 x}}{4}\right )\right ) \ln \left (1-\frac {5 \,{\mathrm e}^{6 x}}{4}\right )}{6}-\frac {\left (\ln \left (\frac {5 \,{\mathrm e}^{6}}{5 \,{\mathrm e}^{6 x}-4}\right )+\ln \left (5 \,{\mathrm e}^{6 x}-4\right )\right ) \ln \left ({\mathrm e}^{6 x}\right )}{6}+\frac {\ln \left (5 \,{\mathrm e}^{6 x}-4\right ) \ln \left (\frac {5 \,{\mathrm e}^{6 x}}{4}\right )}{6}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*exp(6*x)+4)*ln(5*exp(3)^2/(5*exp(6*x)-4))+(30*x+5)*exp(6*x)-4)/(5*exp(6*x)-4),x,method=_RETURNVERBOSE
)

[Out]

x*ln(exp(6*x)-4/5)-5*x

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maxima [A]  time = 0.48, size = 20, normalized size = 0.95 \begin {gather*} -x {\left (\log \relax (5) + 6\right )} + x \log \left (5 \, e^{\left (6 \, x\right )} - 4\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(6*x)+4)*log(5*exp(3)^2/(5*exp(6*x)-4))+(30*x+5)*exp(6*x)-4)/(5*exp(6*x)-4),x, algorithm="ma
xima")

[Out]

-x*(log(5) + 6) + x*log(5*e^(6*x) - 4) + x

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mupad [B]  time = 5.76, size = 18, normalized size = 0.86 \begin {gather*} -x\,\left (\ln \relax (5)+\ln \left (\frac {1}{5\,{\mathrm {e}}^{6\,x}-4}\right )+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((5*exp(6))/(5*exp(6*x) - 4))*(5*exp(6*x) - 4) - exp(6*x)*(30*x + 5) + 4)/(5*exp(6*x) - 4),x)

[Out]

-x*(log(5) + log(1/(5*exp(6*x) - 4)) + 5)

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sympy [A]  time = 0.18, size = 17, normalized size = 0.81 \begin {gather*} - x \log {\left (\frac {5 e^{6}}{5 e^{6 x} - 4} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(6*x)+4)*ln(5*exp(3)**2/(5*exp(6*x)-4))+(30*x+5)*exp(6*x)-4)/(5*exp(6*x)-4),x)

[Out]

-x*log(5*exp(6)/(5*exp(6*x) - 4)) + x

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