3.62.66 \(\int \frac {1+e^4 \log (x)+(6+6 e^4+(1+7 e^4) \log (x)+e^4 \log ^2(x)) \log (\frac {5+\log (2)}{6+\log (x)})}{(6 x+(x+6 e^4 x) \log (x)+e^4 x \log ^2(x)) \log (\frac {5+\log (2)}{6+\log (x)})} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {x+e^4 x \log (x)}{\log \left (\frac {5+\log (2)}{6+\log (x)}\right )}\right ) \]

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Rubi [A]  time = 0.51, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 7, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6, 6688, 6728, 43, 2390, 2302, 29} \begin {gather*} \log (x)+\log \left (e^4 \log (x)+1\right )-\log \left (\log \left (\frac {5+\log (2)}{\log (x)+6}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^4*Log[x] + (6 + 6*E^4 + (1 + 7*E^4)*Log[x] + E^4*Log[x]^2)*Log[(5 + Log[2])/(6 + Log[x])])/((6*x +
(x + 6*E^4*x)*Log[x] + E^4*x*Log[x]^2)*Log[(5 + Log[2])/(6 + Log[x])]),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {1+e^4 x+6 \log \left (\frac {5+\log (2)}{6+x}\right )+6 e^4 \log \left (\frac {5+\log (2)}{6+x}\right )+x \log \left (\frac {5+\log (2)}{6+x}\right )+7 e^4 x \log \left (\frac {5+\log (2)}{6+x}\right )+e^4 x^2 \log \left (\frac {5+\log (2)}{6+x}\right )}{\left (6+x+6 e^4 x+e^4 x^2\right ) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1+e^4 x+6 \log \left (\frac {5+\log (2)}{6+x}\right )+6 e^4 \log \left (\frac {5+\log (2)}{6+x}\right )+x \log \left (\frac {5+\log (2)}{6+x}\right )+7 e^4 x \log \left (\frac {5+\log (2)}{6+x}\right )+e^4 x^2 \log \left (\frac {5+\log (2)}{6+x}\right )}{\left (6+\left (1+6 e^4\right ) x+e^4 x^2\right ) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1+e^4 x+\left (6+6 e^4\right ) \log \left (\frac {5+\log (2)}{6+x}\right )+x \log \left (\frac {5+\log (2)}{6+x}\right )+7 e^4 x \log \left (\frac {5+\log (2)}{6+x}\right )+e^4 x^2 \log \left (\frac {5+\log (2)}{6+x}\right )}{\left (6+\left (1+6 e^4\right ) x+e^4 x^2\right ) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1+e^4 x+\left (6+6 e^4\right ) \log \left (\frac {5+\log (2)}{6+x}\right )+\left (1+7 e^4\right ) x \log \left (\frac {5+\log (2)}{6+x}\right )+e^4 x^2 \log \left (\frac {5+\log (2)}{6+x}\right )}{\left (6+\left (1+6 e^4\right ) x+e^4 x^2\right ) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1+e^4 x+\left (6+x+e^4 \left (6+7 x+x^2\right )\right ) \log \left (\frac {5+\log (2)}{6+x}\right )}{\left (6+\left (1+6 e^4\right ) x+e^4 x^2\right ) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1+e^4+e^4 x}{1+e^4 x}+\frac {1}{(6+x) \log \left (\frac {5+\log (2)}{6+x}\right )}\right ) \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1+e^4+e^4 x}{1+e^4 x} \, dx,x,\log (x)\right )+\operatorname {Subst}\left (\int \frac {1}{(6+x) \log \left (\frac {5+\log (2)}{6+x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \left (1+\frac {e^4}{1+e^4 x}\right ) \, dx,x,\log (x)\right )+\operatorname {Subst}\left (\int \frac {1}{x \log \left (\frac {5+\log (2)}{x}\right )} \, dx,x,6+\log (x)\right )\\ &=\log (x)+\log \left (1+e^4 \log (x)\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {5+\log (2)}{6+\log (x)}\right )\right )\\ &=\log (x)+\log \left (1+e^4 \log (x)\right )-\log \left (\log \left (\frac {5+\log (2)}{6+\log (x)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.12 \begin {gather*} 6+\log (x)+\log \left (1+e^4 \log (x)\right )-\log \left (\log \left (\frac {5+\log (2)}{6+\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^4*Log[x] + (6 + 6*E^4 + (1 + 7*E^4)*Log[x] + E^4*Log[x]^2)*Log[(5 + Log[2])/(6 + Log[x])])/((
6*x + (x + 6*E^4*x)*Log[x] + E^4*x*Log[x]^2)*Log[(5 + Log[2])/(6 + Log[x])]),x]

[Out]

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

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fricas [A]  time = 0.70, size = 26, normalized size = 1.04 \begin {gather*} \log \left (e^{4} \log \relax (x) + 1\right ) + \log \relax (x) - \log \left (\log \left (\frac {\log \relax (2) + 5}{\log \relax (x) + 6}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5)/(log(x)+6))+exp(4)*log(x)+1)/(x*exp
(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x)/log((log(2)+5)/(log(x)+6)),x, algorithm="fricas")

[Out]

log(e^4*log(x) + 1) + log(x) - log(log((log(2) + 5)/(log(x) + 6)))

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giac [A]  time = 0.41, size = 27, normalized size = 1.08 \begin {gather*} \log \left (e^{4} \log \relax (x) + 1\right ) + \log \relax (x) - \log \left (-\log \left (\log \relax (2) + 5\right ) + \log \left (\log \relax (x) + 6\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5)/(log(x)+6))+exp(4)*log(x)+1)/(x*exp
(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x)/log((log(2)+5)/(log(x)+6)),x, algorithm="giac")

[Out]

log(e^4*log(x) + 1) + log(x) - log(-log(log(2) + 5) + log(log(x) + 6))

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maple [A]  time = 0.61, size = 26, normalized size = 1.04

method	result	size
risch	\(\ln \relax (x )+\ln \left (\ln \relax (x )+{\mathrm e}^{-4}\right )-\ln \left (\ln \left (\ln \relax (x )+6\right )-\ln \left (\ln \relax (2)+5\right )\right )\)	\(26\)
norman	\(\ln \relax (x )-\ln \left (\ln \left (\frac {\ln \relax (2)+5}{\ln \relax (x )+6}\right )\right )+\ln \left ({\mathrm e}^{4} \ln \relax (x )+1\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(4)*ln(x)^2+(7*exp(4)+1)*ln(x)+6*exp(4)+6)*ln((ln(2)+5)/(ln(x)+6))+exp(4)*ln(x)+1)/(x*exp(4)*ln(x)^2+
(6*x*exp(4)+x)*ln(x)+6*x)/ln((ln(2)+5)/(ln(x)+6)),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(ln(x)+exp(-4))-ln(ln(ln(x)+6)-ln(ln(2)+5))

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maxima [A]  time = 0.51, size = 30, normalized size = 1.20 \begin {gather*} \log \left ({\left (e^{4} \log \relax (x) + 1\right )} e^{\left (-4\right )}\right ) + \log \relax (x) - \log \left (-\log \left (\log \relax (2) + 5\right ) + \log \left (\log \relax (x) + 6\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5)/(log(x)+6))+exp(4)*log(x)+1)/(x*exp
(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x)/log((log(2)+5)/(log(x)+6)),x, algorithm="maxima")

[Out]

log((e^4*log(x) + 1)*e^(-4)) + log(x) - log(-log(log(2) + 5) + log(log(x) + 6))

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mupad [B]  time = 4.91, size = 24, normalized size = 0.96 \begin {gather*} \ln \left ({\mathrm {e}}^{-4}+\ln \relax (x)\right )-\ln \left (\ln \left (\frac {\ln \relax (2)+5}{\ln \relax (x)+6}\right )\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((log(2) + 5)/(log(x) + 6))*(6*exp(4) + exp(4)*log(x)^2 + log(x)*(7*exp(4) + 1) + 6) + exp(4)*log(x) +
1)/(log((log(2) + 5)/(log(x) + 6))*(6*x + log(x)*(x + 6*x*exp(4)) + x*exp(4)*log(x)^2)),x)

[Out]

log(exp(-4) + log(x)) - log(log((log(2) + 5)/(log(x) + 6))) + log(x)

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sympy [A]  time = 0.36, size = 26, normalized size = 1.04 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\relax (x )} + e^{-4} \right )} - \log {\left (\log {\left (\frac {\log {\relax (2 )} + 5}{\log {\relax (x )} + 6} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(4)*ln(x)**2+(7*exp(4)+1)*ln(x)+6*exp(4)+6)*ln((ln(2)+5)/(ln(x)+6))+exp(4)*ln(x)+1)/(x*exp(4)*l
n(x)**2+(6*x*exp(4)+x)*ln(x)+6*x)/ln((ln(2)+5)/(ln(x)+6)),x)

[Out]

log(x) + log(log(x) + exp(-4)) - log(log((log(2) + 5)/(log(x) + 6)))

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