3.97.23 \(\int \frac {-10+2 e^4-2 x+(5-e^4) \log (x)}{(-5 x+e^4 x-x^2) \log (x)} \, dx\)

Optimal. Leaf size=27 \[ \log \left (\frac {4 e^8 \left (5-e^4+x\right ) \log ^2(3) \log ^2(x)}{225 x}\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps used = 10, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {6, 1593, 6688, 14, 36, 31, 29, 2302} \begin {gather*} -\log (x)+\log \left (x-e^4+5\right )+2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + 2*E^4 - 2*x + (5 - E^4)*Log[x])/((-5*x + E^4*x - x^2)*Log[x]),x]

[Out]

-Log[x] + Log[5 - E^4 + x] + 2*Log[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 \frac {-10+2 e^4-2 x+\left (5-e^4\right ) \log (x)}{\left (\left (-5+e^4\right ) x-x^2\right ) \log (x)} \, dx\\ &=\int \frac {-10+2 e^4-2 x+\left (5-e^4\right ) \log (x)}{\left (-5+e^4-x\right ) x \log (x)} \, dx\\ &=\int \frac {\frac {5-e^4}{-5+e^4-x}+\frac {2}{\log (x)}}{x} \, dx\\ &=\int \left (\frac {5-e^4}{\left (-5+e^4-x\right ) x}+\frac {2}{x \log (x)}\right ) \, dx\\ &=2 \int \frac {1}{x \log (x)} \, dx+\left (5-e^4\right ) \int \frac {1}{\left (-5+e^4-x\right ) x} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )-\int \frac {1}{-5+e^4-x} \, dx-\int \frac {1}{x} \, dx\\ &=-\log (x)+\log \left (5-e^4+x\right )+2 \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 19, normalized size = 0.70 \begin {gather*} -\log (x)+\log \left (5-e^4+x\right )+2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 2*E^4 - 2*x + (5 - E^4)*Log[x])/((-5*x + E^4*x - x^2)*Log[x]),x]

[Out]

-Log[x] + Log[5 - E^4 + x] + 2*Log[Log[x]]

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fricas [A]  time = 0.54, size = 18, normalized size = 0.67 \begin {gather*} \log \left (x - e^{4} + 5\right ) - \log \relax (x) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-exp(4))*log(x)+2*exp(4)-2*x-10)/(x*exp(4)-x^2-5*x)/log(x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 18, normalized size = 0.67 \begin {gather*} \log \left (x - e^{4} + 5\right ) - \log \relax (x) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-exp(4))*log(x)+2*exp(4)-2*x-10)/(x*exp(4)-x^2-5*x)/log(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 19, normalized size = 0.70

method	result	size
norman	\(-\ln \relax (x )+2 \ln \left (\ln \relax (x )\right )+\ln \left ({\mathrm e}^{4}-x -5\right )\)	\(19\)
risch	\(\ln \left (5-{\mathrm e}^{4}+x \right )-\ln \relax (x )+2 \ln \left (\ln \relax (x )\right )\)	\(19\)
default	\(2 \ln \left (\ln \relax (x )\right )+\frac {5 \ln \relax (x )}{{\mathrm e}^{4}-5}-\frac {5 \ln \left (5-{\mathrm e}^{4}+x \right )}{{\mathrm e}^{4}-5}-\frac {{\mathrm e}^{4} \ln \relax (x )}{{\mathrm e}^{4}-5}+\frac {{\mathrm e}^{4} \ln \left (5-{\mathrm e}^{4}+x \right )}{{\mathrm e}^{4}-5}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5-exp(4))*ln(x)+2*exp(4)-2*x-10)/(x*exp(4)-x^2-5*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 18, normalized size = 0.67 \begin {gather*} \log \left (x - e^{4} + 5\right ) - \log \relax (x) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-exp(4))*log(x)+2*exp(4)-2*x-10)/(x*exp(4)-x^2-5*x)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.82, size = 18, normalized size = 0.67 \begin {gather*} \ln \left (x-{\mathrm {e}}^4+5\right )+2\,\ln \left (\ln \relax (x)\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 2*exp(4) + log(x)*(exp(4) - 5) + 10)/(log(x)*(5*x - x*exp(4) + x^2)),x)

[Out]

log(x - exp(4) + 5) + 2*log(log(x)) - log(x)

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sympy [B]  time = 0.31, size = 102, normalized size = 3.78 \begin {gather*} \left (-5 + e^{4}\right ) \left (\frac {\log {\left (x - \frac {e^{8}}{2 \left (-5 + e^{4}\right )} - \frac {e^{4}}{2} - \frac {25}{2 \left (-5 + e^{4}\right )} + \frac {5}{2} + \frac {5 e^{4}}{-5 + e^{4}} \right )}}{-5 + e^{4}} - \frac {\log {\left (x - \frac {e^{4}}{2} - \frac {5 e^{4}}{-5 + e^{4}} + \frac {25}{2 \left (-5 + e^{4}\right )} + \frac {5}{2} + \frac {e^{8}}{2 \left (-5 + e^{4}\right )} \right )}}{-5 + e^{4}}\right ) + 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-exp(4))*ln(x)+2*exp(4)-2*x-10)/(x*exp(4)-x**2-5*x)/ln(x),x)

[Out]

(-5 + exp(4))*(log(x - exp(8)/(2*(-5 + exp(4))) - exp(4)/2 - 25/(2*(-5 + exp(4))) + 5/2 + 5*exp(4)/(-5 + exp(4
)))/(-5 + exp(4)) - log(x - exp(4)/2 - 5*exp(4)/(-5 + exp(4)) + 25/(2*(-5 + exp(4))) + 5/2 + exp(8)/(2*(-5 + e
xp(4))))/(-5 + exp(4))) + 2*log(log(x))

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