3.36.67 \(\int \frac {1}{((-8 x-4 e^5 x) \log (\frac {7}{x})+4 x \log (\frac {7}{x}) \log (\log (\frac {7}{x}))) \log (2+e^5-\log (\log (\frac {7}{x})))} \, dx\)

Optimal. Leaf size=22 \[ 16-\frac {1}{4} \log \left (\log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 2390, 2302, 29} \begin {gather*} -\frac {1}{4} \log \left (\log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(((-8*x - 4*E^5*x)*Log[7/x] + 4*x*Log[7/x]*Log[Log[7/x]])*Log[2 + E^5 - Log[Log[7/x]]]),x]

[Out]

-1/4*Log[Log[2 + E^5 - Log[Log[7/x]]]]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 20, normalized size = 0.91 \begin {gather*} -\frac {1}{4} \log \left (\log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(((-8*x - 4*E^5*x)*Log[7/x] + 4*x*Log[7/x]*Log[Log[7/x]])*Log[2 + E^5 - Log[Log[7/x]]]),x]

[Out]

-1/4*Log[Log[2 + E^5 - Log[Log[7/x]]]]

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fricas [A]  time = 0.56, size = 17, normalized size = 0.77 \begin {gather*} -\frac {1}{4} \, \log \left (\log \left (e^{5} - \log \left (\log \left (\frac {7}{x}\right )\right ) + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-log(log(7/x))+exp(5)+2),x, algorithm=
"fricas")

[Out]

-1/4*log(log(e^5 - log(log(7/x)) + 2))

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giac [A]  time = 0.14, size = 18, normalized size = 0.82 \begin {gather*} -\frac {1}{4} \, \log \left ({\left | \log \left (e^{5} - \log \left (\log \left (\frac {7}{x}\right )\right ) + 2\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-log(log(7/x))+exp(5)+2),x, algorithm=
"giac")

[Out]

-1/4*log(abs(log(e^5 - log(log(7/x)) + 2)))

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (4 x \ln \left (\frac {7}{x}\right ) \ln \left (\ln \left (\frac {7}{x}\right )\right )+\left (-4 x \,{\mathrm e}^{5}-8 x \right ) \ln \left (\frac {7}{x}\right )\right ) \ln \left (-\ln \left (\ln \left (\frac {7}{x}\right )\right )+{\mathrm e}^{5}+2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x*ln(7/x)*ln(ln(7/x))+(-4*x*exp(5)-8*x)*ln(7/x))/ln(-ln(ln(7/x))+exp(5)+2),x)

[Out]

int(1/(4*x*ln(7/x)*ln(ln(7/x))+(-4*x*exp(5)-8*x)*ln(7/x))/ln(-ln(ln(7/x))+exp(5)+2),x)

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maxima [A]  time = 0.48, size = 18, normalized size = 0.82 \begin {gather*} -\frac {1}{4} \, \log \left (\log \left (e^{5} - \log \left (\log \relax (7) - \log \relax (x)\right ) + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-log(log(7/x))+exp(5)+2),x, algorithm=
"maxima")

[Out]

-1/4*log(log(e^5 - log(log(7) - log(x)) + 2))

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mupad [B]  time = 4.20, size = 17, normalized size = 0.77 \begin {gather*} -\frac {\ln \left (\ln \left ({\mathrm {e}}^5-\ln \left (\ln \left (\frac {7}{x}\right )\right )+2\right )\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(log(exp(5) - log(log(7/x)) + 2)*(log(7/x)*(8*x + 4*x*exp(5)) - 4*x*log(log(7/x))*log(7/x))),x)

[Out]

-log(log(exp(5) - log(log(7/x)) + 2))/4

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sympy [A]  time = 0.60, size = 17, normalized size = 0.77 \begin {gather*} - \frac {\log {\left (\log {\left (- \log {\left (\log {\left (\frac {7}{x} \right )} \right )} + 2 + e^{5} \right )} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x*ln(7/x)*ln(ln(7/x))+(-4*x*exp(5)-8*x)*ln(7/x))/ln(-ln(ln(7/x))+exp(5)+2),x)

[Out]

-log(log(-log(log(7/x)) + 2 + exp(5)))/4

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