3.82.3 \(\int \frac {8+2 e^2}{(5+6 x+x^2+e^2 (1+x)) \log (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}) \log (\log (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}))} \, dx\)

Optimal. Leaf size=27 \[ 1+\frac {\log (25)}{4}-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 6688, 6684} \begin {gather*} -\log \left (\log \left (\log \left (\frac {\left (x+e^2+5\right )^2}{(x+1)^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + 2*E^2)/((5 + 6*x + x^2 + E^2*(1 + x))*Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]*L
og[Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]]),x]

[Out]

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

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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} &=\left (2 \left (4+e^2\right )\right ) \int \frac {1}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx\\ &=\left (2 \left (4+e^2\right )\right ) \int \frac {1}{(1+x) \left (5+e^2+x\right ) \log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right ) \log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )} \, dx\\ &=-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 19, normalized size = 0.70 \begin {gather*} -\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + 2*E^2)/((5 + 6*x + x^2 + E^2*(1 + x))*Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x
^2)]*Log[Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]]),x]

[Out]

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

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fricas [A]  time = 0.58, size = 33, normalized size = 1.22 \begin {gather*} -\log \left (\log \left (\log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(1)^2+8)/((x+1)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/
log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))),x, algorithm="fricas")

[Out]

-log(log(log((x^2 + 2*(x + 5)*e^2 + 10*x + e^4 + 25)/(x^2 + 2*x + 1))))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (e^{2} + 4\right )}}{{\left (x^{2} + {\left (x + 1\right )} e^{2} + 6 \, x + 5\right )} \log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right ) \log \left (\log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(1)^2+8)/((x+1)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/
log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))),x, algorithm="giac")

[Out]

integrate(2*(e^2 + 4)/((x^2 + (x + 1)*e^2 + 6*x + 5)*log((x^2 + 2*(x + 5)*e^2 + 10*x + e^4 + 25)/(x^2 + 2*x +
1))*log(log((x^2 + 2*(x + 5)*e^2 + 10*x + e^4 + 25)/(x^2 + 2*x + 1)))), x)

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maple [A]  time = 0.63, size = 39, normalized size = 1.44

method	result	size
norman	\(-\ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(1)^2+8)/((x+1)*exp(1)^2+x^2+6*x+5)/ln((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/ln(ln((
exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(ln((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))))

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maxima [A]  time = 0.49, size = 20, normalized size = 0.74 \begin {gather*} -\log \left (\log \relax (2) + \log \left (\log \left (x + e^{2} + 5\right ) - \log \left (x + 1\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(1)^2+8)/((x+1)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/
log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))),x, algorithm="maxima")

[Out]

-log(log(2) + log(log(x + e^2 + 5) - log(x + 1)))

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mupad [B]  time = 30.87, size = 34, normalized size = 1.26 \begin {gather*} -\ln \left (\ln \left (\ln \left (\frac {10\,x+{\mathrm {e}}^4+x^2+{\mathrm {e}}^2\,\left (2\,x+10\right )+25}{x^2+2\,x+1}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2) + 8)/(log(log((10*x + exp(4) + x^2 + exp(2)*(2*x + 10) + 25)/(2*x + x^2 + 1)))*log((10*x + exp(4
) + x^2 + exp(2)*(2*x + 10) + 25)/(2*x + x^2 + 1))*(6*x + exp(2)*(x + 1) + x^2 + 5)),x)

[Out]

-log(log(log((10*x + exp(4) + x^2 + exp(2)*(2*x + 10) + 25)/(2*x + x^2 + 1))))

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sympy [A]  time = 0.73, size = 34, normalized size = 1.26 \begin {gather*} - \log {\left (\log {\left (\log {\left (\frac {x^{2} + 10 x + \left (2 x + 10\right ) e^{2} + 25 + e^{4}}{x^{2} + 2 x + 1} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(1)**2+8)/((x+1)*exp(1)**2+x**2+6*x+5)/ln((exp(1)**4+(2*x+10)*exp(1)**2+x**2+10*x+25)/(x**2+2*
x+1))/ln(ln((exp(1)**4+(2*x+10)*exp(1)**2+x**2+10*x+25)/(x**2+2*x+1))),x)

[Out]

-log(log(log((x**2 + 10*x + (2*x + 10)*exp(2) + 25 + exp(4))/(x**2 + 2*x + 1))))

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