3.12.16 \(\int \frac {4+4 \log (\frac {5 \log (3)}{e^2 x})+(-4-8 x) \log ^2(\frac {5 \log (3)}{e^2 x})+(-2-2 \log (\frac {5 \log (3)}{e^2 x})+(2+4 x) \log ^2(\frac {5 \log (3)}{e^2 x})) \log (\frac {x+(-x-x^2) \log (\frac {5 \log (3)}{e^2 x})}{\log (\frac {5 \log (3)}{e^2 x})})}{-x \log (\frac {5 \log (3)}{e^2 x})+(x+x^2) \log ^2(\frac {5 \log (3)}{e^2 x})} \, dx\)

Optimal. Leaf size=31 \[ \left (2-\log \left (-x-x^2+\frac {x}{\log \left (\frac {5 \log (3)}{e^2 x}\right )}\right )\right )^2 \]

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Rubi [A]  time = 0.52, antiderivative size = 26, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 3, integrand size = 144, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 12, 6686} \begin {gather*} \left (2-\log \left (-x \left (x+\frac {1}{2-\log \left (\frac {\log (243)}{x}\right )}+1\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 4*Log[(5*Log[3])/(E^2*x)] + (-4 - 8*x)*Log[(5*Log[3])/(E^2*x)]^2 + (-2 - 2*Log[(5*Log[3])/(E^2*x)] +
(2 + 4*x)*Log[(5*Log[3])/(E^2*x)]^2)*Log[(x + (-x - x^2)*Log[(5*Log[3])/(E^2*x)])/Log[(5*Log[3])/(E^2*x)]])/(-
(x*Log[(5*Log[3])/(E^2*x)]) + (x + x^2)*Log[(5*Log[3])/(E^2*x)]^2),x]

[Out]

(2 - Log[-(x*(1 + x + (2 - Log[Log[243]/x])^(-1)))])^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 {2 \left (1+\left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )^2+2 x \left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )^2-\log \left (\frac {\log (243)}{x}\right )\right ) \left (-2+\log \left (x \left (-1-x+\frac {1}{-2+\log \left (\frac {\log (243)}{x}\right )}\right )\right )\right )}{x \left (1-(1+x) \left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )\right ) \left (2-\log \left (\frac {\log (243)}{x}\right )\right )} \, dx\\ &=2 \int \frac {\left (1+\left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )^2+2 x \left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )^2-\log \left (\frac {\log (243)}{x}\right )\right ) \left (-2+\log \left (x \left (-1-x+\frac {1}{-2+\log \left (\frac {\log (243)}{x}\right )}\right )\right )\right )}{x \left (1-(1+x) \left (-2+\log \left (\frac {1}{x}\right )+\log (\log (243))\right )\right ) \left (2-\log \left (\frac {\log (243)}{x}\right )\right )} \, dx\\ &=\left (2-\log \left (-x \left (1+x+\frac {1}{2-\log \left (\frac {\log (243)}{x}\right )}\right )\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.49, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4+4 \log \left (\frac {5 \log (3)}{e^2 x}\right )+(-4-8 x) \log ^2\left (\frac {5 \log (3)}{e^2 x}\right )+\left (-2-2 \log \left (\frac {5 \log (3)}{e^2 x}\right )+(2+4 x) \log ^2\left (\frac {5 \log (3)}{e^2 x}\right )\right ) \log \left (\frac {x+\left (-x-x^2\right ) \log \left (\frac {5 \log (3)}{e^2 x}\right )}{\log \left (\frac {5 \log (3)}{e^2 x}\right )}\right )}{-x \log \left (\frac {5 \log (3)}{e^2 x}\right )+\left (x+x^2\right ) \log ^2\left (\frac {5 \log (3)}{e^2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(4 + 4*Log[(5*Log[3])/(E^2*x)] + (-4 - 8*x)*Log[(5*Log[3])/(E^2*x)]^2 + (-2 - 2*Log[(5*Log[3])/(E^2*
x)] + (2 + 4*x)*Log[(5*Log[3])/(E^2*x)]^2)*Log[(x + (-x - x^2)*Log[(5*Log[3])/(E^2*x)])/Log[(5*Log[3])/(E^2*x)
]])/(-(x*Log[(5*Log[3])/(E^2*x)]) + (x + x^2)*Log[(5*Log[3])/(E^2*x)]^2),x]

[Out]

Integrate[(4 + 4*Log[(5*Log[3])/(E^2*x)] + (-4 - 8*x)*Log[(5*Log[3])/(E^2*x)]^2 + (-2 - 2*Log[(5*Log[3])/(E^2*
x)] + (2 + 4*x)*Log[(5*Log[3])/(E^2*x)]^2)*Log[(x + (-x - x^2)*Log[(5*Log[3])/(E^2*x)])/Log[(5*Log[3])/(E^2*x)
]])/(-(x*Log[(5*Log[3])/(E^2*x)]) + (x + x^2)*Log[(5*Log[3])/(E^2*x)]^2), x]

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fricas [B]  time = 0.88, size = 75, normalized size = 2.42 \begin {gather*} \log \left (-\frac {{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - x}{\log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\right )^{2} - 4 \, \log \left (-\frac {{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - x}{\log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+2)*log(5*log(3)/exp(2)/x)^2-2*log(5*log(3)/exp(2)/x)-2)*log(((-x^2-x)*log(5*log(3)/exp(2)/x)+
x)/log(5*log(3)/exp(2)/x))+(-8*x-4)*log(5*log(3)/exp(2)/x)^2+4*log(5*log(3)/exp(2)/x)+4)/((x^2+x)*log(5*log(3)
/exp(2)/x)^2-x*log(5*log(3)/exp(2)/x)),x, algorithm="fricas")

[Out]

log(-((x^2 + x)*log(5*e^(-2)*log(3)/x) - x)/log(5*e^(-2)*log(3)/x))^2 - 4*log(-((x^2 + x)*log(5*e^(-2)*log(3)/
x) - x)/log(5*e^(-2)*log(3)/x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (2 \, {\left (2 \, x + 1\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - {\left ({\left (2 \, x + 1\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - 1\right )} \log \left (-\frac {{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - x}{\log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\right ) - 2 \, \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - 2\right )}}{{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - x \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+2)*log(5*log(3)/exp(2)/x)^2-2*log(5*log(3)/exp(2)/x)-2)*log(((-x^2-x)*log(5*log(3)/exp(2)/x)+
x)/log(5*log(3)/exp(2)/x))+(-8*x-4)*log(5*log(3)/exp(2)/x)^2+4*log(5*log(3)/exp(2)/x)+4)/((x^2+x)*log(5*log(3)
/exp(2)/x)^2-x*log(5*log(3)/exp(2)/x)),x, algorithm="giac")

[Out]

integrate(-2*(2*(2*x + 1)*log(5*e^(-2)*log(3)/x)^2 - ((2*x + 1)*log(5*e^(-2)*log(3)/x)^2 - log(5*e^(-2)*log(3)
/x) - 1)*log(-((x^2 + x)*log(5*e^(-2)*log(3)/x) - x)/log(5*e^(-2)*log(3)/x)) - 2*log(5*e^(-2)*log(3)/x) - 2)/(
(x^2 + x)*log(5*e^(-2)*log(3)/x)^2 - x*log(5*e^(-2)*log(3)/x)), x)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x +2\right ) \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )^{2}-2 \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )-2\right ) \ln \left (\frac {\left (-x^{2}-x \right ) \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )+x}{\ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )}\right )+\left (-8 x -4\right ) \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )^{2}+4 \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )+4}{\left (x^{2}+x \right ) \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )^{2}-x \ln \left (\frac {5 \ln \relax (3) {\mathrm e}^{-2}}{x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x+2)*ln(5*ln(3)/exp(2)/x)^2-2*ln(5*ln(3)/exp(2)/x)-2)*ln(((-x^2-x)*ln(5*ln(3)/exp(2)/x)+x)/ln(5*ln(3)
/exp(2)/x))+(-8*x-4)*ln(5*ln(3)/exp(2)/x)^2+4*ln(5*ln(3)/exp(2)/x)+4)/((x^2+x)*ln(5*ln(3)/exp(2)/x)^2-x*ln(5*l
n(3)/exp(2)/x)),x)

[Out]

int((((4*x+2)*ln(5*ln(3)/exp(2)/x)^2-2*ln(5*ln(3)/exp(2)/x)-2)*ln(((-x^2-x)*ln(5*ln(3)/exp(2)/x)+x)/ln(5*ln(3)
/exp(2)/x))+(-8*x-4)*ln(5*ln(3)/exp(2)/x)^2+4*ln(5*ln(3)/exp(2)/x)+4)/((x^2+x)*ln(5*ln(3)/exp(2)/x)^2-x*ln(5*l
n(3)/exp(2)/x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {2 \, {\left (2 \, x + 1\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - {\left ({\left (2 \, x + 1\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - 1\right )} \log \left (-\frac {{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - x}{\log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\right ) - 2 \, \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right ) - 2}{{\left (x^{2} + x\right )} \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )^{2} - x \log \left (\frac {5 \, e^{\left (-2\right )} \log \relax (3)}{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+2)*log(5*log(3)/exp(2)/x)^2-2*log(5*log(3)/exp(2)/x)-2)*log(((-x^2-x)*log(5*log(3)/exp(2)/x)+
x)/log(5*log(3)/exp(2)/x))+(-8*x-4)*log(5*log(3)/exp(2)/x)^2+4*log(5*log(3)/exp(2)/x)+4)/((x^2+x)*log(5*log(3)
/exp(2)/x)^2-x*log(5*log(3)/exp(2)/x)),x, algorithm="maxima")

[Out]

-2*integrate((2*(2*x + 1)*log(5*e^(-2)*log(3)/x)^2 - ((2*x + 1)*log(5*e^(-2)*log(3)/x)^2 - log(5*e^(-2)*log(3)
/x) - 1)*log(-((x^2 + x)*log(5*e^(-2)*log(3)/x) - x)/log(5*e^(-2)*log(3)/x)) - 2*log(5*e^(-2)*log(3)/x) - 2)/(
(x^2 + x)*log(5*e^(-2)*log(3)/x)^2 - x*log(5*e^(-2)*log(3)/x)), x)

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mupad [B]  time = 1.48, size = 223, normalized size = 7.19 \begin {gather*} {\ln \left (\frac {x-\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )\,\left (x^2+x\right )}{\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )}\right )}^2+4\,\ln \left (x+1\right )-4\,\ln \left (\frac {4\,\left (x-x\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )+1\right )}{x\,{\left (x+1\right )}^2}-\frac {4\,\left (2\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )-x+3\,x\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )+2\,x^2\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )-1\right )}{x\,{\left (x+1\right )}^2}\right )+4\,\ln \left (\frac {4\,\left (2\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )-x+3\,x\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )+2\,x^2\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )-1\right )}{x\,{\left (x+1\right )}^2}+\frac {4\,\left (x-x\,\ln \left (\frac {5\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x}\right )+1\right )}{x\,{\left (x+1\right )}^2}\right )-4\,\ln \left (x\,\left (x^2+x+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log((5*exp(-2)*log(3))/x) - log((5*exp(-2)*log(3))/x)^2*(8*x + 4) - log((x - log((5*exp(-2)*log(3))/x)
*(x + x^2))/log((5*exp(-2)*log(3))/x))*(2*log((5*exp(-2)*log(3))/x) - log((5*exp(-2)*log(3))/x)^2*(4*x + 2) +
2) + 4)/(x*log((5*exp(-2)*log(3))/x) - log((5*exp(-2)*log(3))/x)^2*(x + x^2)),x)

[Out]

4*log(x + 1) - 4*log((4*(x - x*log((5*exp(-2)*log(3))/x) + 1))/(x*(x + 1)^2) - (4*(2*log((5*exp(-2)*log(3))/x)
- x + 3*x*log((5*exp(-2)*log(3))/x) + 2*x^2*log((5*exp(-2)*log(3))/x) - 1))/(x*(x + 1)^2)) + 4*log((4*(2*log(
(5*exp(-2)*log(3))/x) - x + 3*x*log((5*exp(-2)*log(3))/x) + 2*x^2*log((5*exp(-2)*log(3))/x) - 1))/(x*(x + 1)^2
) + (4*(x - x*log((5*exp(-2)*log(3))/x) + 1))/(x*(x + 1)^2)) - 4*log(x*(x + x^2 + 1)) + log((x - log((5*exp(-2
)*log(3))/x)*(x + x^2))/log((5*exp(-2)*log(3))/x))^2

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+2)*ln(5*ln(3)/exp(2)/x)**2-2*ln(5*ln(3)/exp(2)/x)-2)*ln(((-x**2-x)*ln(5*ln(3)/exp(2)/x)+x)/ln
(5*ln(3)/exp(2)/x))+(-8*x-4)*ln(5*ln(3)/exp(2)/x)**2+4*ln(5*ln(3)/exp(2)/x)+4)/((x**2+x)*ln(5*ln(3)/exp(2)/x)*
*2-x*ln(5*ln(3)/exp(2)/x)),x)

[Out]

Exception raised: PolynomialError

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