\(\int \frac {-4 x \log (x)+(e^2+x) \log (\frac {3}{e^4+2 e^2 x+x^2})}{(e^2 x+x^2) \log (x) \log (\frac {3}{e^4+2 e^2 x+x^2}) \log (\frac {1}{5} \log (x) \log ^2(\frac {3}{e^4+2 e^2 x+x^2}))} \, dx\) [9980]

Optimal result

Integrand size = 95, antiderivative size = 24 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=3+\log \left (2 \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )\right ) \]

[Out]

ln(2*ln(1/5*ln(3/(x+exp(2))^2)^2*ln(x)))+3

Rubi [F]

\[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx \]

[In]

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

[Out]

Defer[Int][1/(x*Log[x]*Log[(Log[x]*Log[3/(E^2 + x)^2]^2)/5]), x] - 4*Defer[Int][1/((E^2 + x)*Log[3/(E^2 + x)^2
]*Log[(Log[x]*Log[3/(E^2 + x)^2]^2)/5]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{x \left (e^2+x\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx \\ & = \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{x \left (e^2+x\right ) \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx \\ & = \int \left (\frac {4 x \log (x)-e^2 \log \left (\frac {3}{\left (e^2+x\right )^2}\right )-x \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{e^2 \left (e^2+x\right ) \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}+\frac {-4 x \log (x)+e^2 \log \left (\frac {3}{\left (e^2+x\right )^2}\right )+x \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{e^2 x \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}\right ) \, dx \\ & = \frac {\int \frac {4 x \log (x)-e^2 \log \left (\frac {3}{\left (e^2+x\right )^2}\right )-x \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{\left (e^2+x\right ) \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\frac {\int \frac {-4 x \log (x)+e^2 \log \left (\frac {3}{\left (e^2+x\right )^2}\right )+x \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{x \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2} \\ & = \frac {\int \frac {4 x \log (x)-\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{\left (e^2+x\right ) \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\frac {\int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right )}{x \log (x) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2} \\ & = \frac {\int \left (\frac {1}{\log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}+\frac {e^2}{x \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}-\frac {4}{\log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}\right ) \, dx}{e^2}+\frac {\int \left (-\frac {e^2}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}-\frac {x}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}+\frac {4 x}{\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}\right ) \, dx}{e^2} \\ & = \frac {\int \frac {1}{\log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}-\frac {\int \frac {x}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}-\frac {4 \int \frac {1}{\log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\frac {4 \int \frac {x}{\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\int \frac {1}{x \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx-\int \frac {1}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx \\ & = -\frac {\int \left (\frac {1}{\log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}-\frac {e^2}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}\right ) \, dx}{e^2}+\frac {\int \frac {1}{\log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\frac {4 \int \left (\frac {1}{\log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}-\frac {e^2}{\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )}\right ) \, dx}{e^2}-\frac {4 \int \frac {1}{\log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx}{e^2}+\int \frac {1}{x \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx-\int \frac {1}{\left (e^2+x\right ) \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx \\ & = -\left (4 \int \frac {1}{\left (e^2+x\right ) \log \left (\frac {3}{\left (e^2+x\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx\right )+\int \frac {1}{x \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (\log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )\right ) \]

[In]

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

[Out]

Log[Log[(Log[x]*Log[3/(E^2 + x)^2]^2)/5]]

Maple [A] (verified)

Time = 94.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(\ln \left (\ln \left (\frac {\ln \left (\frac {3}{2 \,{\mathrm e}^{2} x +x^{2}+{\mathrm e}^{4}}\right )^{2} \ln \left (x \right )}{5}\right )\right )\)	\(27\)
risch	\(\text {Expression too large to display}\)	\(1123\)

[In]

int((-4*x*ln(x)+(x+exp(2))*ln(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2)*x+x^2)/ln(3/(exp(2)^2+2*exp(2)*x+x^2))/ln(
x)/ln(1/5*ln(3/(exp(2)^2+2*exp(2)*x+x^2))^2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/5*ln(3/(exp(2)^2+2*exp(2)*x+x^2))^2*ln(x)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (\log \left (\frac {1}{5} \, \log \left (x\right ) \log \left (\frac {3}{x^{2} + 2 \, x e^{2} + e^{4}}\right )^{2}\right )\right ) \]

[In]

integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2)*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+
x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2*exp(2)*x+x^2))^2*log(x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log {\left (\log {\left (\frac {\log {\left (x \right )} \log {\left (\frac {3}{x^{2} + 2 x e^{2} + e^{4}} \right )}^{2}}{5} \right )} \right )} \]

[In]

integrate((-4*x*ln(x)+(x+exp(2))*ln(3/(exp(2)**2+2*exp(2)*x+x**2)))/(exp(2)*x+x**2)/ln(3/(exp(2)**2+2*exp(2)*x
+x**2))/ln(x)/ln(1/5*ln(3/(exp(2)**2+2*exp(2)*x+x**2))**2*ln(x)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (-\frac {1}{2} \, \log \left (5\right ) + \log \left (-\log \left (3\right ) + 2 \, \log \left (x + e^{2}\right )\right ) + \frac {1}{2} \, \log \left (\log \left (x\right )\right )\right ) \]

[In]

integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2)*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+
x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2*exp(2)*x+x^2))^2*log(x)),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (21) = 42\).

Time = 1.89 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (-\log \left (5\right ) + \log \left (\log \left (3\right )^{2} \log \left (x\right ) - 2 \, \log \left (3\right ) \log \left (x^{2} + 2 \, x e^{2} + e^{4}\right ) \log \left (x\right ) + \log \left (x^{2} + 2 \, x e^{2} + e^{4}\right )^{2} \log \left (x\right )\right )\right ) \]

[In]

integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2)*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+
x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2*exp(2)*x+x^2))^2*log(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\ln \left (\ln \left (\frac {{\ln \left (\frac {3}{x^2+2\,{\mathrm {e}}^2\,x+{\mathrm {e}}^4}\right )}^2\,\ln \left (x\right )}{5}\right )\right ) \]

[In]

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

[Out]

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