\(\int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+(-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+(12 e^{e^{2 x}} x^2-6 x \log (4)) \log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x})) \log (-5 x+\log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x}))}{(-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+(2 e^{e^{2 x}} x^2-x \log (4)) \log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x})) \log (-5 x+\log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x}))} \, dx\) [398]

Optimal result

Integrand size = 241, antiderivative size = 31 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=x+5 \left (x+\log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right )\right ) \]

[Out]

6*x+5*ln(ln(ln(2-2*ln(2)/x/exp(exp(x)^2))-5*x))

Rubi [F]

\[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=\int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx \]

[In]

Int[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + (-60*E^E^(2*x)*x^3 + 30*x^2*Log[4] + (12*E^
E^(2*x)*x^2 - 6*x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])
/(E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^2 - x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4
])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {50 e^{e^{2 x}} x^2-10 e^{2 x} x \log (4)-5 (1+5 x) \log (4)+6 x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx \\ & = \int \left (-\frac {10 e^{2 x} \log (4)}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}+\frac {50 e^{e^{2 x}} x^2-5 \log (4)-25 x \log (4)+60 e^{e^{2 x}} x^3 \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-30 x^2 \log (4) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-12 e^{e^{2 x}} x^2 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )+6 x \log (4) \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx \\ & = -\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \frac {50 e^{e^{2 x}} x^2-5 \log (4)-25 x \log (4)+60 e^{e^{2 x}} x^3 \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-30 x^2 \log (4) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-12 e^{e^{2 x}} x^2 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )+6 x \log (4) \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx \\ & = -\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \frac {50 e^{e^{2 x}} x^2-5 (1+5 x) \log (4)+6 x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx \\ & = -\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \left (-\frac {5 \log (4)}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}+\frac {25+30 x \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-6 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx \\ & = -\left ((5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx+\int \frac {25+30 x \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-6 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx \\ & = -\left ((5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx+\int \left (6+\frac {25}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx \\ & = 6 x+25 \int \frac {1}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx-(5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 x+5 \log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right ) \]

[In]

Integrate[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + (-60*E^E^(2*x)*x^3 + 30*x^2*Log[4] +
(12*E^E^(2*x)*x^2 - 6*x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - L
og[4])/(E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^2 - x*Log[4])*Log[(2*E^E^(2*x)*x -
Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]

[Out]

6*x + 5*Log[Log[-5*x + Log[2 - Log[4]/(E^E^(2*x)*x)]]]

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 317, normalized size of antiderivative = 10.23

\[6 x +5 \ln \left (\ln \left (\ln \left (2\right )+i \pi -\ln \left (x \right )-\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}}\right )+\ln \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )+\operatorname {csgn}\left (i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )}^{2} \left (\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )-1\right )-5 x \right )\right )\]

[In]

int((((12*x^2*exp(exp(x)^2)-12*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-60*x^3*exp(exp(x)^2)+6
0*x^2*ln(2))*ln(ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*ln(2)*exp(x)^2+
2*(25*x+5)*ln(2))/((2*x^2*exp(exp(x)^2)-2*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-10*x^3*exp(
exp(x)^2)+10*x^2*ln(2))/ln(ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-5*x),x)

[Out]

6*x+5*ln(ln(ln(2)+I*Pi-ln(x)-ln(exp(exp(2*x)))+ln(-x*exp(exp(2*x))+ln(2))-1/2*I*Pi*csgn(I*exp(-exp(2*x))*(-x*e
xp(exp(2*x))+ln(2)))*(-csgn(I*exp(-exp(2*x))*(-x*exp(exp(2*x))+ln(2)))+csgn(I*exp(-exp(2*x))))*(-csgn(I*exp(-e
xp(2*x))*(-x*exp(exp(2*x))+ln(2)))+csgn(I*(-x*exp(exp(2*x))+ln(2))))-1/2*I*Pi*csgn(I/x*(-x*exp(exp(2*x))+ln(2)
)*exp(-exp(2*x)))*(-csgn(I/x*(-x*exp(exp(2*x))+ln(2))*exp(-exp(2*x)))+csgn(I/x))*(-csgn(I/x*(-x*exp(exp(2*x))+
ln(2))*exp(-exp(2*x)))+csgn(I*exp(-exp(2*x))*(-x*exp(exp(2*x))+ln(2))))+I*Pi*csgn(I/x*(-x*exp(exp(2*x))+ln(2))
*exp(-exp(2*x)))^2*(csgn(I/x*(-x*exp(exp(2*x))+ln(2))*exp(-exp(2*x)))-1)-5*x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \left (\frac {2 \, {\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}}{x}\right )\right )\right ) \]

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate((((12*x**2*exp(exp(x)**2)-12*x*ln(2))*ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-60*x**3*exp(
exp(x)**2)+60*x**2*ln(2))*ln(ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x)-50*x**2*exp(exp(x)**2)+20*
x*ln(2)*exp(x)**2+2*(25*x+5)*ln(2))/((2*x**2*exp(exp(x)**2)-2*x*ln(2))*ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(e
xp(x)**2))-10*x**3*exp(exp(x)**2)+10*x**2*ln(2))/ln(ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x - e^{\left (2 \, x\right )} + \log \left (2\right ) + \log \left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right ) - \log \left (x\right )\right )\right ) \]

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \left (2\right ) + \log \left ({\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}\right ) - \log \left (x\right )\right )\right ) \]

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.95 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6\,x+5\,\ln \left (\ln \left (\ln \left (\frac {2\,x-2\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,\ln \left (2\right )}{x}\right )-5\,x\right )\right ) \]

[In]

int(-(2*log(2)*(25*x + 5) - log(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x) - 5*x)*(log(-(exp(-exp
(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x)*(12*x*log(2) - 12*x^2*exp(exp(2*x))) - 60*x^2*log(2) + 60*x^3*exp(ex
p(2*x))) - 50*x^2*exp(exp(2*x)) + 20*x*exp(2*x)*log(2))/(log(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x)
)))/x) - 5*x)*(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x)*(2*x*log(2) - 2*x^2*exp(exp(2*x))) - 10
*x^2*log(2) + 10*x^3*exp(exp(2*x)))),x)

[Out]

6*x + 5*log(log(log((2*x - 2*exp(-exp(2*x))*log(2))/x) - 5*x))