\(\int \frac {-20-2 x+(-20 x-2 x^2) \log (x)+((-40 x-2 x^2) \log (\frac {16}{60 x+3 x^2})+(-40 x-2 x^2) \log (x) \log (\frac {16}{60 x+3 x^2})) \log (1+x \log (x))}{(20 x+x^2) \log (\frac {16}{60 x+3 x^2})+(20 x^2+x^3) \log (x) \log (\frac {16}{60 x+3 x^2})} \, dx\) [5542]

Optimal result

Integrand size = 128, antiderivative size = 30 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=\log \left (\log \left (\frac {4}{3 \left (5+\frac {x}{4}\right ) x}\right )\right )-\log ^2(1+x \log (x)) \]

[Out]

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

Rubi [F]

\[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=\int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx \]

[In]

Int[(-20 - 2*x + (-20*x - 2*x^2)*Log[x] + ((-40*x - 2*x^2)*Log[16/(60*x + 3*x^2)] + (-40*x - 2*x^2)*Log[x]*Log
[16/(60*x + 3*x^2)])*Log[1 + x*Log[x]])/((20*x + x^2)*Log[16/(60*x + 3*x^2)] + (20*x^2 + x^3)*Log[x]*Log[16/(6
0*x + 3*x^2)]),x]

[Out]

-Log[1 + x*Log[x]]^2 + 2*Defer[Int][(-10 - x)/(x*(20 + x)*Log[16/(x*(60 + 3*x))]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{x (20+x) (1+x \log (x)) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx \\ & = \int \left (\frac {2 (-10-x)}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )}-\frac {2 (1+\log (x)) \log (1+x \log (x))}{1+x \log (x)}\right ) \, dx \\ & = 2 \int \frac {-10-x}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx-2 \int \frac {(1+\log (x)) \log (1+x \log (x))}{1+x \log (x)} \, dx \\ & = -\log ^2(1+x \log (x))+2 \int \frac {-10-x}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=-2 \left (\frac {1}{2} \log ^2(1+x \log (x))-\frac {1}{2} \log \left (\log \left (\frac {16}{60 x+3 x^2}\right )\right )\right ) \]

[In]

Integrate[(-20 - 2*x + (-20*x - 2*x^2)*Log[x] + ((-40*x - 2*x^2)*Log[16/(60*x + 3*x^2)] + (-40*x - 2*x^2)*Log[
x]*Log[16/(60*x + 3*x^2)])*Log[1 + x*Log[x]])/((20*x + x^2)*Log[16/(60*x + 3*x^2)] + (20*x^2 + x^3)*Log[x]*Log
[16/(60*x + 3*x^2)]),x]

[Out]

-2*(Log[1 + x*Log[x]]^2/2 - Log[Log[16/(60*x + 3*x^2)]]/2)

Maple [A] (verified)

Time = 9.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(-\ln \left (x \ln \left (x \right )+1\right )^{2}+\ln \left (\ln \left (\frac {16}{3 x \left (20+x \right )}\right )\right )\)	\(25\)
default	\(-\ln \left (x \ln \left (x \right )+1\right )^{2}+\ln \left (4 \ln \left (2\right )-\ln \left (3\right )+\ln \left (\frac {1}{x \left (20+x \right )}\right )\right )\)	\(33\)
risch	\(-\ln \left (x \ln \left (x \right )+1\right )^{2}+\ln \left (-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i}{20+x}\right ) \operatorname {csgn}\left (\frac {i}{x \left (20+x \right )}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{20+x}\right ) \operatorname {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{3}+2 i \ln \left (3\right )-8 i \ln \left (2\right )+2 i \ln \left (x \right )+2 i \ln \left (20+x \right )\right )\)	\(133\)

[In]

int((((-2*x^2-40*x)*ln(16/(3*x^2+60*x))*ln(x)+(-2*x^2-40*x)*ln(16/(3*x^2+60*x)))*ln(x*ln(x)+1)+(-2*x^2-20*x)*l
n(x)-2*x-20)/((x^3+20*x^2)*ln(16/(3*x^2+60*x))*ln(x)+(x^2+20*x)*ln(16/(3*x^2+60*x))),x,method=_RETURNVERBOSE)

[Out]

-ln(x*ln(x)+1)^2+ln(ln(16/3/x/(20+x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=-\log \left (x \log \left (x\right ) + 1\right )^{2} + \log \left (\log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )\right ) \]

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="fricas")

[Out]

-log(x*log(x) + 1)^2 + log(log(16/3/(x^2 + 20*x)))

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=- \log {\left (x \log {\left (x \right )} + 1 \right )}^{2} + \log {\left (\log {\left (\frac {16}{3 x^{2} + 60 x} \right )} \right )} \]

[In]

integrate((((-2*x**2-40*x)*ln(16/(3*x**2+60*x))*ln(x)+(-2*x**2-40*x)*ln(16/(3*x**2+60*x)))*ln(x*ln(x)+1)+(-2*x
**2-20*x)*ln(x)-2*x-20)/((x**3+20*x**2)*ln(16/(3*x**2+60*x))*ln(x)+(x**2+20*x)*ln(16/(3*x**2+60*x))),x)

[Out]

-log(x*log(x) + 1)**2 + log(log(16/(3*x**2 + 60*x)))

Maxima [F]

\[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=\int { -\frac {2 \, {\left ({\left ({\left (x^{2} + 20 \, x\right )} \log \left (x\right ) \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right ) + {\left (x^{2} + 20 \, x\right )} \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )\right )} \log \left (x \log \left (x\right ) + 1\right ) + {\left (x^{2} + 10 \, x\right )} \log \left (x\right ) + x + 10\right )}}{{\left (x^{3} + 20 \, x^{2}\right )} \log \left (x\right ) \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right ) + {\left (x^{2} + 20 \, x\right )} \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )} \,d x } \]

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="maxima")

[Out]

-2*integrate((((x^2 + 20*x)*log(x)*log(16/3/(x^2 + 20*x)) + (x^2 + 20*x)*log(16/3/(x^2 + 20*x)))*log(x*log(x)
+ 1) + (x^2 + 10*x)*log(x) + x + 10)/((x^3 + 20*x^2)*log(x)*log(16/3/(x^2 + 20*x)) + (x^2 + 20*x)*log(16/3/(x^
2 + 20*x))), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=-\log \left (x \log \left (x\right ) + 1\right )^{2} + \log \left (-4 \, \log \left (2\right ) + \log \left (3 \, x + 60\right ) + \log \left (x\right )\right ) \]

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="giac")

[Out]

-log(x*log(x) + 1)^2 + log(-4*log(2) + log(3*x + 60) + log(x))

Mupad [B] (verification not implemented)

Time = 12.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx=\ln \left (\ln \left (\frac {16}{3\,x^2+60\,x}\right )\right )-{\ln \left (x\,\ln \left (x\right )+1\right )}^2 \]

[In]

int(-(2*x + log(x)*(20*x + 2*x^2) + log(x*log(x) + 1)*(log(16/(60*x + 3*x^2))*(40*x + 2*x^2) + log(16/(60*x +
3*x^2))*log(x)*(40*x + 2*x^2)) + 20)/(log(16/(60*x + 3*x^2))*(20*x + x^2) + log(16/(60*x + 3*x^2))*log(x)*(20*
x^2 + x^3)),x)

[Out]

log(log(16/(60*x + 3*x^2))) - log(x*log(x) + 1)^2