\(\int \frac {-10 x+5 \log (\log (2))+((-20 x-4 x^2) \log (\frac {5+x}{x})+(5+x) \log (\frac {5+x}{x}) \log (\log (2))) \log (\log (\frac {5+x}{x}))+(150+30 x) \log (\frac {5+x}{x}) \log ^2(\log (\frac {5+x}{x}))}{(25+5 x) \log (\frac {5+x}{x}) \log ^2(\log (\frac {5+x}{x}))} \, dx\) [5591]

Optimal result

Integrand size = 107, antiderivative size = 29 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 x-\frac {x (2 x-\log (\log (2)))}{5 \log \left (\log \left (1+\frac {5}{x}\right )\right )} \]

[Out]

6*x-1/5*x/ln(ln(1+5/x))*(2*x-ln(ln(2)))

Rubi [F]

\[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx \]

[In]

Int[(-10*x + 5*Log[Log[2]] + ((-20*x - 4*x^2)*Log[(5 + x)/x] + (5 + x)*Log[(5 + x)/x]*Log[Log[2]])*Log[Log[(5
+ x)/x]] + (150 + 30*x)*Log[(5 + x)/x]*Log[Log[(5 + x)/x]]^2)/((25 + 5*x)*Log[(5 + x)/x]*Log[Log[(5 + x)/x]]^2
),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\frac {1}{5} x \left (30+\frac {-2 x+\log (\log (2))}{\log \left (\log \left (\frac {5+x}{x}\right )\right )}\right ) \]

[In]

Integrate[(-10*x + 5*Log[Log[2]] + ((-20*x - 4*x^2)*Log[(5 + x)/x] + (5 + x)*Log[(5 + x)/x]*Log[Log[2]])*Log[L
og[(5 + x)/x]] + (150 + 30*x)*Log[(5 + x)/x]*Log[Log[(5 + x)/x]]^2)/((25 + 5*x)*Log[(5 + x)/x]*Log[Log[(5 + x)
/x]]^2),x]

[Out]

(x*(30 + (-2*x + Log[Log[2]])/Log[Log[(5 + x)/x]]))/5

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66

method	result	size
parallelrisch	\(\frac {x \ln \left (\ln \left (2\right )\right )-2 x^{2}+30 \ln \left (\ln \left (\frac {5+x}{x}\right )\right ) x -300 \ln \left (\ln \left (\frac {5+x}{x}\right )\right )}{5 \ln \left (\ln \left (\frac {5+x}{x}\right )\right )}\)	\(48\)

[In]

int(((30*x+150)*ln(1/x*(5+x))*ln(ln(1/x*(5+x)))^2+((5+x)*ln(1/x*(5+x))*ln(ln(2))+(-4*x^2-20*x)*ln(1/x*(5+x)))*
ln(ln(1/x*(5+x)))+5*ln(ln(2))-10*x)/(25+5*x)/ln(1/x*(5+x))/ln(ln(1/x*(5+x)))^2,x,method=_RETURNVERBOSE)

[Out]

1/5*(x*ln(ln(2))-2*x^2+30*ln(ln(1/x*(5+x)))*x-300*ln(ln(1/x*(5+x))))/ln(ln(1/x*(5+x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=-\frac {2 \, x^{2} - x \log \left (\log \left (2\right )\right ) - 30 \, x \log \left (\log \left (\frac {x + 5}{x}\right )\right )}{5 \, \log \left (\log \left (\frac {x + 5}{x}\right )\right )} \]

[In]

integrate(((30*x+150)*log(1/x*(5+x))*log(log(1/x*(5+x)))^2+((5+x)*log(1/x*(5+x))*log(log(2))+(-4*x^2-20*x)*log
(1/x*(5+x)))*log(log(1/x*(5+x)))+5*log(log(2))-10*x)/(25+5*x)/log(1/x*(5+x))/log(log(1/x*(5+x)))^2,x, algorith
m="fricas")

[Out]

-1/5*(2*x^2 - x*log(log(2)) - 30*x*log(log((x + 5)/x)))/log(log((x + 5)/x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 x + \frac {- 2 x^{2} + x \log {\left (\log {\left (2 \right )} \right )}}{5 \log {\left (\log {\left (\frac {x + 5}{x} \right )} \right )}} \]

[In]

integrate(((30*x+150)*ln(1/x*(5+x))*ln(ln(1/x*(5+x)))**2+((5+x)*ln(1/x*(5+x))*ln(ln(2))+(-4*x**2-20*x)*ln(1/x*
(5+x)))*ln(ln(1/x*(5+x)))+5*ln(ln(2))-10*x)/(25+5*x)/ln(1/x*(5+x))/ln(ln(1/x*(5+x)))**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=-\frac {2 \, x^{2} - 30 \, x \log \left (\log \left (x + 5\right ) - \log \left (x\right )\right ) - x \log \left (\log \left (2\right )\right )}{5 \, \log \left (\log \left (x + 5\right ) - \log \left (x\right )\right )} \]

[In]

integrate(((30*x+150)*log(1/x*(5+x))*log(log(1/x*(5+x)))^2+((5+x)*log(1/x*(5+x))*log(log(2))+(-4*x^2-20*x)*log
(1/x*(5+x)))*log(log(1/x*(5+x)))+5*log(log(2))-10*x)/(25+5*x)/log(1/x*(5+x))/log(log(1/x*(5+x)))^2,x, algorith
m="maxima")

[Out]

-1/5*(2*x^2 - 30*x*log(log(x + 5) - log(x)) - x*log(log(2)))/log(log(x + 5) - log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (27) = 54\).

Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 \, x - \frac {2 \, x^{2} \log \left (\frac {x + 5}{x}\right ) - x \log \left (\frac {x + 5}{x}\right ) \log \left (\log \left (2\right )\right )}{5 \, {\left (\log \left (x + 5\right ) \log \left (\log \left (\frac {x + 5}{x}\right )\right ) - \log \left (x\right ) \log \left (\log \left (\frac {x + 5}{x}\right )\right )\right )}} \]

[In]

integrate(((30*x+150)*log(1/x*(5+x))*log(log(1/x*(5+x)))^2+((5+x)*log(1/x*(5+x))*log(log(2))+(-4*x^2-20*x)*log
(1/x*(5+x)))*log(log(1/x*(5+x)))+5*log(log(2))-10*x)/(25+5*x)/log(1/x*(5+x))/log(log(1/x*(5+x)))^2,x, algorith
m="giac")

[Out]

6*x - 1/5*(2*x^2*log((x + 5)/x) - x*log((x + 5)/x)*log(log(2)))/(log(x + 5)*log(log((x + 5)/x)) - log(x)*log(l
og((x + 5)/x)))

Mupad [B] (verification not implemented)

Time = 12.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\frac {x\,\left (\ln \left (\ln \left (2\right )\right )-2\,x+30\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )\right )}{5\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )} \]

[In]

int(-(10*x - 5*log(log(2)) + log(log((x + 5)/x))*(log((x + 5)/x)*(20*x + 4*x^2) - log((x + 5)/x)*log(log(2))*(
x + 5)) - log((x + 5)/x)*log(log((x + 5)/x))^2*(30*x + 150))/(log((x + 5)/x)*log(log((x + 5)/x))^2*(5*x + 25))
,x)

[Out]

(x*(log(log(2)) - 2*x + 30*log(log((x + 5)/x))))/(5*log(log((x + 5)/x)))