\(\int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{((-6 x+8 x^2-2 x \log (2)) \log (x)+(3 x-6 x^2+x \log (2)) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))) \log ^2(\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))})} \, dx\) [8078]

Optimal result

Integrand size = 106, antiderivative size = 22 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log \left (x+\frac {3-2 x+\log (2)}{-2+\log (\log (x))}\right )} \]

[Out]

5/ln((3+ln(2)-2*x)/(ln(ln(x))-2)+x)

Rubi [F]

\[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx \]

[In]

Int[(15 - 10*x + 5*Log[2] - 40*x*Log[x] + 30*x*Log[x]*Log[Log[x]] - 5*x*Log[x]*Log[Log[x]]^2)/(((-6*x + 8*x^2
- 2*x*Log[2])*Log[x] + (3*x - 6*x^2 + x*Log[2])*Log[x]*Log[Log[x]] + x^2*Log[x]*Log[Log[x]]^2)*Log[(3 - 4*x +
Log[2] + x*Log[Log[x]])/(-2 + Log[Log[x]])]^2),x]

[Out]

5*(3 + Log[2])*Defer[Int][1/(x*Log[x]*(2 - Log[Log[x]])*(4*x - 3*(1 + Log[2]/3) - x*Log[Log[x]])*Log[(-4*x + 3
*(1 + Log[2]/3) + x*Log[Log[x]])/(-2 + Log[Log[x]])]^2), x] + 40*Defer[Int][1/((-2 + Log[Log[x]])*(4*x - 3*(1
+ Log[2]/3) - x*Log[Log[x]])*Log[(-4*x + 3*(1 + Log[2]/3) + x*Log[Log[x]])/(-2 + Log[Log[x]])]^2), x] + 10*Def
er[Int][1/(Log[x]*(-2 + Log[Log[x]])*(4*x - 3*(1 + Log[2]/3) - x*Log[Log[x]])*Log[(-4*x + 3*(1 + Log[2]/3) + x
*Log[Log[x]])/(-2 + Log[Log[x]])]^2), x] + 30*Defer[Int][Log[Log[x]]/((2 - Log[Log[x]])*(4*x - 3*(1 + Log[2]/3
) - x*Log[Log[x]])*Log[(-4*x + 3*(1 + Log[2]/3) + x*Log[Log[x]])/(-2 + Log[Log[x]])]^2), x] + 5*Defer[Int][Log
[Log[x]]^2/((-2 + Log[Log[x]])*(4*x - 3*(1 + Log[2]/3) - x*Log[Log[x]])*Log[(-4*x + 3*(1 + Log[2]/3) + x*Log[L
og[x]])/(-2 + Log[Log[x]])]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (-2 x+3 \left (1+\frac {\log (2)}{3}\right )-8 x \log (x)+6 x \log (x) \log (\log (x))-x \log (x) \log ^2(\log (x))\right )}{x \log (x) (2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx \\ & = 5 \int \frac {-2 x+3 \left (1+\frac {\log (2)}{3}\right )-8 x \log (x)+6 x \log (x) \log (\log (x))-x \log (x) \log ^2(\log (x))}{x \log (x) (2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx \\ & = 5 \int \left (\frac {3+\log (2)}{x \log (x) (2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )}+\frac {8}{(-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )}+\frac {2}{\log (x) (-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )}+\frac {6 \log (\log (x))}{(2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )}+\frac {\log ^2(\log (x))}{(-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )}\right ) \, dx \\ & = 5 \int \frac {\log ^2(\log (x))}{(-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx+10 \int \frac {1}{\log (x) (-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx+30 \int \frac {\log (\log (x))}{(2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx+40 \int \frac {1}{(-2+\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx+(5 (3+\log (2))) \int \frac {1}{x \log (x) (2-\log (\log (x))) \left (4 x-3 \left (1+\frac {\log (2)}{3}\right )-x \log (\log (x))\right ) \log ^2\left (\frac {-4 x+3 \left (1+\frac {\log (2)}{3}\right )+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log \left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \]

[In]

Integrate[(15 - 10*x + 5*Log[2] - 40*x*Log[x] + 30*x*Log[x]*Log[Log[x]] - 5*x*Log[x]*Log[Log[x]]^2)/(((-6*x +
8*x^2 - 2*x*Log[2])*Log[x] + (3*x - 6*x^2 + x*Log[2])*Log[x]*Log[Log[x]] + x^2*Log[x]*Log[Log[x]]^2)*Log[(3 -
4*x + Log[2] + x*Log[Log[x]])/(-2 + Log[Log[x]])]^2),x]

[Out]

5/Log[(3 - 4*x + Log[2] + x*Log[Log[x]])/(-2 + Log[Log[x]])]

Maple [A] (verified)

Time = 231.96 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {5}{\ln \left (\frac {x \ln \left (\ln \left (x \right )\right )+\ln \left (2\right )+3-4 x}{\ln \left (\ln \left (x \right )\right )-2}\right )}\)	\(26\)
risch	\(\frac {10 i}{\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (\ln \left (x \right )\right )-2}\right ) \operatorname {csgn}\left (i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )}{\ln \left (\ln \left (x \right )\right )-2}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (\ln \left (x \right )\right )-2}\right ) \operatorname {csgn}\left (\frac {i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )}{\ln \left (\ln \left (x \right )\right )-2}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )}{\ln \left (\ln \left (x \right )\right )-2}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )}{\ln \left (\ln \left (x \right )\right )-2}\right )^{3}-2 i \ln \left (\ln \left (\ln \left (x \right )\right )-2\right )+2 i \ln \left (\left (\ln \left (\ln \left (x \right )\right )-4\right ) x +3+\ln \left (2\right )\right )}\)	\(187\)
default	\(\text {Expression too large to display}\)	\(459\)
parts	\(\text {Expression too large to display}\)	\(459\)

[In]

int((-5*x*ln(x)*ln(ln(x))^2+30*x*ln(x)*ln(ln(x))-40*x*ln(x)+5*ln(2)+15-10*x)/(x^2*ln(x)*ln(ln(x))^2+(x*ln(2)-6
*x^2+3*x)*ln(x)*ln(ln(x))+(-2*x*ln(2)+8*x^2-6*x)*ln(x))/ln((x*ln(ln(x))+ln(2)+3-4*x)/(ln(ln(x))-2))^2,x,method
=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log \left (\frac {x \log \left (\log \left (x\right )\right ) - 4 \, x + \log \left (2\right ) + 3}{\log \left (\log \left (x\right )\right ) - 2}\right )} \]

[In]

integrate((-5*x*log(x)*log(log(x))^2+30*x*log(x)*log(log(x))-40*x*log(x)+5*log(2)+15-10*x)/(x^2*log(x)*log(log
(x))^2+(x*log(2)-6*x^2+3*x)*log(x)*log(log(x))+(-2*x*log(2)+8*x^2-6*x)*log(x))/log((x*log(log(x))+log(2)+3-4*x
)/(log(log(x))-2))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log {\left (\frac {x \log {\left (\log {\left (x \right )} \right )} - 4 x + \log {\left (2 \right )} + 3}{\log {\left (\log {\left (x \right )} \right )} - 2} \right )}} \]

[In]

integrate((-5*x*ln(x)*ln(ln(x))**2+30*x*ln(x)*ln(ln(x))-40*x*ln(x)+5*ln(2)+15-10*x)/(x**2*ln(x)*ln(ln(x))**2+(
x*ln(2)-6*x**2+3*x)*ln(x)*ln(ln(x))+(-2*x*ln(2)+8*x**2-6*x)*ln(x))/ln((x*ln(ln(x))+ln(2)+3-4*x)/(ln(ln(x))-2))
**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log \left (x \log \left (\log \left (x\right )\right ) - 4 \, x + \log \left (2\right ) + 3\right ) - \log \left (\log \left (\log \left (x\right )\right ) - 2\right )} \]

[In]

integrate((-5*x*log(x)*log(log(x))^2+30*x*log(x)*log(log(x))-40*x*log(x)+5*log(2)+15-10*x)/(x^2*log(x)*log(log
(x))^2+(x*log(2)-6*x^2+3*x)*log(x)*log(log(x))+(-2*x*log(2)+8*x^2-6*x)*log(x))/log((x*log(log(x))+log(2)+3-4*x
)/(log(log(x))-2))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {5}{\log \left (x \log \left (\log \left (x\right )\right ) - 4 \, x + \log \left (2\right ) + 3\right ) - \log \left (\log \left (\log \left (x\right )\right ) - 2\right )} \]

[In]

integrate((-5*x*log(x)*log(log(x))^2+30*x*log(x)*log(log(x))-40*x*log(x)+5*log(2)+15-10*x)/(x^2*log(x)*log(log
(x))^2+(x*log(2)-6*x^2+3*x)*log(x)*log(log(x))+(-2*x*log(2)+8*x^2-6*x)*log(x))/log((x*log(log(x))+log(2)+3-4*x
)/(log(log(x))-2))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.90 (sec) , antiderivative size = 141, normalized size of antiderivative = 6.41 \[ \int \frac {15-10 x+5 \log (2)-40 x \log (x)+30 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))}{\left (\left (-6 x+8 x^2-2 x \log (2)\right ) \log (x)+\left (3 x-6 x^2+x \log (2)\right ) \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right ) \log ^2\left (\frac {3-4 x+\log (2)+x \log (\log (x))}{-2+\log (\log (x))}\right )} \, dx=\frac {\left (\ln \left (\ln \left (x\right )\right )-2\right )\,\left (\ln \left (2\right )-4\,x+x\,\ln \left (\ln \left (x\right )\right )+3\right )\,\left (5\,x\,\ln \left (x\right )\,{\ln \left (\ln \left (x\right )\right )}^2-30\,x\,\ln \left (x\right )\,\ln \left (\ln \left (x\right )\right )+10\,x-\ln \left (32\right )+40\,x\,\ln \left (x\right )-15\right )}{\ln \left (\frac {\ln \left (2\right )-4\,x+x\,\ln \left (\ln \left (x\right )\right )+3}{\ln \left (\ln \left (x\right )\right )-2}\right )\,\left (x\,\ln \left (x\right )\,{\ln \left (\ln \left (x\right )\right )}^2-6\,x\,\ln \left (x\right )\,\ln \left (\ln \left (x\right )\right )+2\,x-\ln \left (2\right )+8\,x\,\ln \left (x\right )-3\right )\,\left (8\,x+3\,\ln \left (\ln \left (x\right )\right )-\ln \left (4\right )+\ln \left (\ln \left (x\right )\right )\,\ln \left (2\right )-6\,x\,\ln \left (\ln \left (x\right )\right )+x\,{\ln \left (\ln \left (x\right )\right )}^2-6\right )} \]

[In]

int(-(10*x - 5*log(2) + 40*x*log(x) - 30*x*log(log(x))*log(x) + 5*x*log(log(x))^2*log(x) - 15)/(log((log(2) -
4*x + x*log(log(x)) + 3)/(log(log(x)) - 2))^2*(log(log(x))*log(x)*(3*x + x*log(2) - 6*x^2) - log(x)*(6*x + 2*x
*log(2) - 8*x^2) + x^2*log(log(x))^2*log(x))),x)

[Out]

((log(log(x)) - 2)*(log(2) - 4*x + x*log(log(x)) + 3)*(10*x - log(32) + 40*x*log(x) - 30*x*log(log(x))*log(x)
+ 5*x*log(log(x))^2*log(x) - 15))/(log((log(2) - 4*x + x*log(log(x)) + 3)/(log(log(x)) - 2))*(2*x - log(2) + 8
*x*log(x) - 6*x*log(log(x))*log(x) + x*log(log(x))^2*log(x) - 3)*(8*x + 3*log(log(x)) - log(4) + log(log(x))*l
og(2) - 6*x*log(log(x)) + x*log(log(x))^2 - 6))