\(\int \frac {5+15 x-10 x^2-30 x^3+(20+5 x^2+5 \log (\frac {\log ^4(4)}{x})) \log (4+x^2+\log (\frac {\log ^4(4)}{x}))+(8+48 x+74 x^2+12 x^3+18 x^4+(2+12 x+18 x^2) \log (\frac {\log ^4(4)}{x})) \log ^2(4+x^2+\log (\frac {\log ^4(4)}{x}))}{(8+48 x+74 x^2+12 x^3+18 x^4+(2+12 x+18 x^2) \log (\frac {\log ^4(4)}{x})) \log ^2(4+x^2+\log (\frac {\log ^4(4)}{x}))} \, dx\) [10160]

Optimal result

Integrand size = 167, antiderivative size = 32 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=x+\frac {x}{\left (x+\frac {2+x}{5}\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \]

[Out]

x+x/(2/5+6/5*x)/ln(ln(16*ln(2)^4/x)+x^2+4)

Rubi [F]

\[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=\int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx \]

[In]

Int[(5 + 15*x - 10*x^2 - 30*x^3 + (20 + 5*x^2 + 5*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]] + (8 + 48*x
+ 74*x^2 + 12*x^3 + 18*x^4 + (2 + 12*x + 18*x^2)*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]]^2)/((8 + 48*x
+ 74*x^2 + 12*x^3 + 18*x^4 + (2 + 12*x + 18*x^2)*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]]^2),x]

[Out]

x - (5*Defer[Int][1/((-x^2 - Log[x^(-1)] - 4*(1 + Log[Log[4]]))*Log[x^2 + Log[x^(-1)] + 4*(1 + Log[Log[4]])]^2
), x])/9 - (35*Defer[Int][1/((-1 - 3*x)*(x^2 + Log[x^(-1)] + 4*(1 + Log[Log[4]]))*Log[x^2 + Log[x^(-1)] + 4*(1
+ Log[Log[4]])]^2), x])/18 - (5*Defer[Int][x/((x^2 + Log[x^(-1)] + 4*(1 + Log[Log[4]]))*Log[x^2 + Log[x^(-1)]
+ 4*(1 + Log[Log[4]])]^2), x])/3 + (5*Defer[Int][1/((1 + 3*x)^2*Log[x^2 + Log[x^(-1)] + 4*(1 + Log[Log[4]])])
, x])/2

Rubi steps \begin{align*} \text {integral}& = \int \left (1-\frac {5 \left (-1+2 x^2\right )}{2 (1+3 x) \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )}+\frac {5}{2 (1+3 x)^2 \log \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )}\right ) \, dx \\ & = x-\frac {5}{2} \int \frac {-1+2 x^2}{(1+3 x) \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx+\frac {5}{2} \int \frac {1}{(1+3 x)^2 \log \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx \\ & = x-\frac {5}{2} \int \left (\frac {2}{9 \left (-x^2-\log \left (\frac {1}{x}\right )-4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )}+\frac {7}{9 (-1-3 x) \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )}+\frac {2 x}{3 \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )}\right ) \, dx+\frac {5}{2} \int \frac {1}{(1+3 x)^2 \log \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx \\ & = x-\frac {5}{9} \int \frac {1}{\left (-x^2-\log \left (\frac {1}{x}\right )-4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx-\frac {5}{3} \int \frac {x}{\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx-\frac {35}{18} \int \frac {1}{(-1-3 x) \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right ) \log ^2\left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx+\frac {5}{2} \int \frac {1}{(1+3 x)^2 \log \left (x^2+\log \left (\frac {1}{x}\right )+4 (1+\log (\log (4)))\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=x+\frac {5 x}{2 (1+3 x) \log \left (4+x^2+\log \left (\frac {1}{x}\right )+4 \log (\log (4))\right )} \]

[In]

Integrate[(5 + 15*x - 10*x^2 - 30*x^3 + (20 + 5*x^2 + 5*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]] + (8 +
48*x + 74*x^2 + 12*x^3 + 18*x^4 + (2 + 12*x + 18*x^2)*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]]^2)/((8
+ 48*x + 74*x^2 + 12*x^3 + 18*x^4 + (2 + 12*x + 18*x^2)*Log[Log[4]^4/x])*Log[4 + x^2 + Log[Log[4]^4/x]]^2),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(29)=58\).

Time = 190.87 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.81

method	result	size
parallelrisch	\(\frac {36 \ln \left (\ln \left (\frac {16 \ln \left (2\right )^{4}}{x}\right )+x^{2}+4\right ) x^{2}-420 \ln \left (\ln \left (\frac {16 \ln \left (2\right )^{4}}{x}\right )+x^{2}+4\right ) x +30 x -144 \ln \left (\ln \left (\frac {16 \ln \left (2\right )^{4}}{x}\right )+x^{2}+4\right )}{12 \left (1+3 x \right ) \ln \left (\ln \left (\frac {16 \ln \left (2\right )^{4}}{x}\right )+x^{2}+4\right )}\)	\(90\)

[In]

int((((18*x^2+12*x+2)*ln(16*ln(2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)*ln(ln(16*ln(2)^4/x)+x^2+4)^2+(5*ln(16*ln(2
)^4/x)+5*x^2+20)*ln(ln(16*ln(2)^4/x)+x^2+4)-30*x^3-10*x^2+15*x+5)/((18*x^2+12*x+2)*ln(16*ln(2)^4/x)+18*x^4+12*
x^3+74*x^2+48*x+8)/ln(ln(16*ln(2)^4/x)+x^2+4)^2,x,method=_RETURNVERBOSE)

[Out]

1/12*(36*ln(ln(16*ln(2)^4/x)+x^2+4)*x^2-420*ln(ln(16*ln(2)^4/x)+x^2+4)*x+30*x-144*ln(ln(16*ln(2)^4/x)+x^2+4))/
(1+3*x)/ln(ln(16*ln(2)^4/x)+x^2+4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=\frac {2 \, {\left (3 \, x^{2} + x\right )} \log \left (x^{2} + \log \left (\frac {16 \, \log \left (2\right )^{4}}{x}\right ) + 4\right ) + 5 \, x}{2 \, {\left (3 \, x + 1\right )} \log \left (x^{2} + \log \left (\frac {16 \, \log \left (2\right )^{4}}{x}\right ) + 4\right )} \]

[In]

integrate((((18*x^2+12*x+2)*log(16*log(2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)*log(log(16*log(2)^4/x)+x^2+4)^2+(5
*log(16*log(2)^4/x)+5*x^2+20)*log(log(16*log(2)^4/x)+x^2+4)-30*x^3-10*x^2+15*x+5)/((18*x^2+12*x+2)*log(16*log(
2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)/log(log(16*log(2)^4/x)+x^2+4)^2,x, algorithm="fricas")

[Out]

1/2*(2*(3*x^2 + x)*log(x^2 + log(16*log(2)^4/x) + 4) + 5*x)/((3*x + 1)*log(x^2 + log(16*log(2)^4/x) + 4))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=x + \frac {5 x}{\left (6 x + 2\right ) \log {\left (x^{2} + \log {\left (\frac {16 \log {\left (2 \right )}^{4}}{x} \right )} + 4 \right )}} \]

[In]

integrate((((18*x**2+12*x+2)*ln(16*ln(2)**4/x)+18*x**4+12*x**3+74*x**2+48*x+8)*ln(ln(16*ln(2)**4/x)+x**2+4)**2
+(5*ln(16*ln(2)**4/x)+5*x**2+20)*ln(ln(16*ln(2)**4/x)+x**2+4)-30*x**3-10*x**2+15*x+5)/((18*x**2+12*x+2)*ln(16*
ln(2)**4/x)+18*x**4+12*x**3+74*x**2+48*x+8)/ln(ln(16*ln(2)**4/x)+x**2+4)**2,x)

[Out]

x + 5*x/((6*x + 2)*log(x**2 + log(16*log(2)**4/x) + 4))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (30) = 60\).

Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.94 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=\frac {2 \, {\left (3 \, x^{2} + x\right )} \log \left (x^{2} + 4 \, \log \left (2\right ) - \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right ) + 4\right ) + 5 \, x}{2 \, {\left (3 \, x + 1\right )} \log \left (x^{2} + 4 \, \log \left (2\right ) - \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right ) + 4\right )} \]

[In]

integrate((((18*x^2+12*x+2)*log(16*log(2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)*log(log(16*log(2)^4/x)+x^2+4)^2+(5
*log(16*log(2)^4/x)+5*x^2+20)*log(log(16*log(2)^4/x)+x^2+4)-30*x^3-10*x^2+15*x+5)/((18*x^2+12*x+2)*log(16*log(
2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)/log(log(16*log(2)^4/x)+x^2+4)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=x + \frac {5 \, x}{2 \, {\left (3 \, x \log \left (x^{2} + 4 \, \log \left (2\right ) - \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right ) + 4\right ) + \log \left (x^{2} + 4 \, \log \left (2\right ) - \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right ) + 4\right )\right )}} \]

[In]

integrate((((18*x^2+12*x+2)*log(16*log(2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)*log(log(16*log(2)^4/x)+x^2+4)^2+(5
*log(16*log(2)^4/x)+5*x^2+20)*log(log(16*log(2)^4/x)+x^2+4)-30*x^3-10*x^2+15*x+5)/((18*x^2+12*x+2)*log(16*log(
2)^4/x)+18*x^4+12*x^3+74*x^2+48*x+8)/log(log(16*log(2)^4/x)+x^2+4)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 4.69 \[ \int \frac {5+15 x-10 x^2-30 x^3+\left (20+5 x^2+5 \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log \left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )+\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )}{\left (8+48 x+74 x^2+12 x^3+18 x^4+\left (2+12 x+18 x^2\right ) \log \left (\frac {\log ^4(4)}{x}\right )\right ) \log ^2\left (4+x^2+\log \left (\frac {\log ^4(4)}{x}\right )\right )} \, dx=x-\frac {\frac {5\,x^3}{36}+\frac {5\,x}{9}}{-x^4-\frac {2\,x^3}{3}+\frac {7\,x^2}{18}+\frac {x}{3}+\frac {1}{18}}+\frac {\frac {5\,x}{2\,\left (3\,x+1\right )}-\frac {5\,x\,\ln \left (\ln \left (\frac {16\,{\ln \left (2\right )}^4}{x}\right )+x^2+4\right )\,\left (\ln \left (\frac {16\,{\ln \left (2\right )}^4}{x}\right )+x^2+4\right )}{2\,{\left (3\,x+1\right )}^2\,\left (2\,x^2-1\right )}}{\ln \left (\ln \left (\frac {16\,{\ln \left (2\right )}^4}{x}\right )+x^2+4\right )}-\frac {5\,x\,\ln \left (\frac {16\,{\ln \left (2\right )}^4}{x}\right )}{36\,\left (-x^4-\frac {2\,x^3}{3}+\frac {7\,x^2}{18}+\frac {x}{3}+\frac {1}{18}\right )} \]

[In]

int((15*x + log(log((16*log(2)^4)/x) + x^2 + 4)*(5*log((16*log(2)^4)/x) + 5*x^2 + 20) - 10*x^2 - 30*x^3 + log(
log((16*log(2)^4)/x) + x^2 + 4)^2*(48*x + log((16*log(2)^4)/x)*(12*x + 18*x^2 + 2) + 74*x^2 + 12*x^3 + 18*x^4
+ 8) + 5)/(log(log((16*log(2)^4)/x) + x^2 + 4)^2*(48*x + log((16*log(2)^4)/x)*(12*x + 18*x^2 + 2) + 74*x^2 + 1
2*x^3 + 18*x^4 + 8)),x)

[Out]

x - ((5*x)/9 + (5*x^3)/36)/(x/3 + (7*x^2)/18 - (2*x^3)/3 - x^4 + 1/18) + ((5*x)/(2*(3*x + 1)) - (5*x*log(log((
16*log(2)^4)/x) + x^2 + 4)*(log((16*log(2)^4)/x) + x^2 + 4))/(2*(3*x + 1)^2*(2*x^2 - 1)))/log(log((16*log(2)^4
)/x) + x^2 + 4) - (5*x*log((16*log(2)^4)/x))/(36*(x/3 + (7*x^2)/18 - (2*x^3)/3 - x^4 + 1/18))