Integrand size = 105, antiderivative size = 25 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=2+x^2 (2-\log (5))+\log \left (x-\log \left (x^2+\log (\log (2))\right )\right ) \] Output:
ln(-ln(ln(ln(2))+x^2)+x)+(2-ln(5))*x^2+2
Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=x^2 (2-\log (5))+\log \left (x-\log \left (x^2+\log (\log (2))\right )\right ) \] Input:
Integrate[(2*x - x^2 - 4*x^4 + 2*x^4*Log[5] + (-1 - 4*x^2 + 2*x^2*Log[5])* Log[Log[2]] + (4*x^3 - 2*x^3*Log[5] + (4*x - 2*x*Log[5])*Log[Log[2]])*Log[ x^2 + Log[Log[2]]])/(-x^3 - x*Log[Log[2]] + (x^2 + Log[Log[2]])*Log[x^2 + Log[Log[2]]]),x]
Output:
x^2*(2 - Log[5]) + Log[x - Log[x^2 + Log[Log[2]]]]
Time = 0.78 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6, 7292, 7276, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-4 x^4+2 x^4 \log (5)-x^2+\log (\log (2)) \left (-4 x^2+2 x^2 \log (5)-1\right )+\left (4 x^3-2 x^3 \log (5)+\log (\log (2)) (4 x-2 x \log (5))\right ) \log \left (x^2+\log (\log (2))\right )+2 x}{-x^3+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )-x \log (\log (2))} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^4 (2 \log (5)-4)-x^2+\log (\log (2)) \left (-4 x^2+2 x^2 \log (5)-1\right )+\left (4 x^3-2 x^3 \log (5)+\log (\log (2)) (4 x-2 x \log (5))\right ) \log \left (x^2+\log (\log (2))\right )+2 x}{-x^3+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )-x \log (\log (2))}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (x^4 (2 \log (5)-4)\right )+x^2-\log (\log (2)) \left (-4 x^2+2 x^2 \log (5)-1\right )-\left (4 x^3-2 x^3 \log (5)+\log (\log (2)) (4 x-2 x \log (5))\right ) \log \left (x^2+\log (\log (2))\right )-2 x}{\left (x^2+\log (\log (2))\right ) \left (x-\log \left (x^2+\log (\log (2))\right )\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {x^2-2 x+\log (\log (2))}{\left (x^2+\log (\log (2))\right ) \left (x-\log \left (x^2+\log (\log (2))\right )\right )}-2 x (\log (5)-2)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2 (2-\log (5))+\log \left (x-\log \left (x^2+\log (\log (2))\right )\right )\) |
Input:
Int[(2*x - x^2 - 4*x^4 + 2*x^4*Log[5] + (-1 - 4*x^2 + 2*x^2*Log[5])*Log[Lo g[2]] + (4*x^3 - 2*x^3*Log[5] + (4*x - 2*x*Log[5])*Log[Log[2]])*Log[x^2 + Log[Log[2]]])/(-x^3 - x*Log[Log[2]] + (x^2 + Log[Log[2]])*Log[x^2 + Log[Lo g[2]]]),x]
Output:
x^2*(2 - Log[5]) + Log[x - Log[x^2 + Log[Log[2]]]]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.71 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\left (2-\ln \left (5\right )\right ) x^{2}+\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )+x^{2}\right )+x \right )\) | \(25\) |
default | \(2 x^{2}+\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )+x^{2}\right )+x \right )-x^{2} \ln \left (5\right )\) | \(27\) |
risch | \(-x^{2} \ln \left (5\right )+2 x^{2}+\ln \left (-x +\ln \left (\ln \left (\ln \left (2\right )\right )+x^{2}\right )\right )\) | \(27\) |
parallelrisch | \(-x^{2} \ln \left (5\right )+2 \ln \left (5\right ) \ln \left (\ln \left (2\right )\right )+2 x^{2}+\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )+x^{2}\right )+x \right )-4 \ln \left (\ln \left (2\right )\right )\) | \(39\) |
Input:
int((((-2*x*ln(5)+4*x)*ln(ln(2))-2*x^3*ln(5)+4*x^3)*ln(ln(ln(2))+x^2)+(2*x ^2*ln(5)-4*x^2-1)*ln(ln(2))+2*x^4*ln(5)-4*x^4-x^2+2*x)/((ln(ln(2))+x^2)*ln (ln(ln(2))+x^2)-x*ln(ln(2))-x^3),x,method=_RETURNVERBOSE)
Output:
(2-ln(5))*x^2+ln(-ln(ln(ln(2))+x^2)+x)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=-x^{2} \log \left (5\right ) + 2 \, x^{2} + \log \left (-x + \log \left (x^{2} + \log \left (\log \left (2\right )\right )\right )\right ) \] Input:
integrate((((-2*x*log(5)+4*x)*log(log(2))-2*x^3*log(5)+4*x^3)*log(log(log( 2))+x^2)+(2*x^2*log(5)-4*x^2-1)*log(log(2))+2*x^4*log(5)-4*x^4-x^2+2*x)/(( log(log(2))+x^2)*log(log(log(2))+x^2)-x*log(log(2))-x^3),x, algorithm="fri cas")
Output:
-x^2*log(5) + 2*x^2 + log(-x + log(x^2 + log(log(2))))
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=x^{2} \cdot \left (2 - \log {\left (5 \right )}\right ) + \log {\left (- x + \log {\left (x^{2} + \log {\left (\log {\left (2 \right )} \right )} \right )} \right )} \] Input:
integrate((((-2*x*ln(5)+4*x)*ln(ln(2))-2*x**3*ln(5)+4*x**3)*ln(ln(ln(2))+x **2)+(2*x**2*ln(5)-4*x**2-1)*ln(ln(2))+2*x**4*ln(5)-4*x**4-x**2+2*x)/((ln( ln(2))+x**2)*ln(ln(ln(2))+x**2)-x*ln(ln(2))-x**3),x)
Output:
x**2*(2 - log(5)) + log(-x + log(x**2 + log(log(2))))
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=-x^{2} {\left (\log \left (5\right ) - 2\right )} + \log \left (-x + \log \left (x^{2} + \log \left (\log \left (2\right )\right )\right )\right ) \] Input:
integrate((((-2*x*log(5)+4*x)*log(log(2))-2*x^3*log(5)+4*x^3)*log(log(log( 2))+x^2)+(2*x^2*log(5)-4*x^2-1)*log(log(2))+2*x^4*log(5)-4*x^4-x^2+2*x)/(( log(log(2))+x^2)*log(log(log(2))+x^2)-x*log(log(2))-x^3),x, algorithm="max ima")
Output:
-x^2*(log(5) - 2) + log(-x + log(x^2 + log(log(2))))
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=-x^{2} {\left (\log \left (5\right ) - 2\right )} + \log \left (x - \log \left (x^{2} + \log \left (\log \left (2\right )\right )\right )\right ) \] Input:
integrate((((-2*x*log(5)+4*x)*log(log(2))-2*x^3*log(5)+4*x^3)*log(log(log( 2))+x^2)+(2*x^2*log(5)-4*x^2-1)*log(log(2))+2*x^4*log(5)-4*x^4-x^2+2*x)/(( log(log(2))+x^2)*log(log(log(2))+x^2)-x*log(log(2))-x^3),x, algorithm="gia c")
Output:
-x^2*(log(5) - 2) + log(x - log(x^2 + log(log(2))))
Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=\ln \left (\ln \left (x^2+\ln \left (\ln \left (2\right )\right )\right )-x\right )-x^2\,\left (\ln \left (5\right )-2\right ) \] Input:
int(-(2*x - log(log(2))*(4*x^2 - 2*x^2*log(5) + 1) + 2*x^4*log(5) + log(lo g(log(2)) + x^2)*(log(log(2))*(4*x - 2*x*log(5)) - 2*x^3*log(5) + 4*x^3) - x^2 - 4*x^4)/(x*log(log(2)) + x^3 - log(log(log(2)) + x^2)*(log(log(2)) + x^2)),x)
Output:
log(log(log(log(2)) + x^2) - x) - x^2*(log(5) - 2)
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {2 x-x^2-4 x^4+2 x^4 \log (5)+\left (-1-4 x^2+2 x^2 \log (5)\right ) \log (\log (2))+\left (4 x^3-2 x^3 \log (5)+(4 x-2 x \log (5)) \log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )}{-x^3-x \log (\log (2))+\left (x^2+\log (\log (2))\right ) \log \left (x^2+\log (\log (2))\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )\right )+x^{2}\right )-x \right )-\mathrm {log}\left (5\right ) x^{2}+2 x^{2} \] Input:
int((((-2*x*log(5)+4*x)*log(log(2))-2*x^3*log(5)+4*x^3)*log(log(log(2))+x^ 2)+(2*x^2*log(5)-4*x^2-1)*log(log(2))+2*x^4*log(5)-4*x^4-x^2+2*x)/((log(lo g(2))+x^2)*log(log(log(2))+x^2)-x*log(log(2))-x^3),x)
Output:
log(log(log(log(2)) + x**2) - x) - log(5)*x**2 + 2*x**2