Integrand size = 60, antiderivative size = 29 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-1-x+\frac {x}{\log (2)}+\frac {-4+x}{5 x \log \left (\frac {8}{x^2}\right )} \] Output:
x/ln(2)+1/5*(-4+x)/x/ln(8/x^2)-x-1
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {-5 x (-1+\log (2))+\frac {(-4+x) \log (2)}{x \log \left (\frac {8}{x^2}\right )}}{5 \log (2)} \] Input:
Integrate[((-8 + 2*x)*Log[2] + 4*Log[2]*Log[8/x^2] + (5*x^2 - 5*x^2*Log[2] )*Log[8/x^2]^2)/(5*x^2*Log[2]*Log[8/x^2]^2),x]
Output:
(-5*x*(-1 + Log[2]) + ((-4 + x)*Log[2])/(x*Log[8/x^2]))/(5*Log[2])
Time = 0.49 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 25, 7293, 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.
\(\displaystyle \int \frac {\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )+4 \log (2) \log \left (\frac {8}{x^2}\right )+(2 x-8) \log (2)}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {-5 \left (x^2-x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )-4 \log (2) \log \left (\frac {8}{x^2}\right )+2 (4-x) \log (2)}{x^2 \log ^2\left (\frac {8}{x^2}\right )}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-5 \left (x^2-x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )-4 \log (2) \log \left (\frac {8}{x^2}\right )+2 (4-x) \log (2)}{x^2 \log ^2\left (\frac {8}{x^2}\right )}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-\frac {2 \log (2) (x-4)}{x^2 \log ^2\left (\frac {8}{x^2}\right )}-\frac {4 \log (2)}{x^2 \log \left (\frac {8}{x^2}\right )}-5 (1-\log (2))\right )dx}{5 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\log (2)}{\log \left (\frac {8}{x^2}\right )}+\frac {4 \log (2)}{x \log \left (\frac {8}{x^2}\right )}-5 x (1-\log (2))}{5 \log (2)}\) |
Input:
Int[((-8 + 2*x)*Log[2] + 4*Log[2]*Log[8/x^2] + (5*x^2 - 5*x^2*Log[2])*Log[ 8/x^2]^2)/(5*x^2*Log[2]*Log[8/x^2]^2),x]
Output:
-1/5*(-5*x*(1 - Log[2]) - Log[2]/Log[8/x^2] + (4*Log[2])/(x*Log[8/x^2]))/L og[2]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-x +\frac {x}{\ln \left (2\right )}+\frac {x -4}{5 x \ln \left (\frac {8}{x^{2}}\right )}\) | \(27\) |
norman | \(\frac {-\frac {4}{5}+\frac {x}{5}-\frac {\left (\ln \left (2\right )-1\right ) x^{2} \ln \left (\frac {8}{x^{2}}\right )}{\ln \left (2\right )}}{\ln \left (\frac {8}{x^{2}}\right ) x}\) | \(37\) |
parallelrisch | \(-\frac {10 \ln \left (2\right ) x^{2} \ln \left (\frac {8}{x^{2}}\right )-10 x^{2} \ln \left (\frac {8}{x^{2}}\right )-2 x \ln \left (2\right )+8 \ln \left (2\right )}{10 \ln \left (2\right ) x \ln \left (\frac {8}{x^{2}}\right )}\) | \(52\) |
Input:
int(1/5*((-5*x^2*ln(2)+5*x^2)*ln(8/x^2)^2+4*ln(2)*ln(8/x^2)+(2*x-8)*ln(2)) /x^2/ln(2)/ln(8/x^2)^2,x,method=_RETURNVERBOSE)
Output:
-x+x/ln(2)+1/5*(x-4)/x/ln(8/x^2)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {{\left (x - 4\right )} \log \left (2\right ) - 5 \, {\left (x^{2} \log \left (2\right ) - x^{2}\right )} \log \left (\frac {8}{x^{2}}\right )}{5 \, x \log \left (2\right ) \log \left (\frac {8}{x^{2}}\right )} \] Input:
integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x -8)*log(2))/x^2/log(2)/log(8/x^2)^2,x, algorithm="fricas")
Output:
1/5*((x - 4)*log(2) - 5*(x^2*log(2) - x^2)*log(8/x^2))/(x*log(2)*log(8/x^2 ))
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {x \left (1 - \log {\left (2 \right )}\right )}{\log {\left (2 \right )}} + \frac {x - 4}{5 x \log {\left (\frac {8}{x^{2}} \right )}} \] Input:
integrate(1/5*((-5*x**2*ln(2)+5*x**2)*ln(8/x**2)**2+4*ln(2)*ln(8/x**2)+(2* x-8)*ln(2))/x**2/ln(2)/ln(8/x**2)**2,x)
Output:
x*(1 - log(2))/log(2) + (x - 4)/(5*x*log(8/x**2))
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {5 \, x \log \left (2\right ) - 5 \, x - \frac {x \log \left (2\right ) - 4 \, \log \left (2\right )}{3 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )}}{5 \, \log \left (2\right )} \] Input:
integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x -8)*log(2))/x^2/log(2)/log(8/x^2)^2,x, algorithm="maxima")
Output:
-1/5*(5*x*log(2) - 5*x - (x*log(2) - 4*log(2))/(3*x*log(2) - 2*x*log(x)))/ log(2)
Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {5 \, x {\left (\log \left (2\right ) - 1\right )} - \frac {x \log \left (2\right ) - 4 \, \log \left (2\right )}{3 \, x \log \left (2\right ) - x \log \left (x^{2}\right )}}{5 \, \log \left (2\right )} \] Input:
integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x -8)*log(2))/x^2/log(2)/log(8/x^2)^2,x, algorithm="giac")
Output:
-1/5*(5*x*(log(2) - 1) - (x*log(2) - 4*log(2))/(3*x*log(2) - x*log(x^2)))/ log(2)
Time = 2.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {x\,\left (\ln \left (32\right )-5\right )}{5\,\ln \left (2\right )}-\frac {\frac {x\,\ln \left (16\right )}{5}-\frac {x^2\,\ln \left (2\right )}{5}}{x^2\,\ln \left (2\right )\,\ln \left (\frac {8}{x^2}\right )} \] Input:
int(((log(2)*(2*x - 8))/5 + (4*log(2)*log(8/x^2))/5 - (log(8/x^2)^2*(5*x^2 *log(2) - 5*x^2))/5)/(x^2*log(2)*log(8/x^2)^2),x)
Output:
- (x*(log(32) - 5))/(5*log(2)) - ((x*log(16))/5 - (x^2*log(2))/5)/(x^2*log (2)*log(8/x^2))
Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {-5 \,\mathrm {log}\left (\frac {8}{x^{2}}\right ) \mathrm {log}\left (2\right ) x^{2}+5 \,\mathrm {log}\left (\frac {8}{x^{2}}\right ) x^{2}+\mathrm {log}\left (2\right ) x -4 \,\mathrm {log}\left (2\right )}{5 \,\mathrm {log}\left (\frac {8}{x^{2}}\right ) \mathrm {log}\left (2\right ) x} \] Input:
int(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x-8)*lo g(2))/x^2/log(2)/log(8/x^2)^2,x)
Output:
( - 5*log(8/x**2)*log(2)*x**2 + 5*log(8/x**2)*x**2 + log(2)*x - 4*log(2))/ (5*log(8/x**2)*log(2)*x)