Integrand size = 61, antiderivative size = 32 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x \left (x-\frac {\left (2+x+\log (x)-\log \left (\frac {x^2}{x-x \log (5)}\right )\right )^2}{x^2}\right ) \] Output:
x*(x-(x-ln(x^2/(-x*ln(5)+x))+2+ln(x))^2/x^2)
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=-\frac {4+x^2-x^3+\log ^2(x)-2 \log (x) \left (-2+\log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x} \] Input:
Integrate[(4 - x^2 + 2*x^3 + Log[x]^2 + Log[x]*(4 - 2*Log[-(x/(-1 + Log[5] ))]) - 4*Log[-(x/(-1 + Log[5]))] + Log[-(x/(-1 + Log[5]))]^2)/x^2,x]
Output:
-((4 + x^2 - x^3 + Log[x]^2 - 2*Log[x]*(-2 + Log[-(x/(-1 + Log[5]))]) - 4* Log[-(x/(-1 + Log[5]))] + Log[-(x/(-1 + Log[5]))]^2)/x)
Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(32)=64\).
Time = 0.58 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2010, 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.
\(\displaystyle \int \frac {2 x^3-x^2+\log ^2(x)+\log ^2\left (-\frac {x}{\log (5)-1}\right )+\log (x) \left (4-2 \log \left (-\frac {x}{\log (5)-1}\right )\right )-4 \log \left (-\frac {x}{\log (5)-1}\right )+4}{x^2} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {\log ^2\left (-\frac {x}{\log (5)-1}\right )}{x^2}-\frac {2 (\log (x)+2) \log \left (-\frac {x}{\log (5)-1}\right )}{x^2}+\frac {2 x^3-x^2+\log ^2(x)+4 \log (x)+4}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2-x-\frac {10}{x}-\frac {\log ^2(x)}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x}-\frac {6 \log (x)}{x}+\frac {2 (\log (x)+3)}{x}+\frac {2 (\log (x)+2) \log \left (\frac {x}{1-\log (5)}\right )}{x}\) |
Input:
Int[(4 - x^2 + 2*x^3 + Log[x]^2 + Log[x]*(4 - 2*Log[-(x/(-1 + Log[5]))]) - 4*Log[-(x/(-1 + Log[5]))] + Log[-(x/(-1 + Log[5]))]^2)/x^2,x]
Output:
-10/x - x + x^2 - (6*Log[x])/x - Log[x]^2/x + (2*(3 + Log[x]))/x + (2*(2 + Log[x])*Log[x/(1 - Log[5])])/x - Log[x/(1 - Log[5])]^2/x
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(32)=64\).
Time = 0.64 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12
method | result | size |
parallelrisch | \(-\frac {8-2 x^{3}+2 x^{2}+2 \ln \left (x \right )^{2}-4 \ln \left (x \right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )+2 \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )^{2}+8 \ln \left (x \right )-8 \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )}{2 x}\) | \(68\) |
default | \(-\frac {\ln \left (-x \right )^{2}}{x}+\frac {2 \ln \left (-x \right )}{x}-\frac {4}{x}-\frac {\ln \left (\ln \left (5\right )-1\right )^{2}}{x}+2 \ln \left (\ln \left (5\right )-1\right ) \left (\frac {\ln \left (-x \right )}{x}+\frac {1}{x}\right )+\frac {\ln \left (x \right )^{2}}{x}-2 \left (\ln \left (-x \right )-\ln \left (x \right )\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+2 \ln \left (\ln \left (5\right )-1\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x^{2}-x -\frac {2 \ln \left (x \right )}{x}-\frac {4 \ln \left (\ln \left (5\right )-1\right )}{x}\) | \(133\) |
risch | \(x^{2}-x +\frac {-4+\pi ^{2}+\pi ^{2} \operatorname {csgn}\left (i x \right )^{6}-2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2}+\pi ^{2} \operatorname {csgn}\left (i x \right )^{4}-2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{5}+2 i \ln \left (\ln \left (5\right )-1\right ) \pi \operatorname {csgn}\left (i x \right )^{3}-2 i \ln \left (\ln \left (5\right )-1\right ) \pi \operatorname {csgn}\left (i x \right )^{2}+2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3}+2 i \pi \ln \left (\ln \left (5\right )-1\right )-4 \ln \left (\ln \left (5\right )-1\right )-\ln \left (\ln \left (5\right )-1\right )^{2}-4 i \pi \operatorname {csgn}\left (i x \right )^{2}+4 i \pi \operatorname {csgn}\left (i x \right )^{3}+4 i \pi }{x}\) | \(158\) |
parts | \(\frac {\left (-\ln \left (5\right )+1\right ) \left (\frac {\left (\ln \left (5\right )-1\right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )^{2}}{x}+\frac {2 \ln \left (-\frac {x}{\ln \left (5\right )-1}\right ) \left (\ln \left (5\right )-1\right )}{x}+\frac {2 \ln \left (5\right )-2}{x}\right )}{\left (\ln \left (5\right )-1\right )^{2}}+\frac {\ln \left (x \right )^{2}}{x}-2 \left (\ln \left (-x \right )-\ln \left (x \right )\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+2 \ln \left (\ln \left (5\right )-1\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x^{2}-x -\frac {6}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {4 \left (-\ln \left (5\right )+1\right ) \left (\frac {\ln \left (-\frac {x}{\ln \left (5\right )-1}\right ) \left (\ln \left (5\right )-1\right )}{x}+\frac {\ln \left (5\right )-1}{x}\right )}{\left (\ln \left (5\right )-1\right )^{2}}\) | \(175\) |
Input:
int((ln(x)^2+(-2*ln(-x/(ln(5)-1))+4)*ln(x)+ln(-x/(ln(5)-1))^2-4*ln(-x/(ln( 5)-1))+2*x^3-x^2+4)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/2/x*(8-2*x^3+2*x^2+2*ln(x)^2-4*ln(x)*ln(-x/(ln(5)-1))+2*ln(-x/(ln(5)-1) )^2+8*ln(x)-8*ln(-x/(ln(5)-1)))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=\frac {x^{3} - x^{2} - \log \left (-\log \left (5\right ) + 1\right )^{2} - 4 \, \log \left (-\log \left (5\right ) + 1\right ) - 4}{x} \] Input:
integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2- 4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^2,x, algorithm="fricas")
Output:
(x^3 - x^2 - log(-log(5) + 1)^2 - 4*log(-log(5) + 1) - 4)/x
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} - x + \frac {-4 - \log {\left (-1 + \log {\left (5 \right )} \right )}^{2} - 4 \log {\left (-1 + \log {\left (5 \right )} \right )} + \pi ^{2} + 2 i \pi \log {\left (-1 + \log {\left (5 \right )} \right )} + 4 i \pi }{x} \] Input:
integrate((ln(x)**2+(-2*ln(-x/(ln(5)-1))+4)*ln(x)+ln(-x/(ln(5)-1))**2-4*ln (-x/(ln(5)-1))+2*x**3-x**2+4)/x**2,x)
Output:
x**2 - x + (-4 - log(-1 + log(5))**2 - 4*log(-1 + log(5)) + pi**2 + 2*I*pi *log(-1 + log(5)) + 4*I*pi)/x
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (34) = 68\).
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.75 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} + 2 \, {\left (\frac {\log \left (-\frac {x}{\log \left (5\right ) - 1}\right )}{x} + \frac {1}{x}\right )} \log \left (x\right ) - x - \frac {\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 2}{x} - \frac {\log \left (-\frac {x}{\log \left (5\right ) - 1}\right )^{2} + 2 \, \log \left (-\frac {x}{\log \left (5\right ) - 1}\right ) + 2}{x} + \frac {2 \, {\left (\log \left (x\right ) - \log \left (-\log \left (5\right ) + 1\right ) + 2\right )}}{x} - \frac {4 \, \log \left (x\right )}{x} + \frac {4 \, \log \left (-\frac {x}{\log \left (5\right ) - 1}\right )}{x} - \frac {4}{x} \] Input:
integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2- 4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^2,x, algorithm="maxima")
Output:
x^2 + 2*(log(-x/(log(5) - 1))/x + 1/x)*log(x) - x - (log(x)^2 + 2*log(x) + 2)/x - (log(-x/(log(5) - 1))^2 + 2*log(-x/(log(5) - 1)) + 2)/x + 2*(log(x ) - log(-log(5) + 1) + 2)/x - 4*log(x)/x + 4*log(-x/(log(5) - 1))/x - 4/x
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} - x - \frac {-4 i \, \pi - \pi ^{2} - 2 i \, \pi \log \left (\log \left (5\right ) - 1\right ) + \log \left (\log \left (5\right ) - 1\right )^{2} + 4 \, \log \left (\log \left (5\right ) - 1\right ) + 4}{x} \] Input:
integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2- 4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^2,x, algorithm="giac")
Output:
x^2 - x - (-4*I*pi - pi^2 - 2*I*pi*log(log(5) - 1) + log(log(5) - 1)^2 + 4 *log(log(5) - 1) + 4)/x
Time = 1.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x\,\left (x-1\right )-\frac {{\left (\ln \left (\ln \left (5\right )-1\right )-\ln \left (-x\right )+\ln \left (x\right )+2\right )}^2}{x} \] Input:
int((log(x)^2 - 4*log(-x/(log(5) - 1)) - log(x)*(2*log(-x/(log(5) - 1)) - 4) + log(-x/(log(5) - 1))^2 - x^2 + 2*x^3 + 4)/x^2,x)
Output:
x*(x - 1) - (log(log(5) - 1) - log(-x) + log(x) + 2)^2/x
Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=\frac {-\mathrm {log}\left (-\frac {x}{\mathrm {log}\left (5\right )-1}\right )^{2}+2 \,\mathrm {log}\left (-\frac {x}{\mathrm {log}\left (5\right )-1}\right ) \mathrm {log}\left (x \right )+4 \,\mathrm {log}\left (-\frac {x}{\mathrm {log}\left (5\right )-1}\right )-\mathrm {log}\left (x \right )^{2}-4 \,\mathrm {log}\left (x \right )+x^{3}-x^{2}-4}{x} \] Input:
int((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2-4*log( -x/(log(5)-1))+2*x^3-x^2+4)/x^2,x)
Output:
( - log(( - x)/(log(5) - 1))**2 + 2*log(( - x)/(log(5) - 1))*log(x) + 4*lo g(( - x)/(log(5) - 1)) - log(x)**2 - 4*log(x) + x**3 - x**2 - 4)/x