Integrand size = 159, antiderivative size = 34 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=2+\frac {\left (-\frac {2}{5 x}+\log (x)\right ) \left (3-\log ^2\left (5-\frac {x^2}{\log (4)}\right )\right )}{x^2} \]
\[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=\int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx \]
Integrate[(-18*x^2 - 15*x^3 + (90 + 75*x)*Log[4] - 8*x^2*Log[(-x^2 + 5*Log [4])/Log[4]] + (6*x^2 + 5*x^3 + (-30 - 25*x)*Log[4])*Log[(-x^2 + 5*Log[4]) /Log[4]]^2 + Log[x]*(30*x^3 - 150*x*Log[4] + 20*x^3*Log[(-x^2 + 5*Log[4])/ Log[4]] + (-10*x^3 + 50*x*Log[4])*Log[(-x^2 + 5*Log[4])/Log[4]]^2))/(-5*x^ 6 + 25*x^4*Log[4]),x]
Integrate[(-18*x^2 - 15*x^3 + (90 + 75*x)*Log[4] - 8*x^2*Log[(-x^2 + 5*Log [4])/Log[4]] + (6*x^2 + 5*x^3 + (-30 - 25*x)*Log[4])*Log[(-x^2 + 5*Log[4]) /Log[4]]^2 + Log[x]*(30*x^3 - 150*x*Log[4] + 20*x^3*Log[(-x^2 + 5*Log[4])/ Log[4]] + (-10*x^3 + 50*x*Log[4])*Log[(-x^2 + 5*Log[4])/Log[4]]^2))/(-5*x^ 6 + 25*x^4*Log[4]), x]
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.
| \(\displaystyle \int \frac {-15 x^3-18 x^2-8 x^2 \log \left (\frac {5 \log (4)-x^2}{\log (4)}\right )+\left (5 x^3+6 x^2+(-25 x-30) \log (4)\right ) \log ^2\left (\frac {5 \log (4)-x^2}{\log (4)}\right )+\log (x) \left (30 x^3+\left (50 x \log (4)-10 x^3\right ) \log ^2\left (\frac {5 \log (4)-x^2}{\log (4)}\right )+20 x^3 \log \left (\frac {5 \log (4)-x^2}{\log (4)}\right )-150 x \log (4)\right )+(75 x+90) \log (4)}{25 x^4 \log (4)-5 x^6} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-15 x^3-18 x^2-8 x^2 \log \left (\frac {5 \log (4)-x^2}{\log (4)}\right )+\left (5 x^3+6 x^2+(-25 x-30) \log (4)\right ) \log ^2\left (\frac {5 \log (4)-x^2}{\log (4)}\right )+\log (x) \left (30 x^3+\left (50 x \log (4)-10 x^3\right ) \log ^2\left (\frac {5 \log (4)-x^2}{\log (4)}\right )+20 x^3 \log \left (\frac {5 \log (4)-x^2}{\log (4)}\right )-150 x \log (4)\right )+(75 x+90) \log (4)}{x^4 \left (25 \log (4)-5 x^2\right )}dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (-\frac {3 (-5 x+10 x \log (x)-6)}{5 x^4}-\frac {4 (5 x \log (x)-2) \log \left (5-\frac {x^2}{\log (4)}\right )}{5 x^2 \left (x^2-5 \log (4)\right )}+\frac {(-5 x+10 x \log (x)-6) \log ^2\left (5-\frac {x^2}{\log (4)}\right )}{5 x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \int \frac {\log (x) \log \left (5-\frac {x^2}{\log (4)}\right )}{\sqrt {5 \log (4)}-x}dx}{5 \log (4)}-\frac {2 \int \frac {\log (x) \log \left (5-\frac {x^2}{\log (4)}\right )}{x+\sqrt {5 \log (4)}}dx}{5 \log (4)}+2 \int \frac {\log (x) \log ^2\left (5-\frac {x^2}{\log (4)}\right )}{x^3}dx-\frac {2 \log (x) \operatorname {PolyLog}\left (2,\frac {x^2}{5 \log (4)}\right )}{5 \log (4)}-\frac {\operatorname {PolyLog}\left (2,\frac {x^2}{\log (1024)}\right )}{5 \log (4)}+\frac {\operatorname {PolyLog}\left (3,\frac {x^2}{5 \log (4)}\right )}{5 \log (4)}-\frac {6}{5 x^3}+\frac {2 \log \left (5-\frac {x^2}{\log (4)}\right ) \log ^2(x)}{5 \log (4)}-\frac {2 \log \left (1-\frac {x^2}{5 \log (4)}\right ) \log ^2(x)}{5 \log (4)}+\frac {\left (5-\frac {x^2}{\log (4)}\right ) \log ^2\left (5-\frac {x^2}{\log (4)}\right )}{10 x^2}+\frac {3 \log (x)}{x^2}+\frac {2 \log ^2\left (5-\frac {x^2}{\log (4)}\right )}{5 x^3}+\frac {2 \log (5) \log (x)}{5 \log (4)}\) |
Int[(-18*x^2 - 15*x^3 + (90 + 75*x)*Log[4] - 8*x^2*Log[(-x^2 + 5*Log[4])/L og[4]] + (6*x^2 + 5*x^3 + (-30 - 25*x)*Log[4])*Log[(-x^2 + 5*Log[4])/Log[4 ]]^2 + Log[x]*(30*x^3 - 150*x*Log[4] + 20*x^3*Log[(-x^2 + 5*Log[4])/Log[4] ] + (-10*x^3 + 50*x*Log[4])*Log[(-x^2 + 5*Log[4])/Log[4]]^2))/(-5*x^6 + 25 *x^4*Log[4]),x]
3.17.79.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, 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 = 1.71 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32
| method | result | size |
| risch | \(-\frac {\left (5 x \ln \left (x \right )-2\right ) \ln \left (\frac {10 \ln \left (2\right )-x^{2}}{2 \ln \left (2\right )}\right )^{2}}{5 x^{3}}+\frac {3 x \ln \left (x \right )-\frac {6}{5}}{x^{3}}\) | \(45\) |
| parallelrisch | \(\frac {-6-5 \ln \left (\frac {10 \ln \left (2\right )-x^{2}}{2 \ln \left (2\right )}\right )^{2} x \ln \left (x \right )+15 x \ln \left (x \right )+2 \ln \left (\frac {10 \ln \left (2\right )-x^{2}}{2 \ln \left (2\right )}\right )^{2}}{5 x^{3}}\) | \(58\) |
int((((100*x*ln(2)-10*x^3)*ln(1/2*(10*ln(2)-x^2)/ln(2))^2+20*x^3*ln(1/2*(1 0*ln(2)-x^2)/ln(2))-300*x*ln(2)+30*x^3)*ln(x)+(2*(-25*x-30)*ln(2)+5*x^3+6* x^2)*ln(1/2*(10*ln(2)-x^2)/ln(2))^2-8*x^2*ln(1/2*(10*ln(2)-x^2)/ln(2))+2*( 75*x+90)*ln(2)-15*x^3-18*x^2)/(50*x^4*ln(2)-5*x^6),x,method=_RETURNVERBOSE )
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=-\frac {5 \, {\left (x \log \left (-\frac {x^{2} - 10 \, \log \left (2\right )}{2 \, \log \left (2\right )}\right )^{2} - 3 \, x\right )} \log \left (x\right ) - 2 \, \log \left (-\frac {x^{2} - 10 \, \log \left (2\right )}{2 \, \log \left (2\right )}\right )^{2} + 6}{5 \, x^{3}} \]
integrate((((100*x*log(2)-10*x^3)*log(1/2*(10*log(2)-x^2)/log(2))^2+20*x^3 *log(1/2*(10*log(2)-x^2)/log(2))-300*x*log(2)+30*x^3)*log(x)+(2*(-25*x-30) *log(2)+5*x^3+6*x^2)*log(1/2*(10*log(2)-x^2)/log(2))^2-8*x^2*log(1/2*(10*l og(2)-x^2)/log(2))+2*(75*x+90)*log(2)-15*x^3-18*x^2)/(50*x^4*log(2)-5*x^6) ,x, algorithm=\
-1/5*(5*(x*log(-1/2*(x^2 - 10*log(2))/log(2))^2 - 3*x)*log(x) - 2*log(-1/2 *(x^2 - 10*log(2))/log(2))^2 + 6)/x^3
Time = 0.52 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=\frac {3 \log {\left (x \right )}}{x^{2}} + \frac {\left (- 5 x \log {\left (x \right )} + 2\right ) \log {\left (\frac {- \frac {x^{2}}{2} + 5 \log {\left (2 \right )}}{\log {\left (2 \right )}} \right )}^{2}}{5 x^{3}} - \frac {6}{5 x^{3}} \]
integrate((((100*x*ln(2)-10*x**3)*ln(1/2*(10*ln(2)-x**2)/ln(2))**2+20*x**3 *ln(1/2*(10*ln(2)-x**2)/ln(2))-300*x*ln(2)+30*x**3)*ln(x)+(2*(-25*x-30)*ln (2)+5*x**3+6*x**2)*ln(1/2*(10*ln(2)-x**2)/ln(2))**2-8*x**2*ln(1/2*(10*ln(2 )-x**2)/ln(2))+2*(75*x+90)*ln(2)-15*x**3-18*x**2)/(50*x**4*ln(2)-5*x**6),x )
Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (33) = 66\).
Time = 0.30 (sec) , antiderivative size = 273, normalized size of antiderivative = 8.03 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=-\frac {3}{20} \, {\left (\frac {\log \left (x^{2} - 10 \, \log \left (2\right )\right )}{\log \left (2\right )^{2}} - \frac {\log \left (x^{2}\right )}{\log \left (2\right )^{2}} + \frac {10}{x^{2} \log \left (2\right )}\right )} \log \left (2\right ) - \frac {3}{500} \, {\left (\frac {3 \, \sqrt {10} \log \left (\frac {x - \sqrt {10} \sqrt {\log \left (2\right )}}{x + \sqrt {10} \sqrt {\log \left (2\right )}}\right )}{\log \left (2\right )^{\frac {5}{2}}} + \frac {20 \, {\left (3 \, x^{2} + 10 \, \log \left (2\right )\right )}}{x^{3} \log \left (2\right )^{2}}\right )} \log \left (2\right ) + \frac {3 \, \log \left (x^{2} - 10 \, \log \left (2\right )\right )}{20 \, \log \left (2\right )} - \frac {3 \, \log \left (x^{2}\right )}{20 \, \log \left (2\right )} + \frac {9 \, \sqrt {10} \log \left (\frac {x - \sqrt {10} \sqrt {\log \left (2\right )}}{x + \sqrt {10} \sqrt {\log \left (2\right )}}\right )}{500 \, \log \left (2\right )^{\frac {3}{2}}} - \frac {2 \, {\left (5 \, x \log \left (x\right ) - 2\right )} \log \left (-x^{2} + 10 \, \log \left (2\right )\right )^{2} + 10 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2} - 3\right )} x \log \left (x\right ) - 4 \, \log \left (2\right )^{2} - 4 \, {\left (5 \, x {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} \log \left (x\right ) - 2 \, \log \left (2\right ) - 2 \, \log \left (\log \left (2\right )\right )\right )} \log \left (-x^{2} + 10 \, \log \left (2\right )\right ) - 8 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) - 4 \, \log \left (\log \left (2\right )\right )^{2} - 15 \, x}{10 \, x^{3}} + \frac {9}{25 \, x \log \left (2\right )} \]
integrate((((100*x*log(2)-10*x^3)*log(1/2*(10*log(2)-x^2)/log(2))^2+20*x^3 *log(1/2*(10*log(2)-x^2)/log(2))-300*x*log(2)+30*x^3)*log(x)+(2*(-25*x-30) *log(2)+5*x^3+6*x^2)*log(1/2*(10*log(2)-x^2)/log(2))^2-8*x^2*log(1/2*(10*l og(2)-x^2)/log(2))+2*(75*x+90)*log(2)-15*x^3-18*x^2)/(50*x^4*log(2)-5*x^6) ,x, algorithm=\
-3/20*(log(x^2 - 10*log(2))/log(2)^2 - log(x^2)/log(2)^2 + 10/(x^2*log(2)) )*log(2) - 3/500*(3*sqrt(10)*log((x - sqrt(10)*sqrt(log(2)))/(x + sqrt(10) *sqrt(log(2))))/log(2)^(5/2) + 20*(3*x^2 + 10*log(2))/(x^3*log(2)^2))*log( 2) + 3/20*log(x^2 - 10*log(2))/log(2) - 3/20*log(x^2)/log(2) + 9/500*sqrt( 10)*log((x - sqrt(10)*sqrt(log(2)))/(x + sqrt(10)*sqrt(log(2))))/log(2)^(3 /2) - 1/10*(2*(5*x*log(x) - 2)*log(-x^2 + 10*log(2))^2 + 10*(log(2)^2 + 2* log(2)*log(log(2)) + log(log(2))^2 - 3)*x*log(x) - 4*log(2)^2 - 4*(5*x*(lo g(2) + log(log(2)))*log(x) - 2*log(2) - 2*log(log(2)))*log(-x^2 + 10*log(2 )) - 8*log(2)*log(log(2)) - 4*log(log(2))^2 - 15*x)/x^3 + 9/25/(x*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (33) = 66\).
Time = 0.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.38 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=-\frac {1}{5} \, {\left (\frac {5 \, \log \left (x\right )}{x^{2}} - \frac {2}{x^{3}}\right )} \log \left (-x^{2} + 10 \, \log \left (2\right )\right )^{2} + \frac {2}{5} \, {\left (\frac {5 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} \log \left (x\right )}{x^{2}} - \frac {2 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )}}{x^{3}}\right )} \log \left (-x^{2} + 10 \, \log \left (2\right )\right ) - \frac {{\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2} - 3\right )} \log \left (x\right )}{x^{2}} + \frac {2 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2} - 3\right )}}{5 \, x^{3}} \]
integrate((((100*x*log(2)-10*x^3)*log(1/2*(10*log(2)-x^2)/log(2))^2+20*x^3 *log(1/2*(10*log(2)-x^2)/log(2))-300*x*log(2)+30*x^3)*log(x)+(2*(-25*x-30) *log(2)+5*x^3+6*x^2)*log(1/2*(10*log(2)-x^2)/log(2))^2-8*x^2*log(1/2*(10*l og(2)-x^2)/log(2))+2*(75*x+90)*log(2)-15*x^3-18*x^2)/(50*x^4*log(2)-5*x^6) ,x, algorithm=\
-1/5*(5*log(x)/x^2 - 2/x^3)*log(-x^2 + 10*log(2))^2 + 2/5*(5*(log(2) + log (log(2)))*log(x)/x^2 - 2*(log(2) + log(log(2)))/x^3)*log(-x^2 + 10*log(2)) - (log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2 - 3)*log(x)/x^2 + 2/5* (log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2 - 3)/x^3
Time = 10.83 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {-18 x^2-15 x^3+(90+75 x) \log (4)-8 x^2 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (6 x^2+5 x^3+(-30-25 x) \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\log (x) \left (30 x^3-150 x \log (4)+20 x^3 \log \left (\frac {-x^2+5 \log (4)}{\log (4)}\right )+\left (-10 x^3+50 x \log (4)\right ) \log ^2\left (\frac {-x^2+5 \log (4)}{\log (4)}\right )\right )}{-5 x^6+25 x^4 \log (4)} \, dx=\frac {3\,\ln \left (x\right )}{x^2}-{\ln \left (\frac {\ln \left (32\right )-\frac {x^2}{2}}{\ln \left (2\right )}\right )}^2\,\left (\frac {\ln \left (x\right )}{x^2}-\frac {\frac {x}{2}+\frac {2}{5}}{x^3}+\frac {1}{2\,x^2}\right )-\frac {6}{5\,x^3} \]
int((2*log(2)*(75*x + 90) - 8*x^2*log((5*log(2) - x^2/2)/log(2)) + log((5* log(2) - x^2/2)/log(2))^2*(6*x^2 - 2*log(2)*(25*x + 30) + 5*x^3) - 18*x^2 - 15*x^3 + log(x)*(20*x^3*log((5*log(2) - x^2/2)/log(2)) - 300*x*log(2) + log((5*log(2) - x^2/2)/log(2))^2*(100*x*log(2) - 10*x^3) + 30*x^3))/(50*x^ 4*log(2) - 5*x^6),x)