Integrand size = 68, antiderivative size = 26 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log ^2\left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right ) \]
Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log ^2\left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right ) \]
Integrate[((-25*x + 9*x^2 - 8*x^3 + 16*x^4 - 50*Log[x])*Log[E^(25/(16*x^2) )/(x - x^2 + 2*Log[x])])/(4*x^4 - 4*x^5 + 8*x^3*Log[x]),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.
\(\displaystyle \int \frac {\left (16 x^4-8 x^3+9 x^2-25 x-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{-4 x^5+4 x^4+8 x^3 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (16 x^4-8 x^3+9 x^2-25 x-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{4 x^3 \left (-x^2+x+2 \log (x)\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {\left (-16 x^4+8 x^3-9 x^2+25 x+50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^3 \left (-x^2+x+2 \log (x)\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {\left (-16 x^4+8 x^3-9 x^2+25 x+50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^3 \left (-x^2+x+2 \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-\frac {50 \log (x) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^3 \left (x^2-x-2 \log (x)\right )}+\frac {16 x \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2-x-2 \log (x)}+\frac {9 \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x \left (x^2-x-2 \log (x)\right )}-\frac {25 \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2 \left (x^2-x-2 \log (x)\right )}-\frac {8 \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2-x-2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (8 \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2-x-2 \log (x)}dx+25 \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2 \left (x^2-x-2 \log (x)\right )}dx-9 \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x \left (x^2-x-2 \log (x)\right )}dx-16 \int \frac {x \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^2-x-2 \log (x)}dx+50 \int \frac {\log (x) \log \left (\frac {e^{\frac {25}{16 x^2}}}{-x^2+x+2 \log (x)}\right )}{x^3 \left (x^2-x-2 \log (x)\right )}dx\right )\) |
Int[((-25*x + 9*x^2 - 8*x^3 + 16*x^4 - 50*Log[x])*Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])])/(4*x^4 - 4*x^5 + 8*x^3*Log[x]),x]
3.10.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 10.91 (sec) , antiderivative size = 569, normalized size of antiderivative = 21.88
method | result | size |
risch | \(-\frac {\left (16 \ln \left (-2 \ln \left (x \right )+x^{2}-x \right ) x^{2}-25\right ) \ln \left ({\mathrm e}^{\frac {25}{16 x^{2}}}\right )}{8 x^{2}}+\frac {-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{3}-256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )+800 i \pi \,x^{2}-800 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}+400 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}-512 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{3} x^{4}-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}+512 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right ) x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}-625+256 \ln \left (-2 \ln \left (x \right )+x^{2}-x \right )^{2} x^{4}}{256 x^{4}}\) | \(569\) |
int((-50*ln(x)+16*x^4-8*x^3+9*x^2-25*x)*ln(exp(25/16/x^2)/(2*ln(x)-x^2+x)) /(8*x^3*ln(x)-4*x^5+4*x^4),x,method=_RETURNVERBOSE)
-1/8*(16*ln(-2*ln(x)+x^2-x)*x^2-25)/x^2*ln(exp(25/16/x^2))+1/256*(-400*I*P i*x^2*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^3-256*I*Pi*ln(-1/2*x^2+1/2*x+ ln(x))*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2*x^4 -400*I*Pi*x^2*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2))*csgn(I*exp(25 /16/x^2)/(2*ln(x)-x^2+x))+800*I*Pi*x^2-800*I*Pi*x^2*csgn(I*exp(25/16/x^2)/ (2*ln(x)-x^2+x))^2+400*I*Pi*x^2*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/16/x^ 2)/(2*ln(x)-x^2+x))^2-512*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*x^4+256*I*Pi*ln(-1 /2*x^2+1/2*x+ln(x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^3*x^4-400*I*Pi* x^2*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2+512*I *Pi*ln(-1/2*x^2+1/2*x+ln(x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2*x^4+ 256*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16 /x^2))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))*x^4+256*I*Pi*ln(-1/2*x^2+1/2 *x+ln(x))*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2 *x^4-625+256*ln(-2*ln(x)+x^2-x)^2*x^4)/x^4
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log \left (-\frac {e^{\left (\frac {25}{16 \, x^{2}}\right )}}{x^{2} - x - 2 \, \log \left (x\right )}\right )^{2} \]
integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x )-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log {\left (\frac {e^{\frac {25}{16 x^{2}}}}{- x^{2} + x + 2 \log {\left (x \right )}} \right )}^{2} \]
integrate((-50*ln(x)+16*x**4-8*x**3+9*x**2-25*x)*ln(exp(25/16/x**2)/(2*ln( x)-x**2+x))/(8*x**3*ln(x)-4*x**5+4*x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (24) = 48\).
Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.04 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\frac {1}{8} \, {\left (\frac {25}{x^{2}} - 16 \, \log \left (-\frac {1}{2} \, x^{2} + \frac {1}{2} \, x + \log \left (x\right )\right )\right )} \log \left (-\frac {e^{\left (\frac {25}{16 \, x^{2}}\right )}}{x^{2} - x - 2 \, \log \left (x\right )}\right ) - \frac {256 \, x^{4} \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right )^{2} + 800 \, x^{2} \log \left (2\right ) - 32 \, {\left (16 \, x^{4} \log \left (2\right ) + 25 \, x^{2}\right )} \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right ) + 625}{256 \, x^{4}} \]
integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x )-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4),x, algorithm=\
1/8*(25/x^2 - 16*log(-1/2*x^2 + 1/2*x + log(x)))*log(-e^(25/16/x^2)/(x^2 - x - 2*log(x))) - 1/256*(256*x^4*log(-x^2 + x + 2*log(x))^2 + 800*x^2*log( 2) - 32*(16*x^4*log(2) + 25*x^2)*log(-x^2 + x + 2*log(x)) + 625)/x^4
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log \left (x^{2} - x - 2 \, \log \left (x\right )\right )^{2} - 2 i \, \pi \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right ) - \frac {25 \, \log \left (x^{2} - x - 2 \, \log \left (x\right )\right )}{8 \, x^{2}} + \frac {25 \, {\left (32 i \, \pi x^{2} + 25\right )}}{256 \, x^{4}} \]
integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x )-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4),x, algorithm=\
log(x^2 - x - 2*log(x))^2 - 2*I*pi*log(-x^2 + x + 2*log(x)) - 25/8*log(x^2 - x - 2*log(x))/x^2 + 25/256*(32*I*pi*x^2 + 25)/x^4
Time = 9.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\frac {{\left (16\,x^2\,\ln \left (\frac {1}{x+2\,\ln \left (x\right )-x^2}\right )+25\right )}^2}{256\,x^4} \]