Integrand size = 68, antiderivative size = 23 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=-5+x (2+x) \log \left (\frac {e^{25} \left (5-\log ^2(x)\right )}{x}\right ) \]
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=x (2+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right ) \]
Integrate[(10 + 5*x + (4 + 2*x)*Log[x] + (-2 - x)*Log[x]^2 + (-10 - 10*x + (2 + 2*x)*Log[x]^2)*Log[(5*E^25 - E^25*Log[x]^2)/x])/(-5 + Log[x]^2),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.
\(\displaystyle \int \frac {5 x+(-x-2) \log ^2(x)+\left (-10 x+(2 x+2) \log ^2(x)-10\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )+(2 x+4) \log (x)+10}{\log ^2(x)-5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 (x+1) \left (\log \left (-\frac {\log ^2(x)-5}{x}\right )+25\right )-\frac {(x+2) \left (\log ^2(x)-2 \log (x)-5\right )}{\log ^2(x)-5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {x \log (x)}{\log ^2(x)-5}dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )dx+2 e^{\sqrt {5}} \operatorname {ExpIntegralEi}\left (\log (x)-\sqrt {5}\right )+2 e^{-\sqrt {5}} \operatorname {ExpIntegralEi}\left (\log (x)+\sqrt {5}\right )-\frac {x^2}{2}-2 x+25 (x+1)^2\) |
Int[(10 + 5*x + (4 + 2*x)*Log[x] + (-2 - x)*Log[x]^2 + (-10 - 10*x + (2 + 2*x)*Log[x]^2)*Log[(5*E^25 - E^25*Log[x]^2)/x])/(-5 + Log[x]^2),x]
3.7.54.3.1 Defintions of rubi rules used
Time = 3.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(x^{2} \ln \left (-\frac {{\mathrm e}^{25} \left (\ln \left (x \right )^{2}-5\right )}{x}\right )+2 \ln \left (-\frac {{\mathrm e}^{25} \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) x\) | \(37\) |
norman | \(x^{2} \ln \left (\frac {-{\mathrm e}^{25} \ln \left (x \right )^{2}+5 \,{\mathrm e}^{25}}{x}\right )+2 x \ln \left (\frac {-{\mathrm e}^{25} \ln \left (x \right )^{2}+5 \,{\mathrm e}^{25}}{x}\right )\) | \(45\) |
risch | \(\left (x^{2}+2 x \right ) \ln \left (\ln \left (x \right )^{2}-5\right )-x^{2} \ln \left (x \right )-2 x \ln \left (x \right )+i \pi \,x^{2}-2 i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i x \pi +\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}}{2}+i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{3}+i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{3}}{2}+25 x^{2}+50 x\) | \(323\) |
int((((2+2*x)*ln(x)^2-10*x-10)*ln((-exp(25)*ln(x)^2+5*exp(25))/x)+(-2-x)*l n(x)^2+(4+2*x)*ln(x)+5*x+10)/(ln(x)^2-5),x,method=_RETURNVERBOSE)
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx={\left (x^{2} + 2 \, x\right )} \log \left (-\frac {e^{25} \log \left (x\right )^{2} - 5 \, e^{25}}{x}\right ) \]
integrate((((2+2*x)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x) +(-2-x)*log(x)^2+(4+2*x)*log(x)+5*x+10)/(log(x)^2-5),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=\left (x^{2} + 2 x\right ) \log {\left (\frac {- e^{25} \log {\left (x \right )}^{2} + 5 e^{25}}{x} \right )} \]
integrate((((2+2*x)*ln(x)**2-10*x-10)*ln((-exp(25)*ln(x)**2+5*exp(25))/x)+ (-2-x)*ln(x)**2+(4+2*x)*ln(x)+5*x+10)/(ln(x)**2-5),x)
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \left (x\right )^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + 50 \, x \]
integrate((((2+2*x)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x) +(-2-x)*log(x)^2+(4+2*x)*log(x)+5*x+10)/(log(x)^2-5),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \left (x\right )^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + 50 \, x \]
integrate((((2+2*x)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x) +(-2-x)*log(x)^2+(4+2*x)*log(x)+5*x+10)/(log(x)^2-5),x, algorithm=\
Time = 12.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=x\,\ln \left (\frac {5\,{\mathrm {e}}^{25}-{\mathrm {e}}^{25}\,{\ln \left (x\right )}^2}{x}\right )\,\left (x+2\right ) \]