Integrand size = 118, antiderivative size = 29 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\frac {25 \left (-2+4 x^4\right )}{8 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \]
Time = 6.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\frac {25 \left (-1+2 x^4\right )}{4 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \]
Integrate[(100*x + 800*x^5 + (-125 - 750*x^4)*Log[2] + (-25*x - 150*x^5 + (25 + 150*x^4)*Log[2])*Log[(-x + Log[2])/5])/(-100*x^3 + 100*x^2*Log[2] + (40*x^3 - 40*x^2*Log[2])*Log[(-x + Log[2])/5] + (-4*x^3 + 4*x^2*Log[2])*Lo g[(-x + Log[2])/5]^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.
| \(\displaystyle \int \frac {800 x^5+\left (-750 x^4-125\right ) \log (2)+\left (-150 x^5+\left (150 x^4+25\right ) \log (2)-25 x\right ) \log \left (\frac {1}{5} (\log (2)-x)\right )+100 x}{-100 x^3+100 x^2 \log (2)+\left (4 x^2 \log (2)-4 x^3\right ) \log ^2\left (\frac {1}{5} (\log (2)-x)\right )+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (\log (2)-x)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-800 x^5-\left (-750 x^4-125\right ) \log (2)-\left (-150 x^5+\left (150 x^4+25\right ) \log (2)-25 x\right ) \log \left (\frac {1}{5} (\log (2)-x)\right )-100 x}{4 x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int -\frac {25 \left (32 x^5+4 x-\left (6 x^5+x-\left (6 x^4+1\right ) \log (2)\right ) \log \left (\frac {1}{5} (\log (2)-x)\right )-5 \left (6 x^4+1\right ) \log (2)\right )}{x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {25}{4} \int \frac {32 x^5+4 x-\left (6 x^5+x-\left (6 x^4+1\right ) \log (2)\right ) \log \left (\frac {1}{5} (\log (2)-x)\right )-5 \left (6 x^4+1\right ) \log (2)}{x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {25}{4} \int \left (\frac {-6 x^4-1}{x^2 \left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )}+\frac {2 x^4-1}{x (x-\log (2)) \left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {25}{4} \left (-\int \frac {1}{x^2 \left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )}dx+\frac {\int \frac {1}{x \left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )^2}dx}{\log (2)}+10 e^5 \log ^2(2) \operatorname {ExpIntegralEi}\left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )+50 e^{10} \log (4) \operatorname {ExpIntegralEi}\left (-2 \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )\right )-5 e^5 \log (2) \log (4) \operatorname {ExpIntegralEi}\left (\log \left (\frac {1}{5} (\log (2)-x)\right )-5\right )-100 e^{10} \log (2) \operatorname {ExpIntegralEi}\left (-2 \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )\right )+\frac {2 x^2 (x-\log (2))}{5-\log \left (\frac {1}{5} (\log (2)-x)\right )}-\frac {1-2 \log ^4(2)}{\log (2) \left (5-\log \left (\frac {1}{5} (\log (2)-x)\right )\right )}+\frac {2 \log ^2(2) (x-\log (2))}{5-\log \left (\frac {1}{5} (\log (2)-x)\right )}+\frac {x \log (4) (x-\log (2))}{5-\log \left (\frac {1}{5} (\log (2)-x)\right )}\right )\) |
Int[(100*x + 800*x^5 + (-125 - 750*x^4)*Log[2] + (-25*x - 150*x^5 + (25 + 150*x^4)*Log[2])*Log[(-x + Log[2])/5])/(-100*x^3 + 100*x^2*Log[2] + (40*x^ 3 - 40*x^2*Log[2])*Log[(-x + Log[2])/5] + (-4*x^3 + 4*x^2*Log[2])*Log[(-x + Log[2])/5]^2),x]
3.28.67.3.1 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]]
Time = 1.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
| method | result | size |
| norman | \(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-5\right )}\) | \(25\) |
| risch | \(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-5\right )}\) | \(26\) |
| parallelrisch | \(\frac {50 x^{4}-25}{4 x \left (\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-5\right )}\) | \(26\) |
| derivativedivides | \(-\frac {25 \left (40 \ln \left (2\right )^{3} \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-1250 \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{4}+1000 \ln \left (2\right ) \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{3}-300 \ln \left (2\right )^{2} \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{2}+1-2 \ln \left (2\right )^{4}\right )}{4 \left (\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-5\right ) x}\) | \(83\) |
| default | \(-\frac {25 \left (40 \ln \left (2\right )^{3} \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-1250 \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{4}+1000 \ln \left (2\right ) \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{3}-300 \ln \left (2\right )^{2} \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )^{2}+1-2 \ln \left (2\right )^{4}\right )}{4 \left (\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )-5\right ) x}\) | \(83\) |
int((((150*x^4+25)*ln(2)-150*x^5-25*x)*ln(1/5*ln(2)-1/5*x)+(-750*x^4-125)* ln(2)+800*x^5+100*x)/((4*x^2*ln(2)-4*x^3)*ln(1/5*ln(2)-1/5*x)^2+(-40*x^2*l n(2)+40*x^3)*ln(1/5*ln(2)-1/5*x)+100*x^2*ln(2)-100*x^3),x,method=_RETURNVE RBOSE)
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, \log \left (2\right )\right ) - 5 \, x\right )}} \]
integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750* x^4-125)*log(2)+800*x^5+100*x)/((4*x^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x) ^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3),x , algorithm=\
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\frac {50 x^{4} - 25}{4 x \log {\left (- \frac {x}{5} + \frac {\log {\left (2 \right )}}{5} \right )} - 20 x} \]
integrate((((150*x**4+25)*ln(2)-150*x**5-25*x)*ln(1/5*ln(2)-1/5*x)+(-750*x **4-125)*ln(2)+800*x**5+100*x)/((4*x**2*ln(2)-4*x**3)*ln(1/5*ln(2)-1/5*x)* *2+(-40*x**2*ln(2)+40*x**3)*ln(1/5*ln(2)-1/5*x)+100*x**2*ln(2)-100*x**3),x )
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=-\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x {\left (\log \left (5\right ) + 5\right )} - x \log \left (-x + \log \left (2\right )\right )\right )}} \]
integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750* x^4-125)*log(2)+800*x^5+100*x)/((4*x^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x) ^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3),x , algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.17 \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\frac {25 \, {\left (2 \, {\left (x - \log \left (2\right )\right )}^{4} + 8 \, {\left (x - \log \left (2\right )\right )}^{3} \log \left (2\right ) + 12 \, {\left (x - \log \left (2\right )\right )}^{2} \log \left (2\right )^{2} + 8 \, {\left (x - \log \left (2\right )\right )} \log \left (2\right )^{3} + 2 \, \log \left (2\right )^{4} - 1\right )}}{4 \, {\left ({\left (x - \log \left (2\right )\right )} \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, \log \left (2\right )\right ) + \log \left (2\right ) \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, \log \left (2\right )\right ) - 5 \, x\right )}} \]
integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750* x^4-125)*log(2)+800*x^5+100*x)/((4*x^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x) ^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3),x , algorithm=\
25/4*(2*(x - log(2))^4 + 8*(x - log(2))^3*log(2) + 12*(x - log(2))^2*log(2 )^2 + 8*(x - log(2))*log(2)^3 + 2*log(2)^4 - 1)/((x - log(2))*log(-1/5*x + 1/5*log(2)) + log(2)*log(-1/5*x + 1/5*log(2)) - 5*x)
Timed out. \[ \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx=\int -\frac {100\,x-\ln \left (2\right )\,\left (750\,x^4+125\right )-\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )\,\left (25\,x-\ln \left (2\right )\,\left (150\,x^4+25\right )+150\,x^5\right )+800\,x^5}{\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )\,\left (40\,x^2\,\ln \left (2\right )-40\,x^3\right )-{\ln \left (\frac {\ln \left (2\right )}{5}-\frac {x}{5}\right )}^2\,\left (4\,x^2\,\ln \left (2\right )-4\,x^3\right )-100\,x^2\,\ln \left (2\right )+100\,x^3} \,d x \]
int(-(100*x - log(2)*(750*x^4 + 125) - log(log(2)/5 - x/5)*(25*x - log(2)* (150*x^4 + 25) + 150*x^5) + 800*x^5)/(log(log(2)/5 - x/5)*(40*x^2*log(2) - 40*x^3) - log(log(2)/5 - x/5)^2*(4*x^2*log(2) - 4*x^3) - 100*x^2*log(2) + 100*x^3),x)