Integrand size = 129, antiderivative size = 22 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=\frac {x (-2+5 x \log (25))^2}{1+\log \left (-\frac {7}{2}+\log (x)\right )} \]
Time = 0.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=\frac {x (2-5 x \log (25))^2}{1+\log \left (-\frac {7}{2}+\log (x)\right )} \]
Integrate[(-36 + 320*x*Log[25] - 575*x^2*Log[25]^2 + (8 - 80*x*Log[25] + 1 50*x^2*Log[25]^2)*Log[x] + (-28 + 280*x*Log[25] - 525*x^2*Log[25]^2 + (8 - 80*x*Log[25] + 150*x^2*Log[25]^2)*Log[x])*Log[(-7 + 2*Log[x])/2])/(-7 + 2 *Log[x] + (-14 + 4*Log[x])*Log[(-7 + 2*Log[x])/2] + (-7 + 2*Log[x])*Log[(- 7 + 2*Log[x])/2]^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.
| \(\displaystyle \int \frac {-575 x^2 \log ^2(25)+\left (150 x^2 \log ^2(25)-80 x \log (25)+8\right ) \log (x)+\left (-525 x^2 \log ^2(25)+\left (150 x^2 \log ^2(25)-80 x \log (25)+8\right ) \log (x)+280 x \log (25)-28\right ) \log \left (\frac {1}{2} (2 \log (x)-7)\right )+320 x \log (25)-36}{(2 \log (x)-7) \log ^2\left (\frac {1}{2} (2 \log (x)-7)\right )+(4 \log (x)-14) \log \left (\frac {1}{2} (2 \log (x)-7)\right )+2 \log (x)-7} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {(2-5 x \log (25)) \left (-115 x \log (25)-7 (15 x \log (25)-2) \log \left (\log (x)-\frac {7}{2}\right )+2 (15 x \log (25)-2) \log (x) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )+18\right )}{(7-2 \log (x)) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {75 x^2 \log ^2(25)-40 x \log (25)+4}{\log \left (\log (x)-\frac {7}{2}\right )+1}-\frac {2 (5 x \log (25)-2)^2}{(2 \log (x)-7) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -50 \log ^2(25) \int \frac {x^2}{(2 \log (x)-7) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )^2}dx+75 \log ^2(25) \int \frac {x^2}{\log \left (\log (x)-\frac {7}{2}\right )+1}dx-8 \int \frac {1}{(2 \log (x)-7) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )^2}dx+40 \log (25) \int \frac {x}{(2 \log (x)-7) \left (\log \left (\log (x)-\frac {7}{2}\right )+1\right )^2}dx+4 \int \frac {1}{\log \left (\log (x)-\frac {7}{2}\right )+1}dx-40 \log (25) \int \frac {x}{\log \left (\log (x)-\frac {7}{2}\right )+1}dx\) |
Int[(-36 + 320*x*Log[25] - 575*x^2*Log[25]^2 + (8 - 80*x*Log[25] + 150*x^2 *Log[25]^2)*Log[x] + (-28 + 280*x*Log[25] - 525*x^2*Log[25]^2 + (8 - 80*x* Log[25] + 150*x^2*Log[25]^2)*Log[x])*Log[(-7 + 2*Log[x])/2])/(-7 + 2*Log[x ] + (-14 + 4*Log[x])*Log[(-7 + 2*Log[x])/2] + (-7 + 2*Log[x])*Log[(-7 + 2* Log[x])/2]^2),x]
3.23.67.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
| method | result | size |
| risch | \(\frac {4 x \left (25 x^{2} \ln \left (5\right )^{2}-10 x \ln \left (5\right )+1\right )}{1+\ln \left (\ln \left (x \right )-\frac {7}{2}\right )}\) | \(29\) |
| parallelrisch | \(\frac {400 x^{3} \ln \left (5\right )^{2}-160 x^{2} \ln \left (5\right )+16 x}{4+4 \ln \left (\ln \left (x \right )-\frac {7}{2}\right )}\) | \(32\) |
| default | \(-\frac {4 x}{\ln \left (2\right )-\ln \left (2 \ln \left (x \right )-7\right )-1}+\frac {40 \ln \left (5\right ) x^{2}}{\ln \left (2\right )-\ln \left (2 \ln \left (x \right )-7\right )-1}-\frac {100 \ln \left (5\right )^{2} x^{3}}{\ln \left (2\right )-\ln \left (2 \ln \left (x \right )-7\right )-1}\) | \(66\) |
int((((600*x^2*ln(5)^2-160*x*ln(5)+8)*ln(x)-2100*x^2*ln(5)^2+560*x*ln(5)-2 8)*ln(ln(x)-7/2)+(600*x^2*ln(5)^2-160*x*ln(5)+8)*ln(x)-2300*x^2*ln(5)^2+64 0*x*ln(5)-36)/((2*ln(x)-7)*ln(ln(x)-7/2)^2+(4*ln(x)-14)*ln(ln(x)-7/2)+2*ln (x)-7),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=\frac {4 \, {\left (25 \, x^{3} \log \left (5\right )^{2} - 10 \, x^{2} \log \left (5\right ) + x\right )}}{\log \left (\log \left (x\right ) - \frac {7}{2}\right ) + 1} \]
integrate((((600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-2100*x^2*log(5)^2+560 *x*log(5)-28)*log(log(x)-7/2)+(600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-230 0*x^2*log(5)^2+640*x*log(5)-36)/((2*log(x)-7)*log(log(x)-7/2)^2+(4*log(x)- 14)*log(log(x)-7/2)+2*log(x)-7),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=\frac {100 x^{3} \log {\left (5 \right )}^{2} - 40 x^{2} \log {\left (5 \right )} + 4 x}{\log {\left (\log {\left (x \right )} - \frac {7}{2} \right )} + 1} \]
integrate((((600*x**2*ln(5)**2-160*x*ln(5)+8)*ln(x)-2100*x**2*ln(5)**2+560 *x*ln(5)-28)*ln(ln(x)-7/2)+(600*x**2*ln(5)**2-160*x*ln(5)+8)*ln(x)-2300*x* *2*ln(5)**2+640*x*ln(5)-36)/((2*ln(x)-7)*ln(ln(x)-7/2)**2+(4*ln(x)-14)*ln( ln(x)-7/2)+2*ln(x)-7),x)
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=-\frac {4 \, {\left (25 \, x^{3} \log \left (5\right )^{2} - 10 \, x^{2} \log \left (5\right ) + x\right )}}{\log \left (2\right ) - \log \left (2 \, \log \left (x\right ) - 7\right ) - 1} \]
integrate((((600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-2100*x^2*log(5)^2+560 *x*log(5)-28)*log(log(x)-7/2)+(600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-230 0*x^2*log(5)^2+640*x*log(5)-36)/((2*log(x)-7)*log(log(x)-7/2)^2+(4*log(x)- 14)*log(log(x)-7/2)+2*log(x)-7),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (21) = 42\).
Time = 0.36 (sec) , antiderivative size = 454, normalized size of antiderivative = 20.64 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx =\text {Too large to display} \]
integrate((((600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-2100*x^2*log(5)^2+560 *x*log(5)-28)*log(log(x)-7/2)+(600*x^2*log(5)^2-160*x*log(5)+8)*log(x)-230 0*x^2*log(5)^2+640*x*log(5)-36)/((2*log(x)-7)*log(log(x)-7/2)^2+(4*log(x)- 14)*log(log(x)-7/2)+2*log(x)-7),x, algorithm=\
-100*x^3*log(5)^2*log(2)/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2*lo g(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) + 100*x^3*log(5)^2*log(2 *log(x) - 7)/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2*log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) + 100*x^3*log(5)^2/(log(2)^2 - 2*log (2)*log(2*log(x) - 7) + log(2*log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) + 40*x^2*log(5)*log(2)/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + lo g(2*log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) - 40*x^2*log(5)*lo g(2*log(x) - 7)/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2*log(x) - 7) ^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) - 40*x^2*log(5)/(log(2)^2 - 2*log (2)*log(2*log(x) - 7) + log(2*log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) - 4*x*log(2)/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2*log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1) + 4*x*log(2*log(x) - 7)/(log (2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2*log(x) - 7)^2 - 2*log(2) + 2*lo g(2*log(x) - 7) + 1) + 4*x/(log(2)^2 - 2*log(2)*log(2*log(x) - 7) + log(2* log(x) - 7)^2 - 2*log(2) + 2*log(2*log(x) - 7) + 1)
Time = 14.75 (sec) , antiderivative size = 112, normalized size of antiderivative = 5.09 \[ \int \frac {-36+320 x \log (25)-575 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)+\left (-28+280 x \log (25)-525 x^2 \log ^2(25)+\left (8-80 x \log (25)+150 x^2 \log ^2(25)\right ) \log (x)\right ) \log \left (\frac {1}{2} (-7+2 \log (x))\right )}{-7+2 \log (x)+(-14+4 \log (x)) \log \left (\frac {1}{2} (-7+2 \log (x))\right )+(-7+2 \log (x)) \log ^2\left (\frac {1}{2} (-7+2 \log (x))\right )} \, dx=\ln \left (x\right )\,\left (300\,{\ln \left (5\right )}^2\,x^3-80\,\ln \left (5\right )\,x^2+4\,x\right )-1050\,x^3\,{\ln \left (5\right )}^2-14\,x+\frac {2\,x\,\left (5\,x\,\ln \left (5\right )-1\right )\,\left (2\,\ln \left (x\right )+115\,x\,\ln \left (5\right )-30\,x\,\ln \left (5\right )\,\ln \left (x\right )-9\right )-2\,x\,\ln \left (\ln \left (x\right )-\frac {7}{2}\right )\,\left (2\,\ln \left (x\right )-7\right )\,\left (75\,{\ln \left (5\right )}^2\,x^2-20\,\ln \left (5\right )\,x+1\right )}{\ln \left (\ln \left (x\right )-\frac {7}{2}\right )+1}+280\,x^2\,\ln \left (5\right ) \]
int(-(2300*x^2*log(5)^2 - 640*x*log(5) + log(log(x) - 7/2)*(2100*x^2*log(5 )^2 - 560*x*log(5) - log(x)*(600*x^2*log(5)^2 - 160*x*log(5) + 8) + 28) - log(x)*(600*x^2*log(5)^2 - 160*x*log(5) + 8) + 36)/(2*log(x) + log(log(x) - 7/2)^2*(2*log(x) - 7) + log(log(x) - 7/2)*(4*log(x) - 14) - 7),x)