Integrand size = 379, antiderivative size = 31 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=-x+\frac {x}{\log (4)+\frac {2 \left (1+e^5\right )}{x+\log \left (x+\frac {\log (3)}{5}\right )}} \]
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(31)=62\).
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.61 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=-\frac {x (-1+\log (4)) \log (4)+\frac {\left (1+e^5\right ) x \left (10 x \log (4)+\log (4) \log (9) (1+\log (16))-\log ^2(4) \log (81)+\log (1048576)\right )}{(5+5 x+\log (3)) \left (2+2 e^5+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )}}{\log ^2(4)} \]
Integrate[(-10*x - 20*E^10*x + 20*x^2 + E^5*(-30*x + 20*x^2) + (-4 - 4*E^1 0 + 4*x + E^5*(-8 + 4*x))*Log[3] + (-20*x^2 - 20*E^5*x^2 + 5*x^3 + (-4*x - 4*E^5*x + x^2)*Log[3])*Log[4] + (-5*x^3 - x^2*Log[3])*Log[4]^2 + (10*x + 10*E^5*x + (2 + 2*E^5)*Log[3] + (-20*x - 20*E^5*x + 10*x^2 + (-4 - 4*E^5 + 2*x)*Log[3])*Log[4] + (-10*x^2 - 2*x*Log[3])*Log[4]^2)*Log[(5*x + Log[3]) /5] + ((5*x + Log[3])*Log[4] + (-5*x - Log[3])*Log[4]^2)*Log[(5*x + Log[3] )/5]^2)/(20*x + 40*E^5*x + 20*E^10*x + (4 + 8*E^5 + 4*E^10)*Log[3] + (20*x ^2 + 20*E^5*x^2 + (4*x + 4*E^5*x)*Log[3])*Log[4] + (5*x^3 + x^2*Log[3])*Lo g[4]^2 + ((20*x + 20*E^5*x + (4 + 4*E^5)*Log[3])*Log[4] + (10*x^2 + 2*x*Lo g[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + (5*x + Log[3])*Log[4]^2*Log[(5*x + Log[3])/5]^2),x]
-((x*(-1 + Log[4])*Log[4] + ((1 + E^5)*x*(10*x*Log[4] + Log[4]*Log[9]*(1 + Log[16]) - Log[4]^2*Log[81] + Log[1048576]))/((5 + 5*x + Log[3])*(2 + 2*E ^5 + x*Log[4] + Log[4]*Log[x + Log[3]/5])))/Log[4]^2)
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.
\(\displaystyle \int \frac {20 x^2+e^5 \left (20 x^2-30 x\right )+\left (\log ^2(4) \left (-10 x^2-2 x \log (3)\right )+\log (4) \left (10 x^2-20 e^5 x-20 x+\left (2 x-4 e^5-4\right ) \log (3)\right )+10 e^5 x+10 x+\left (2+2 e^5\right ) \log (3)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log ^2(4) \left (-5 x^3-x^2 \log (3)\right )+\log (4) \left (5 x^3-20 e^5 x^2-20 x^2+\left (x^2-4 e^5 x-4 x\right ) \log (3)\right )-20 e^{10} x-10 x+\left (\log ^2(4) (-5 x-\log (3))+\log (4) (5 x+\log (3))\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4 x+e^5 (4 x-8)-4 e^{10}-4\right ) \log (3)}{\left (\log ^2(4) \left (10 x^2+2 x \log (3)\right )+\log (4) \left (20 e^5 x+20 x+\left (4+4 e^5\right ) \log (3)\right )\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log (4) \left (20 e^5 x^2+20 x^2+\left (4 e^5 x+4 x\right ) \log (3)\right )+\log ^2(4) \left (5 x^3+x^2 \log (3)\right )+20 e^{10} x+40 e^5 x+20 x+\log ^2(4) (5 x+\log (3)) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4+8 e^5+4 e^{10}\right ) \log (3)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {20 x^2+e^5 \left (20 x^2-30 x\right )+\left (\log ^2(4) \left (-10 x^2-2 x \log (3)\right )+\log (4) \left (10 x^2-20 e^5 x-20 x+\left (2 x-4 e^5-4\right ) \log (3)\right )+10 e^5 x+10 x+\left (2+2 e^5\right ) \log (3)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log ^2(4) \left (-5 x^3-x^2 \log (3)\right )+\log (4) \left (5 x^3-20 e^5 x^2-20 x^2+\left (x^2-4 e^5 x-4 x\right ) \log (3)\right )-20 e^{10} x-10 x+\left (\log ^2(4) (-5 x-\log (3))+\log (4) (5 x+\log (3))\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4 x+e^5 (4 x-8)-4 e^{10}-4\right ) \log (3)}{\left (\log ^2(4) \left (10 x^2+2 x \log (3)\right )+\log (4) \left (20 e^5 x+20 x+\left (4+4 e^5\right ) \log (3)\right )\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log (4) \left (20 e^5 x^2+20 x^2+\left (4 e^5 x+4 x\right ) \log (3)\right )+\log ^2(4) \left (5 x^3+x^2 \log (3)\right )+\left (20+40 e^5\right ) x+20 e^{10} x+\log ^2(4) (5 x+\log (3)) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4+8 e^5+4 e^{10}\right ) \log (3)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {20 x^2+e^5 \left (20 x^2-30 x\right )+\left (\log ^2(4) \left (-10 x^2-2 x \log (3)\right )+\log (4) \left (10 x^2-20 e^5 x-20 x+\left (2 x-4 e^5-4\right ) \log (3)\right )+10 e^5 x+10 x+\left (2+2 e^5\right ) \log (3)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log ^2(4) \left (-5 x^3-x^2 \log (3)\right )+\log (4) \left (5 x^3-20 e^5 x^2-20 x^2+\left (x^2-4 e^5 x-4 x\right ) \log (3)\right )-20 e^{10} x-10 x+\left (\log ^2(4) (-5 x-\log (3))+\log (4) (5 x+\log (3))\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4 x+e^5 (4 x-8)-4 e^{10}-4\right ) \log (3)}{\left (\log ^2(4) \left (10 x^2+2 x \log (3)\right )+\log (4) \left (20 e^5 x+20 x+\left (4+4 e^5\right ) \log (3)\right )\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log (4) \left (20 e^5 x^2+20 x^2+\left (4 e^5 x+4 x\right ) \log (3)\right )+\log ^2(4) \left (5 x^3+x^2 \log (3)\right )+\left (20+40 e^5+20 e^{10}\right ) x+\log ^2(4) (5 x+\log (3)) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4+8 e^5+4 e^{10}\right ) \log (3)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {20 x^2+e^5 \left (20 x^2-30 x\right )+\left (\log ^2(4) \left (-10 x^2-2 x \log (3)\right )+\log (4) \left (10 x^2-20 e^5 x-20 x+\left (2 x-4 e^5-4\right ) \log (3)\right )+10 e^5 x+10 x+\left (2+2 e^5\right ) \log (3)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log ^2(4) \left (-5 x^3-x^2 \log (3)\right )+\log (4) \left (5 x^3-20 e^5 x^2-20 x^2+\left (x^2-4 e^5 x-4 x\right ) \log (3)\right )+\left (-10-20 e^{10}\right ) x+\left (\log ^2(4) (-5 x-\log (3))+\log (4) (5 x+\log (3))\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4 x+e^5 (4 x-8)-4 e^{10}-4\right ) \log (3)}{\left (\log ^2(4) \left (10 x^2+2 x \log (3)\right )+\log (4) \left (20 e^5 x+20 x+\left (4+4 e^5\right ) \log (3)\right )\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\log (4) \left (20 e^5 x^2+20 x^2+\left (4 e^5 x+4 x\right ) \log (3)\right )+\log ^2(4) \left (5 x^3+x^2 \log (3)\right )+\left (20+40 e^5+20 e^{10}\right ) x+\log ^2(4) (5 x+\log (3)) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )+\left (4+8 e^5+4 e^{10}\right ) \log (3)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-5 x^3 (\log (4)-1) \log (4)-x^2 (\log (4)-1) \left (20+20 e^5+\log (3) \log (4)\right )-(\log (4)-1) \log (4) (5 x+\log (3)) \log ^2\left (x+\frac {\log (3)}{5}\right )-\left (1+e^5\right ) x \left (10+20 e^5+(\log (4)-1) \log (81)\right )-2 (5 x+\log (3)) \left (x (\log (4)-1) \log (4)-1+\log (16)+e^5 (\log (16)-1)\right ) \log \left (x+\frac {\log (3)}{5}\right )-4 \left (1+e^5\right )^2 \log (3)}{(5 x+\log (3)) \left (x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (1+e^5\right ) x \left (10 x \log (4)-\log ^2(4) \log (81)+\log (1048576)+\log (4) \log (9) \log (16)+\log (4) \log (9)\right )}{\log (4) (5 x+\log (3)) \left (x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )^2}+\frac {2 \left (-1-e^5\right )}{\log (4) \left (x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )}+\frac {1-\log (4)}{\log (4)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (1+e^5\right ) \left (-\log ^2(4) \log (81)+\log (4) \log (9) \log (16)+\log (1048576)\right ) \int \frac {1}{\left (\log (4) x+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )^2}dx}{5 \log (4)}-\frac {\left (1+e^5\right ) \log (3) \left (-\log ^2(4) \log (81)+\log (4) \log (9) \log (16)+\log (1048576)\right ) \int \frac {1}{(5 x+\log (3)) \left (\log (4) x+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )^2}dx}{5 \log (4)}+2 \left (1+e^5\right ) \int \frac {x}{\left (\log (4) x+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )\right )^2}dx-\frac {2 \left (1+e^5\right ) \int \frac {1}{\log (4) x+\log (4) \log \left (x+\frac {\log (3)}{5}\right )+2 \left (1+e^5\right )}dx}{\log (4)}+\frac {x (1-\log (4))}{\log (4)}\) |
Int[(-10*x - 20*E^10*x + 20*x^2 + E^5*(-30*x + 20*x^2) + (-4 - 4*E^10 + 4* x + E^5*(-8 + 4*x))*Log[3] + (-20*x^2 - 20*E^5*x^2 + 5*x^3 + (-4*x - 4*E^5 *x + x^2)*Log[3])*Log[4] + (-5*x^3 - x^2*Log[3])*Log[4]^2 + (10*x + 10*E^5 *x + (2 + 2*E^5)*Log[3] + (-20*x - 20*E^5*x + 10*x^2 + (-4 - 4*E^5 + 2*x)* Log[3])*Log[4] + (-10*x^2 - 2*x*Log[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + ((5*x + Log[3])*Log[4] + (-5*x - Log[3])*Log[4]^2)*Log[(5*x + Log[3])/5]^2 )/(20*x + 40*E^5*x + 20*E^10*x + (4 + 8*E^5 + 4*E^10)*Log[3] + (20*x^2 + 2 0*E^5*x^2 + (4*x + 4*E^5*x)*Log[3])*Log[4] + (5*x^3 + x^2*Log[3])*Log[4]^2 + ((20*x + 20*E^5*x + (4 + 4*E^5)*Log[3])*Log[4] + (10*x^2 + 2*x*Log[3])* Log[4]^2)*Log[(5*x + Log[3])/5] + (5*x + Log[3])*Log[4]^2*Log[(5*x + Log[3 ])/5]^2),x]
3.26.11.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39
method | result | size |
risch | \(-x +\frac {x}{2 \ln \left (2\right )}-\frac {\left ({\mathrm e}^{5}+1\right ) x}{2 \ln \left (2\right ) \left (\ln \left (2\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+x \ln \left (2\right )+{\mathrm e}^{5}+1\right )}\) | \(43\) |
norman | \(\frac {\left (-\ln \left (2\right )+\frac {1}{2}\right ) x^{2}+\left ({\mathrm e}^{5}+1\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+\left (-\ln \left (2\right )+\frac {1}{2}\right ) x \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+\frac {{\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1}{\ln \left (2\right )}}{\ln \left (2\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+x \ln \left (2\right )+{\mathrm e}^{5}+1}\) | \(75\) |
derivativedivides | \(-\frac {\left (10 \ln \left (2\right )-5\right ) \left (\frac {\ln \left (3\right )}{5}+x \right )^{2}+\left (-2 \ln \left (2\right ) \ln \left (3\right )+10 \,{\mathrm e}^{5}+\ln \left (3\right )+10\right ) \left (\frac {\ln \left (3\right )}{5}+x \right )+\left (10 \ln \left (2\right )-5\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) \left (\frac {\ln \left (3\right )}{5}+x \right )-\frac {\ln \left (3\right ) \left ({\mathrm e}^{5}+1\right )}{\ln \left (2\right )}}{2 \left (5 \left (\frac {\ln \left (3\right )}{5}+x \right ) \ln \left (2\right )-\ln \left (2\right ) \ln \left (3\right )+5 \ln \left (2\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+5 \,{\mathrm e}^{5}+5\right )}\) | \(107\) |
default | \(-\frac {\left (10 \ln \left (2\right )-5\right ) \left (\frac {\ln \left (3\right )}{5}+x \right )^{2}+\left (-2 \ln \left (2\right ) \ln \left (3\right )+10 \,{\mathrm e}^{5}+\ln \left (3\right )+10\right ) \left (\frac {\ln \left (3\right )}{5}+x \right )+\left (10 \ln \left (2\right )-5\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) \left (\frac {\ln \left (3\right )}{5}+x \right )-\frac {\ln \left (3\right ) \left ({\mathrm e}^{5}+1\right )}{\ln \left (2\right )}}{2 \left (5 \left (\frac {\ln \left (3\right )}{5}+x \right ) \ln \left (2\right )-\ln \left (2\right ) \ln \left (3\right )+5 \ln \left (2\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+5 \,{\mathrm e}^{5}+5\right )}\) | \(107\) |
parallelrisch | \(\frac {-50 x \,{\mathrm e}^{5} \ln \left (2\right )-50 x^{2} \ln \left (2\right )^{2}-50 x \ln \left (2\right )+25 x^{2} \ln \left (2\right )-10 \,{\mathrm e}^{5} \ln \left (3\right )+20 \ln \left (2\right ) \ln \left (3\right )-10 x \ln \left (2\right ) \ln \left (3\right )-10 \ln \left (3\right )-10 \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) \ln \left (2\right ) \ln \left (3\right )+25 \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) \ln \left (2\right ) x +20 \ln \left (2\right )^{2} \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) \ln \left (3\right )-50 \ln \left (2\right )^{2} \ln \left (\frac {\ln \left (3\right )}{5}+x \right ) x +20 \,{\mathrm e}^{5} \ln \left (2\right ) \ln \left (3\right )+20 \ln \left (3\right ) \ln \left (2\right )^{2} x}{50 \ln \left (2\right ) \left (\ln \left (2\right ) \ln \left (\frac {\ln \left (3\right )}{5}+x \right )+x \ln \left (2\right )+{\mathrm e}^{5}+1\right )}\) | \(150\) |
int(((4*(-ln(3)-5*x)*ln(2)^2+2*(ln(3)+5*x)*ln(2))*ln(1/5*ln(3)+x)^2+(4*(-2 *x*ln(3)-10*x^2)*ln(2)^2+2*((-4*exp(5)+2*x-4)*ln(3)-20*x*exp(5)+10*x^2-20* x)*ln(2)+(2*exp(5)+2)*ln(3)+10*x*exp(5)+10*x)*ln(1/5*ln(3)+x)+4*(-x^2*ln(3 )-5*x^3)*ln(2)^2+2*((-4*x*exp(5)+x^2-4*x)*ln(3)-20*x^2*exp(5)+5*x^3-20*x^2 )*ln(2)+(-4*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*ln(3)-20*x*exp(5)^2+(20*x^2-30* x)*exp(5)+20*x^2-10*x)/(4*(ln(3)+5*x)*ln(2)^2*ln(1/5*ln(3)+x)^2+(4*(2*x*ln (3)+10*x^2)*ln(2)^2+2*((4*exp(5)+4)*ln(3)+20*x*exp(5)+20*x)*ln(2))*ln(1/5* ln(3)+x)+4*(x^2*ln(3)+5*x^3)*ln(2)^2+2*((4*x*exp(5)+4*x)*ln(3)+20*x^2*exp( 5)+20*x^2)*ln(2)+(4*exp(5)^2+8*exp(5)+4)*ln(3)+20*x*exp(5)^2+40*x*exp(5)+2 0*x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=-\frac {2 \, x^{2} \log \left (2\right ) - x^{2} + 2 \, x e^{5} + {\left (2 \, x \log \left (2\right ) - x\right )} \log \left (x + \frac {1}{5} \, \log \left (3\right )\right ) + 2 \, x}{2 \, {\left (x \log \left (2\right ) + \log \left (2\right ) \log \left (x + \frac {1}{5} \, \log \left (3\right )\right ) + e^{5} + 1\right )}} \]
integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3) +x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*ex p(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*log (3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x ^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-2 0*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^2*lo g(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+2 0*x*exp(5)+20*x)*log(2))*log(1/5*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2 *((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8*exp(5 )+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm=\
-1/2*(2*x^2*log(2) - x^2 + 2*x*e^5 + (2*x*log(2) - x)*log(x + 1/5*log(3)) + 2*x)/(x*log(2) + log(2)*log(x + 1/5*log(3)) + e^5 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=\frac {x \left (1 - 2 \log {\left (2 \right )}\right )}{2 \log {\left (2 \right )}} + \frac {- x e^{5} - x}{2 x \log {\left (2 \right )}^{2} + 2 \log {\left (2 \right )}^{2} \log {\left (x + \frac {\log {\left (3 \right )}}{5} \right )} + 2 \log {\left (2 \right )} + 2 e^{5} \log {\left (2 \right )}} \]
integrate(((4*(-ln(3)-5*x)*ln(2)**2+2*(ln(3)+5*x)*ln(2))*ln(1/5*ln(3)+x)** 2+(4*(-2*x*ln(3)-10*x**2)*ln(2)**2+2*((-4*exp(5)+2*x-4)*ln(3)-20*x*exp(5)+ 10*x**2-20*x)*ln(2)+(2*exp(5)+2)*ln(3)+10*x*exp(5)+10*x)*ln(1/5*ln(3)+x)+4 *(-x**2*ln(3)-5*x**3)*ln(2)**2+2*((-4*x*exp(5)+x**2-4*x)*ln(3)-20*x**2*exp (5)+5*x**3-20*x**2)*ln(2)+(-4*exp(5)**2+(4*x-8)*exp(5)+4*x-4)*ln(3)-20*x*e xp(5)**2+(20*x**2-30*x)*exp(5)+20*x**2-10*x)/(4*(ln(3)+5*x)*ln(2)**2*ln(1/ 5*ln(3)+x)**2+(4*(2*x*ln(3)+10*x**2)*ln(2)**2+2*((4*exp(5)+4)*ln(3)+20*x*e xp(5)+20*x)*ln(2))*ln(1/5*ln(3)+x)+4*(x**2*ln(3)+5*x**3)*ln(2)**2+2*((4*x* exp(5)+4*x)*ln(3)+20*x**2*exp(5)+20*x**2)*ln(2)+(4*exp(5)**2+8*exp(5)+4)*l n(3)+20*x*exp(5)**2+40*x*exp(5)+20*x),x)
x*(1 - 2*log(2))/(2*log(2)) + (-x*exp(5) - x)/(2*x*log(2)**2 + 2*log(2)**2 *log(x + log(3)/5) + 2*log(2) + 2*exp(5)*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (28) = 56\).
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.35 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=-\frac {x^{2} {\left (2 \, \log \left (2\right ) - 1\right )} + x {\left (2 \, \log \left (2\right ) - 1\right )} \log \left (5 \, x + \log \left (3\right )\right ) - {\left (2 \, \log \left (5\right ) \log \left (2\right ) - 2 \, e^{5} - \log \left (5\right ) - 2\right )} x}{2 \, {\left (x \log \left (2\right ) - \log \left (5\right ) \log \left (2\right ) + \log \left (2\right ) \log \left (5 \, x + \log \left (3\right )\right ) + e^{5} + 1\right )}} \]
integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3) +x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*ex p(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*log (3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x ^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-2 0*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^2*lo g(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+2 0*x*exp(5)+20*x)*log(2))*log(1/5*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2 *((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8*exp(5 )+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm=\
-1/2*(x^2*(2*log(2) - 1) + x*(2*log(2) - 1)*log(5*x + log(3)) - (2*log(5)* log(2) - 2*e^5 - log(5) - 2)*x)/(x*log(2) - log(5)*log(2) + log(2)*log(5*x + log(3)) + e^5 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (28) = 56\).
Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=-\frac {2 \, x^{2} \log \left (2\right ) - 2 \, x \log \left (5\right ) \log \left (2\right ) + 2 \, x \log \left (2\right ) \log \left (5 \, x + \log \left (3\right )\right ) - x^{2} + 2 \, x e^{5} + x \log \left (5\right ) - x \log \left (5 \, x + \log \left (3\right )\right ) + 2 \, x}{2 \, {\left (x \log \left (2\right ) - \log \left (5\right ) \log \left (2\right ) + \log \left (2\right ) \log \left (5 \, x + \log \left (3\right )\right ) + e^{5} + 1\right )}} \]
integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3) +x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*ex p(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*log (3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x ^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-2 0*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^2*lo g(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+2 0*x*exp(5)+20*x)*log(2))*log(1/5*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2 *((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8*exp(5 )+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm=\
-1/2*(2*x^2*log(2) - 2*x*log(5)*log(2) + 2*x*log(2)*log(5*x + log(3)) - x^ 2 + 2*x*e^5 + x*log(5) - x*log(5*x + log(3)) + 2*x)/(x*log(2) - log(5)*log (2) + log(2)*log(5*x + log(3)) + e^5 + 1)
Time = 14.99 (sec) , antiderivative size = 126, normalized size of antiderivative = 4.06 \[ \int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx=\frac {2\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+x^2\,{\ln \left (2\right )}^2-2\,x^2\,{\ln \left (2\right )}^3+x\,\ln \left (2\right )-2\,x\,{\ln \left (2\right )}^2+\ln \left (x+\frac {\ln \left (3\right )}{5}\right )\,\ln \left (2\right )-2\,x\,{\mathrm {e}}^5\,{\ln \left (2\right )}^2+\ln \left (x+\frac {\ln \left (3\right )}{5}\right )\,{\mathrm {e}}^5\,\ln \left (2\right )+x\,{\mathrm {e}}^5\,\ln \left (2\right )+x\,\ln \left (x+\frac {\ln \left (3\right )}{5}\right )\,{\ln \left (2\right )}^2-2\,x\,\ln \left (x+\frac {\ln \left (3\right )}{5}\right )\,{\ln \left (2\right )}^3+1}{2\,{\ln \left (2\right )}^2\,\left ({\mathrm {e}}^5+x\,\ln \left (2\right )+\ln \left (x+\frac {\ln \left (3\right )}{5}\right )\,\ln \left (2\right )+1\right )} \]
int(-(10*x - log(x + log(3)/5)*(10*x - 4*log(2)^2*(2*x*log(3) + 10*x^2) + 10*x*exp(5) - 2*log(2)*(20*x + 20*x*exp(5) + log(3)*(4*exp(5) - 2*x + 4) - 10*x^2) + log(3)*(2*exp(5) + 2)) + exp(5)*(30*x - 20*x^2) + 20*x*exp(10) + 4*log(2)^2*(x^2*log(3) + 5*x^3) - log(x + log(3)/5)^2*(2*log(2)*(5*x + l og(3)) - 4*log(2)^2*(5*x + log(3))) + 2*log(2)*(log(3)*(4*x + 4*x*exp(5) - x^2) + 20*x^2*exp(5) + 20*x^2 - 5*x^3) - log(3)*(4*x - 4*exp(10) + exp(5) *(4*x - 8) - 4) - 20*x^2)/(20*x + log(x + log(3)/5)*(4*log(2)^2*(2*x*log(3 ) + 10*x^2) + 2*log(2)*(20*x + 20*x*exp(5) + log(3)*(4*exp(5) + 4))) + 40* x*exp(5) + 20*x*exp(10) + 2*log(2)*(log(3)*(4*x + 4*x*exp(5)) + 20*x^2*exp (5) + 20*x^2) + 4*log(2)^2*(x^2*log(3) + 5*x^3) + log(3)*(8*exp(5) + 4*exp (10) + 4) + 4*log(x + log(3)/5)^2*log(2)^2*(5*x + log(3))),x)
(2*exp(5) + exp(10) + x^2*log(2)^2 - 2*x^2*log(2)^3 + x*log(2) - 2*x*log(2 )^2 + log(x + log(3)/5)*log(2) - 2*x*exp(5)*log(2)^2 + log(x + log(3)/5)*e xp(5)*log(2) + x*exp(5)*log(2) + x*log(x + log(3)/5)*log(2)^2 - 2*x*log(x + log(3)/5)*log(2)^3 + 1)/(2*log(2)^2*(exp(5) + x*log(2) + log(x + log(3)/ 5)*log(2) + 1))