Integrand size = 232, antiderivative size = 29 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 (-2+x) \left (1-2 e^2+2 x\right ) (5+x \log (3))} \]
Leaf count is larger than twice the leaf count of optimal. \(151\) vs. \(2(29)=58\).
Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 5.21 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5} \left (\frac {25-e^2 (10+\log (81))+\log (59049)}{\left (5-2 e^2\right )^2 (-2+x) (5+\log (9))^2}-\frac {4 \left (-50+e^4 \log (81)+\log (243)-e^2 (-20+\log (531441))\right )}{\left (5-2 e^2\right )^2 \left (-1+2 e^2-2 x\right ) \left (10-\log (3)+e^2 \log (9)\right )^2}+\frac {\log ^2(3) \left (50-\log (3) \log (9)+e^2 \left (4 \log ^2(3)+\log (59049)\right )+\log (14348907)\right )}{(5+x \log (3)) (5+\log (9))^2 \left (10-\log (3)+e^2 \log (9)\right )^2}\right ) \]
Integrate[(15 + 10*E^2 - 20*x + (2 + 6*x - 6*x^2 + E^2*(-4 + 4*x))*Log[3]) /(500 + 1500*x + 125*x^2 - 1500*x^3 + 500*x^4 + E^4*(2000 - 2000*x + 500*x ^2) + E^2*(-2000 - 2000*x + 3500*x^2 - 1000*x^3) + (200*x + 600*x^2 + 50*x ^3 - 600*x^4 + 200*x^5 + E^4*(800*x - 800*x^2 + 200*x^3) + E^2*(-800*x - 8 00*x^2 + 1400*x^3 - 400*x^4))*Log[3] + (20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^6 + E^4*(80*x^2 - 80*x^3 + 20*x^4) + E^2*(-80*x^2 - 80*x^3 + 140*x^4 - 40*x^5))*Log[3]^2),x]
((25 - E^2*(10 + Log[81]) + Log[59049])/((5 - 2*E^2)^2*(-2 + x)*(5 + Log[9 ])^2) - (4*(-50 + E^4*Log[81] + Log[243] - E^2*(-20 + Log[531441])))/((5 - 2*E^2)^2*(-1 + 2*E^2 - 2*x)*(10 - Log[3] + E^2*Log[9])^2) + (Log[3]^2*(50 - Log[3]*Log[9] + E^2*(4*Log[3]^2 + Log[59049]) + Log[14348907]))/((5 + x *Log[3])*(5 + Log[9])^2*(10 - Log[3] + E^2*Log[9])^2))/5
Leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(29)=58\).
Time = 0.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.52, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2462, 2009}
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.
| \(\displaystyle \int \frac {\left (-6 x^2+6 x+e^2 (4 x-4)+2\right ) \log (3)-20 x+10 e^2+15}{500 x^4-1500 x^3+125 x^2+e^4 \left (500 x^2-2000 x+2000\right )+e^2 \left (-1000 x^3+3500 x^2-2000 x-2000\right )+\left (200 x^5-600 x^4+50 x^3+600 x^2+e^4 \left (200 x^3-800 x^2+800 x\right )+e^2 \left (-400 x^4+1400 x^3-800 x^2-800 x\right )+200 x\right ) \log (3)+\left (20 x^6-60 x^5+5 x^4+60 x^3+20 x^2+e^4 \left (20 x^4-80 x^3+80 x^2\right )+e^2 \left (-40 x^5+140 x^4-80 x^3-80 x^2\right )\right ) \log ^2(3)+1500 x+500} \, dx\) |
| \(\Big \downarrow \) 2462 |
| \(\displaystyle \int \left (\frac {\log ^3(3)}{5 (5+\log (9)) \left (-10+\log (3)-e^2 \log (9)\right ) (x \log (3)+5)^2}+\frac {1}{5 \left (2 e^2-5\right ) (x-2)^2 (5+\log (9))}-\frac {8}{5 \left (2 e^2-5\right ) \left (-2 x+2 e^2-1\right )^2 \left (10-\log (3)+e^2 \log (9)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log ^2(3)}{5 (5+\log (9)) \left (10-\log (3)+e^2 \log (9)\right ) (x \log (3)+5)}-\frac {4}{5 \left (5-2 e^2\right ) \left (2 x-2 e^2+1\right ) \left (10-\log (3)+e^2 \log (9)\right )}-\frac {1}{5 \left (5-2 e^2\right ) (2-x) (5+\log (9))}\) |
Int[(15 + 10*E^2 - 20*x + (2 + 6*x - 6*x^2 + E^2*(-4 + 4*x))*Log[3])/(500 + 1500*x + 125*x^2 - 1500*x^3 + 500*x^4 + E^4*(2000 - 2000*x + 500*x^2) + E^2*(-2000 - 2000*x + 3500*x^2 - 1000*x^3) + (200*x + 600*x^2 + 50*x^3 - 6 00*x^4 + 200*x^5 + E^4*(800*x - 800*x^2 + 200*x^3) + E^2*(-800*x - 800*x^2 + 1400*x^3 - 400*x^4))*Log[3] + (20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^ 6 + E^4*(80*x^2 - 80*x^3 + 20*x^4) + E^2*(-80*x^2 - 80*x^3 + 140*x^4 - 40* x^5))*Log[3]^2),x]
-1/5*1/((5 - 2*E^2)*(2 - x)*(5 + Log[9])) - 4/(5*(5 - 2*E^2)*(1 - 2*E^2 + 2*x)*(10 - Log[3] + E^2*Log[9])) + Log[3]^2/(5*(5 + x*Log[3])*(5 + Log[9]) *(10 - Log[3] + E^2*Log[9]))
3.21.47.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
| method | result | size |
| norman | \(-\frac {1}{5 \left (-2+x \right ) \left (x \ln \left (3\right )+5\right ) \left (2 \,{\mathrm e}^{2}-2 x -1\right )}\) | \(27\) |
| risch | \(-\frac {1}{10 \left (x^{2} {\mathrm e}^{2} \ln \left (3\right )-x^{3} \ln \left (3\right )-2 x \,{\mathrm e}^{2} \ln \left (3\right )+\frac {3 x^{2} \ln \left (3\right )}{2}+5 \,{\mathrm e}^{2} x +x \ln \left (3\right )-5 x^{2}-10 \,{\mathrm e}^{2}+\frac {15 x}{2}+5\right )}\) | \(57\) |
| gosper | \(-\frac {1}{5 \left (2 x^{2} {\mathrm e}^{2} \ln \left (3\right )-2 x^{3} \ln \left (3\right )-4 x \,{\mathrm e}^{2} \ln \left (3\right )+3 x^{2} \ln \left (3\right )+10 \,{\mathrm e}^{2} x +2 x \ln \left (3\right )-10 x^{2}-20 \,{\mathrm e}^{2}+15 x +10\right )}\) | \(59\) |
| parallelrisch | \(-\frac {1}{5 \left (2 x^{2} {\mathrm e}^{2} \ln \left (3\right )-2 x^{3} \ln \left (3\right )-4 x \,{\mathrm e}^{2} \ln \left (3\right )+3 x^{2} \ln \left (3\right )+10 \,{\mathrm e}^{2} x +2 x \ln \left (3\right )-10 x^{2}-20 \,{\mathrm e}^{2}+15 x +10\right )}\) | \(59\) |
int((((-4+4*x)*exp(2)-6*x^2+6*x+2)*ln(3)+10*exp(2)-20*x+15)/(((20*x^4-80*x ^3+80*x^2)*exp(2)^2+(-40*x^5+140*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-60*x^5+5 *x^4+60*x^3+20*x^2)*ln(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-400*x^4+14 00*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x)*ln(3)+( 500*x^2-2000*x+2000)*exp(2)^2+(-1000*x^3+3500*x^2-2000*x-2000)*exp(2)+500* x^4-1500*x^3+125*x^2+1500*x+500),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (10 \, x^{2} - 10 \, {\left (x - 2\right )} e^{2} + {\left (2 \, x^{3} - 3 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, x\right )} \log \left (3\right ) - 15 \, x - 10\right )}} \]
integrate((((-4+4*x)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x ^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+140*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-6 0*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-40 0*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x) *log(3)+(500*x^2-2000*x+2000)*exp(2)^2+(-1000*x^3+3500*x^2-2000*x-2000)*ex p(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm=\
1/5/(10*x^2 - 10*(x - 2)*e^2 + (2*x^3 - 3*x^2 - 2*(x^2 - 2*x)*e^2 - 2*x)*l og(3) - 15*x - 10)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 5.99 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{10 x^{3} \log {\left (3 \right )} + x^{2} \left (- 10 e^{2} \log {\left (3 \right )} - 15 \log {\left (3 \right )} + 50\right ) + x \left (- 50 e^{2} - 75 - 10 \log {\left (3 \right )} + 20 e^{2} \log {\left (3 \right )}\right ) - 50 + 100 e^{2}} \]
integrate((((-4+4*x)*exp(2)-6*x**2+6*x+2)*ln(3)+10*exp(2)-20*x+15)/(((20*x **4-80*x**3+80*x**2)*exp(2)**2+(-40*x**5+140*x**4-80*x**3-80*x**2)*exp(2)+ 20*x**6-60*x**5+5*x**4+60*x**3+20*x**2)*ln(3)**2+((200*x**3-800*x**2+800*x )*exp(2)**2+(-400*x**4+1400*x**3-800*x**2-800*x)*exp(2)+200*x**5-600*x**4+ 50*x**3+600*x**2+200*x)*ln(3)+(500*x**2-2000*x+2000)*exp(2)**2+(-1000*x**3 +3500*x**2-2000*x-2000)*exp(2)+500*x**4-1500*x**3+125*x**2+1500*x+500),x)
1/(10*x**3*log(3) + x**2*(-10*exp(2)*log(3) - 15*log(3) + 50) + x*(-50*exp (2) - 75 - 10*log(3) + 20*exp(2)*log(3)) - 50 + 100*exp(2))
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (2 \, x^{3} \log \left (3\right ) - {\left ({\left (2 \, e^{2} + 3\right )} \log \left (3\right ) - 10\right )} x^{2} + {\left (2 \, {\left (2 \, e^{2} - 1\right )} \log \left (3\right ) - 10 \, e^{2} - 15\right )} x + 20 \, e^{2} - 10\right )}} \]
integrate((((-4+4*x)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x ^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+140*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-6 0*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-40 0*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x) *log(3)+(500*x^2-2000*x+2000)*exp(2)^2+(-1000*x^3+3500*x^2-2000*x-2000)*ex p(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm=\
1/5/(2*x^3*log(3) - ((2*e^2 + 3)*log(3) - 10)*x^2 + (2*(2*e^2 - 1)*log(3) - 10*e^2 - 15)*x + 20*e^2 - 10)
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (10 \, x^{2} - 10 \, x e^{2} + {\left (2 \, x^{3} - 3 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, x\right )} \log \left (3\right ) - 15 \, x + 20 \, e^{2} - 10\right )}} \]
integrate((((-4+4*x)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x ^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+140*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-6 0*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-40 0*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x) *log(3)+(500*x^2-2000*x+2000)*exp(2)^2+(-1000*x^3+3500*x^2-2000*x-2000)*ex p(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm=\
1/5/(10*x^2 - 10*x*e^2 + (2*x^3 - 3*x^2 - 2*(x^2 - 2*x)*e^2 - 2*x)*log(3) - 15*x + 20*e^2 - 10)
Time = 12.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.28 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {4}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \left (3\right )-\ln \left (3\right )+10\right )\,\left (2\,x-2\,{\mathrm {e}}^2+1\right )}-\frac {1}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,\ln \left (3\right )+5\right )\,\left (x-2\right )}+\frac {{\ln \left (3\right )}^2}{5\,\left (2\,\ln \left (3\right )+5\right )\,\left (x\,\ln \left (3\right )+5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \left (3\right )-\ln \left (3\right )+10\right )} \]
int((10*exp(2) - 20*x + log(3)*(6*x - 6*x^2 + exp(2)*(4*x - 4) + 2) + 15)/ (1500*x + exp(4)*(500*x^2 - 2000*x + 2000) + log(3)^2*(exp(4)*(80*x^2 - 80 *x^3 + 20*x^4) + 20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^6 - exp(2)*(80*x^ 2 + 80*x^3 - 140*x^4 + 40*x^5)) - exp(2)*(2000*x - 3500*x^2 + 1000*x^3 + 2 000) + 125*x^2 - 1500*x^3 + 500*x^4 + log(3)*(200*x + exp(4)*(800*x - 800* x^2 + 200*x^3) - exp(2)*(800*x + 800*x^2 - 1400*x^3 + 400*x^4) + 600*x^2 + 50*x^3 - 600*x^4 + 200*x^5) + 500),x)