Integrand size = 232, antiderivative size = 29 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 (-2+x) \left (1-2 e^2+2 x\right ) (5+x \log (3))} \]
[Out]
Leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(29)=58\).
Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.52, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.004, Rules used = {2099} \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {\log ^2(3)}{5 (5+\log (9)) \left (10-\log (3)+e^2 \log (9)\right ) (x \log (3)+5)}-\frac {4}{5 \left (5-2 e^2\right ) \left (2 x-2 e^2+1\right ) \left (10-\log (3)+e^2 \log (9)\right )}-\frac {1}{5 \left (5-2 e^2\right ) (2-x) (5+\log (9))} \]
[In]
[Out]
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{5 \left (-5+2 e^2\right ) (-2+x)^2 (5+\log (9))}+\frac {\log ^3(3)}{5 (5+x \log (3))^2 (5+\log (9)) \left (-10+\log (3)-e^2 \log (9)\right )}-\frac {8}{5 \left (-5+2 e^2\right ) \left (-1+2 e^2-2 x\right )^2 \left (10-\log (3)+e^2 \log (9)\right )}\right ) \, dx \\ & = -\frac {1}{5 \left (5-2 e^2\right ) (2-x) (5+\log (9))}-\frac {4}{5 \left (5-2 e^2\right ) \left (1-2 e^2+2 x\right ) \left (10-\log (3)+e^2 \log (9)\right )}+\frac {\log ^2(3)}{5 (5+x \log (3)) (5+\log (9)) \left (10-\log (3)+e^2 \log (9)\right )} \\ \end{align*}
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 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5} \left (\frac {25-e^2 (10+\log (81))+\log (59049)}{\left (5-2 e^2\right )^2 (-2+x) (5+\log (9))^2}-\frac {4 \left (-50+e^4 \log (81)+\log (243)-e^2 (-20+\log (531441))\right )}{\left (5-2 e^2\right )^2 \left (-1+2 e^2-2 x\right ) \left (10-\log (3)+e^2 \log (9)\right )^2}+\frac {\log ^2(3) \left (50-\log (3) \log (9)+e^2 \left (4 \log ^2(3)+\log (59049)\right )+\log (14348907)\right )}{(5+x \log (3)) (5+\log (9))^2 \left (10-\log (3)+e^2 \log (9)\right )^2}\right ) \]
[In]
[Out]
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 | \(-\frac {1}{5 \left (2 x^{2} {\mathrm e}^{2} \ln \left (3\right )-2 x^{3} \ln \left (3\right )-4 x \,{\mathrm e}^{2} \ln \left (3\right )+3 x^{2} \ln \left (3\right )+10 \,{\mathrm e}^{2} x +2 x \ln \left (3\right )-10 x^{2}-20 \,{\mathrm e}^{2}+15 x +10\right )}\) | \(59\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (10 \, x^{2} - 10 \, {\left (x - 2\right )} e^{2} + {\left (2 \, x^{3} - 3 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, x\right )} \log \left (3\right ) - 15 \, x - 10\right )}} \]
[In]
[Out]
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 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{10 x^{3} \log {\left (3 \right )} + x^{2} \left (- 10 e^{2} \log {\left (3 \right )} - 15 \log {\left (3 \right )} + 50\right ) + x \left (- 50 e^{2} - 75 - 10 \log {\left (3 \right )} + 20 e^{2} \log {\left (3 \right )}\right ) - 50 + 100 e^{2}} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (2 \, x^{3} \log \left (3\right ) - {\left ({\left (2 \, e^{2} + 3\right )} \log \left (3\right ) - 10\right )} x^{2} + {\left (2 \, {\left (2 \, e^{2} - 1\right )} \log \left (3\right ) - 10 \, e^{2} - 15\right )} x + 20 \, e^{2} - 10\right )}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {1}{5 \, {\left (10 \, x^{2} - 10 \, x e^{2} + {\left (2 \, x^{3} - 3 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, x\right )} \log \left (3\right ) - 15 \, x + 20 \, e^{2} - 10\right )}} \]
[In]
[Out]
Time = 12.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.28 \[ \int \frac {15+10 e^2-20 x+\left (2+6 x-6 x^2+e^2 (-4+4 x)\right ) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 \left (2000-2000 x+500 x^2\right )+e^2 \left (-2000-2000 x+3500 x^2-1000 x^3\right )+\left (200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 \left (800 x-800 x^2+200 x^3\right )+e^2 \left (-800 x-800 x^2+1400 x^3-400 x^4\right )\right ) \log (3)+\left (20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 \left (80 x^2-80 x^3+20 x^4\right )+e^2 \left (-80 x^2-80 x^3+140 x^4-40 x^5\right )\right ) \log ^2(3)} \, dx=\frac {4}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \left (3\right )-\ln \left (3\right )+10\right )\,\left (2\,x-2\,{\mathrm {e}}^2+1\right )}-\frac {1}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,\ln \left (3\right )+5\right )\,\left (x-2\right )}+\frac {{\ln \left (3\right )}^2}{5\,\left (2\,\ln \left (3\right )+5\right )\,\left (x\,\ln \left (3\right )+5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \left (3\right )-\ln \left (3\right )+10\right )} \]
[In]
[Out]