Integrand size = 138, antiderivative size = 26 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=\left (-1-x+\frac {e^{e^{2 x}}}{5+x}-\log ^4(3)\right )^2 \] Output:
(-1-x-ln(3)^4+exp(exp(x)^2)/(5+x))^2
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=\frac {\left (-e^{e^{2 x}}+x (5+x)\right ) \left (-e^{e^{2 x}}+(5+x) \left (2+x+2 \log ^4(3)\right )\right )}{(5+x)^2} \] Input:
Integrate[(250 + 400*x + 180*x^2 + 32*x^3 + 2*x^4 + E^(2*E^(2*x))*(-2 + E^ (2*x)*(20 + 4*x)) + (250 + 150*x + 30*x^2 + 2*x^3)*Log[3]^4 + E^E^(2*x)*(- 40 - 8*x + (10 + 2*x)*Log[3]^4 + E^(2*x)*(-100 - 140*x - 44*x^2 - 4*x^3 + (-100 - 40*x - 4*x^2)*Log[3]^4)))/(125 + 75*x + 15*x^2 + x^3),x]
Output:
((-E^E^(2*x) + x*(5 + x))*(-E^E^(2*x) + (5 + x)*(2 + x + 2*Log[3]^4)))/(5 + x)^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.
| \(\displaystyle \int \frac {2 x^4+32 x^3+180 x^2+e^{e^{2 x}} \left (e^{2 x} \left (-4 x^3-44 x^2+\left (-4 x^2-40 x-100\right ) \log ^4(3)-140 x-100\right )-8 x+(2 x+10) \log ^4(3)-40\right )+\left (2 x^3+30 x^2+150 x+250\right ) \log ^4(3)+400 x+e^{2 e^{2 x}} \left (e^{2 x} (4 x+20)-2\right )+250}{x^3+15 x^2+75 x+125} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {2 x^4+32 x^3+180 x^2+e^{e^{2 x}} \left (e^{2 x} \left (-4 x^3-44 x^2+\left (-4 x^2-40 x-100\right ) \log ^4(3)-140 x-100\right )-8 x+(2 x+10) \log ^4(3)-40\right )+\left (2 x^3+30 x^2+150 x+250\right ) \log ^4(3)+400 x+e^{2 e^{2 x}} \left (e^{2 x} (4 x+20)-2\right )+250}{(x+5)^3}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left ((x+5)^2-2 e^{2 x+e^{2 x}} (x+5)+e^{e^{2 x}}\right ) \left ((x+5) \left (x+1+\log ^4(3)\right )-e^{e^{2 x}}\right )}{(x+5)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int -\frac {\left ((x+5)^2-2 e^{2 x+e^{2 x}} (x+5)+e^{e^{2 x}}\right ) \left (e^{e^{2 x}}-(x+5) \left (x+\log ^4(3)+1\right )\right )}{(x+5)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {\left ((x+5)^2-2 e^{2 x+e^{2 x}} (x+5)+e^{e^{2 x}}\right ) \left (e^{e^{2 x}}-(x+5) \left (x+\log ^4(3)+1\right )\right )}{(x+5)^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 \int \left (\frac {\left (x^2+10 x+e^{e^{2 x}}+25\right ) \left (-x^2-6 \left (1+\frac {\log ^4(3)}{6}\right ) x+e^{e^{2 x}}-5 \left (1+\log ^4(3)\right )\right )}{(x+5)^3}+\frac {2 e^{2 x+e^{2 x}} \left (x^2+6 \left (1+\frac {\log ^4(3)}{6}\right ) x-e^{e^{2 x}}+5 \left (1+\log ^4(3)\right )\right )}{(x+5)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \left (\int \frac {e^{2 e^{2 x}}}{(x+5)^3}dx-2 \int \frac {e^{2 \left (x+e^{2 x}\right )}}{(x+5)^2}dx+\left (4-\log ^4(3)\right ) \int \frac {e^{e^{2 x}}}{(x+5)^2}dx-2 \left (4-\log ^4(3)\right ) \int \frac {e^{2 x+e^{2 x}}}{x+5}dx-\frac {x^2}{2}-\frac {5 x^2 \left (8+3 \log ^4(3)\right )}{2 (x+5)^2}+15 x+e^{e^{2 x}}-\frac {500}{x+5}+\frac {625}{2 (x+5)^2}-x \left (16+\log ^4(3)\right )+15 \left (16+\log ^4(3)\right ) \log (x+5)-15 \left (6+\log ^4(3)\right ) \log (x+5)+\frac {75 \left (16+\log ^4(3)\right )}{x+5}-\frac {125 \left (16+\log ^4(3)\right )}{2 (x+5)^2}-\frac {150 \left (6+\log ^4(3)\right )}{x+5}+\frac {375 \left (6+\log ^4(3)\right )}{2 (x+5)^2}+\frac {125 \left (1+\log ^4(3)\right )}{2 (x+5)^2}-150 \log (x+5)\right )\) |
Input:
Int[(250 + 400*x + 180*x^2 + 32*x^3 + 2*x^4 + E^(2*E^(2*x))*(-2 + E^(2*x)* (20 + 4*x)) + (250 + 150*x + 30*x^2 + 2*x^3)*Log[3]^4 + E^E^(2*x)*(-40 - 8 *x + (10 + 2*x)*Log[3]^4 + E^(2*x)*(-100 - 140*x - 44*x^2 - 4*x^3 + (-100 - 40*x - 4*x^2)*Log[3]^4)))/(125 + 75*x + 15*x^2 + x^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(24)=48\).
Time = 0.52 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00
| method | result | size |
| risch | \(2 x \ln \left (3\right )^{4}+x^{2}+2 x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{2 x}}}{x^{2}+10 x +25}-\frac {2 \left (\ln \left (3\right )^{4}+x +1\right ) {\mathrm e}^{{\mathrm e}^{2 x}}}{5+x}\) | \(52\) |
| parallelrisch | \(\frac {2 \ln \left (3\right )^{4} x^{3}-2 \ln \left (3\right )^{4} {\mathrm e}^{{\mathrm e}^{2 x}} x -150 x \ln \left (3\right )^{4}-10 \ln \left (3\right )^{4} {\mathrm e}^{{\mathrm e}^{2 x}}-500 \ln \left (3\right )^{4}+x^{4}+12 x^{3}-2 \,{\mathrm e}^{{\mathrm e}^{2 x}} x^{2}-1125-12 x \,{\mathrm e}^{{\mathrm e}^{2 x}}+{\mathrm e}^{2 \,{\mathrm e}^{2 x}}-400 x -10 \,{\mathrm e}^{{\mathrm e}^{2 x}}}{x^{2}+10 x +25}\) | \(102\) |
Input:
int((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*ln(3)^4-4*x ^3-44*x^2-140*x-100)*exp(x)^2+(2*x+10)*ln(3)^4-8*x-40)*exp(exp(x)^2)+(2*x^ 3+30*x^2+150*x+250)*ln(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+15*x^2+75 *x+125),x,method=_RETURNVERBOSE)
Output:
2*x*ln(3)^4+x^2+2*x+1/(x^2+10*x+25)*exp(2*exp(2*x))-2*(ln(3)^4+x+1)/(5+x)* exp(exp(2*x))
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (21) = 42\).
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.92 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=\frac {2 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} \log \left (3\right )^{4} + x^{4} + 12 \, x^{3} + 45 \, x^{2} - 2 \, {\left ({\left (x + 5\right )} \log \left (3\right )^{4} + x^{2} + 6 \, x + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \] Input:
integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3 )^4-4*x^3-44*x^2-140*x-100)*exp(x)^2+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^ 2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+ 15*x^2+75*x+125),x, algorithm="fricas")
Output:
(2*(x^3 + 10*x^2 + 25*x)*log(3)^4 + x^4 + 12*x^3 + 45*x^2 - 2*((x + 5)*log (3)^4 + x^2 + 6*x + 5)*e^(e^(2*x)) + 50*x + e^(2*e^(2*x)))/(x^2 + 10*x + 2 5)
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (19) = 38\).
Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.35 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=x^{2} + x \left (2 + 2 \log {\left (3 \right )}^{4}\right ) + \frac {\left (x + 5\right ) e^{2 e^{2 x}} + \left (- 2 x^{3} - 22 x^{2} - 2 x^{2} \log {\left (3 \right )}^{4} - 70 x - 20 x \log {\left (3 \right )}^{4} - 50 \log {\left (3 \right )}^{4} - 50\right ) e^{e^{2 x}}}{x^{3} + 15 x^{2} + 75 x + 125} \] Input:
integrate((((20+4*x)*exp(x)**2-2)*exp(exp(x)**2)**2+(((-4*x**2-40*x-100)*l n(3)**4-4*x**3-44*x**2-140*x-100)*exp(x)**2+(2*x+10)*ln(3)**4-8*x-40)*exp( exp(x)**2)+(2*x**3+30*x**2+150*x+250)*ln(3)**4+2*x**4+32*x**3+180*x**2+400 *x+250)/(x**3+15*x**2+75*x+125),x)
Output:
x**2 + x*(2 + 2*log(3)**4) + ((x + 5)*exp(2*exp(2*x)) + (-2*x**3 - 22*x**2 - 2*x**2*log(3)**4 - 70*x - 20*x*log(3)**4 - 50*log(3)**4 - 50)*exp(exp(2 *x)))/(x**3 + 15*x**2 + 75*x + 125)
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (21) = 42\).
Time = 0.16 (sec) , antiderivative size = 234, normalized size of antiderivative = 9.00 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx={\left (2 \, x - \frac {25 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} - 30 \, \log \left (x + 5\right )\right )} \log \left (3\right )^{4} + 15 \, {\left (\frac {5 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} + 2 \, \log \left (x + 5\right )\right )} \log \left (3\right )^{4} - \frac {75 \, {\left (2 \, x + 5\right )} \log \left (3\right )^{4}}{x^{2} + 10 \, x + 25} - \frac {125 \, \log \left (3\right )^{4}}{x^{2} + 10 \, x + 25} + x^{2} + 2 \, x - \frac {2 \, {\left (5 \, \log \left (3\right )^{4} + {\left (\log \left (3\right )^{4} + 6\right )} x + x^{2} + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} + \frac {125 \, {\left (8 \, x + 35\right )}}{x^{2} + 10 \, x + 25} - \frac {400 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} + \frac {450 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} - \frac {200 \, {\left (2 \, x + 5\right )}}{x^{2} + 10 \, x + 25} - \frac {125}{x^{2} + 10 \, x + 25} \] Input:
integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3 )^4-4*x^3-44*x^2-140*x-100)*exp(x)^2+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^ 2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+ 15*x^2+75*x+125),x, algorithm="maxima")
Output:
(2*x - 25*(6*x + 25)/(x^2 + 10*x + 25) - 30*log(x + 5))*log(3)^4 + 15*(5*( 4*x + 15)/(x^2 + 10*x + 25) + 2*log(x + 5))*log(3)^4 - 75*(2*x + 5)*log(3) ^4/(x^2 + 10*x + 25) - 125*log(3)^4/(x^2 + 10*x + 25) + x^2 + 2*x - (2*(5* log(3)^4 + (log(3)^4 + 6)*x + x^2 + 5)*e^(e^(2*x)) - e^(2*e^(2*x)))/(x^2 + 10*x + 25) + 125*(8*x + 35)/(x^2 + 10*x + 25) - 400*(6*x + 25)/(x^2 + 10* x + 25) + 450*(4*x + 15)/(x^2 + 10*x + 25) - 200*(2*x + 5)/(x^2 + 10*x + 2 5) - 125/(x^2 + 10*x + 25)
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (21) = 42\).
Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.15 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=\frac {2 \, x^{3} \log \left (3\right )^{4} + 20 \, x^{2} \log \left (3\right )^{4} - 2 \, x e^{\left (e^{\left (2 \, x\right )}\right )} \log \left (3\right )^{4} + 50 \, x \log \left (3\right )^{4} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )} \log \left (3\right )^{4} + x^{4} + 12 \, x^{3} - 2 \, x^{2} e^{\left (e^{\left (2 \, x\right )}\right )} + 45 \, x^{2} - 12 \, x e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \] Input:
integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3 )^4-4*x^3-44*x^2-140*x-100)*exp(x)^2+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^ 2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+ 15*x^2+75*x+125),x, algorithm="giac")
Output:
(2*x^3*log(3)^4 + 20*x^2*log(3)^4 - 2*x*e^(e^(2*x))*log(3)^4 + 50*x*log(3) ^4 - 10*e^(e^(2*x))*log(3)^4 + x^4 + 12*x^3 - 2*x^2*e^(e^(2*x)) + 45*x^2 - 12*x*e^(e^(2*x)) + 50*x + e^(2*e^(2*x)) - 10*e^(e^(2*x)))/(x^2 + 10*x + 2 5)
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=x\,\left (2\,{\ln \left (3\right )}^4+2\right )+\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x^2+10\,x+25}+x^2-\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\left (2\,x+2\,{\ln \left (3\right )}^4+2\right )}{x+5} \] Input:
int((400*x - exp(exp(2*x))*(8*x - log(3)^4*(2*x + 10) + exp(2*x)*(140*x + log(3)^4*(40*x + 4*x^2 + 100) + 44*x^2 + 4*x^3 + 100) + 40) + exp(2*exp(2* x))*(exp(2*x)*(4*x + 20) - 2) + log(3)^4*(150*x + 30*x^2 + 2*x^3 + 250) + 180*x^2 + 32*x^3 + 2*x^4 + 250)/(75*x + 15*x^2 + x^3 + 125),x)
Output:
x*(2*log(3)^4 + 2) + exp(2*exp(2*x))/(10*x + x^2 + 25) + x^2 - (exp(exp(2* x))*(2*x + 2*log(3)^4 + 2))/(x + 5)
Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.62 \[ \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx=\frac {e^{2 e^{2 x}}-2 e^{e^{2 x}} \mathrm {log}\left (3\right )^{4} x -10 e^{e^{2 x}} \mathrm {log}\left (3\right )^{4}-2 e^{e^{2 x}} x^{2}-12 e^{e^{2 x}} x -10 e^{e^{2 x}}+2 \mathrm {log}\left (3\right )^{4} x^{3}+20 \mathrm {log}\left (3\right )^{4} x^{2}+50 \mathrm {log}\left (3\right )^{4} x +x^{4}+12 x^{3}+45 x^{2}+50 x}{x^{2}+10 x +25} \] Input:
int((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3)^4-4* x^3-44*x^2-140*x-100)*exp(x)^2+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^2)+(2* x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+15*x^2 +75*x+125),x)
Output:
(e**(2*e**(2*x)) - 2*e**(e**(2*x))*log(3)**4*x - 10*e**(e**(2*x))*log(3)** 4 - 2*e**(e**(2*x))*x**2 - 12*e**(e**(2*x))*x - 10*e**(e**(2*x)) + 2*log(3 )**4*x**3 + 20*log(3)**4*x**2 + 50*log(3)**4*x + x**4 + 12*x**3 + 45*x**2 + 50*x)/(x**2 + 10*x + 25)