Integrand size = 221, antiderivative size = 35 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=\frac {2+x+\frac {x}{3+x+e^3 (5+x)}}{1+e^3+\frac {x}{5}}-\log (x) \]
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=-\frac {5 \left (9+2 x+5 e^6 (5+x)+e^3 (30+8 x)\right )}{15+8 x+x^2+5 e^6 (5+x)+e^3 \left (40+15 x+x^2\right )}-\log (x) \]
Integrate[(-225 - 30*x - 4*x^2 - 6*x^3 - x^4 + E^12*(-625 - 250*x - 25*x^2 ) + E^9*(-2000 - 525*x + 50*x^2 + 15*x^3) + E^3*(-1200 - 215*x + 40*x^2 + 4*x^3 - 2*x^4) + E^6*(-2350 - 500*x + 115*x^2 + 25*x^3 - x^4))/(225*x + 24 0*x^2 + 94*x^3 + 16*x^4 + x^5 + E^12*(625*x + 250*x^2 + 25*x^3) + E^9*(200 0*x + 1150*x^2 + 200*x^3 + 10*x^4) + E^6*(2350*x + 1750*x^2 + 435*x^3 + 40 *x^4 + x^5) + E^3*(1200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5)),x]
(-5*(9 + 2*x + 5*E^6*(5 + x) + E^3*(30 + 8*x)))/(15 + 8*x + x^2 + 5*E^6*(5 + x) + E^3*(40 + 15*x + x^2)) - Log[x]
Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(35)=70\).
Time = 0.70 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2026, 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.
\(\displaystyle \int \frac {-x^4-6 x^3-4 x^2+e^{12} \left (-25 x^2-250 x-625\right )+e^9 \left (15 x^3+50 x^2-525 x-2000\right )+e^3 \left (-2 x^4+4 x^3+40 x^2-215 x-1200\right )+e^6 \left (-x^4+25 x^3+115 x^2-500 x-2350\right )-30 x-225}{x^5+16 x^4+94 x^3+240 x^2+e^{12} \left (25 x^3+250 x^2+625 x\right )+e^9 \left (10 x^4+200 x^3+1150 x^2+2000 x\right )+e^6 \left (x^5+40 x^4+435 x^3+1750 x^2+2350 x\right )+e^3 \left (2 x^5+46 x^4+350 x^3+1090 x^2+1200 x\right )+225 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^4-6 x^3-4 x^2+e^{12} \left (-25 x^2-250 x-625\right )+e^9 \left (15 x^3+50 x^2-525 x-2000\right )+e^3 \left (-2 x^4+4 x^3+40 x^2-215 x-1200\right )+e^6 \left (-x^4+25 x^3+115 x^2-500 x-2350\right )-30 x-225}{x \left (\left (1+e^3\right )^2 x^4+2 \left (8+23 e^3+20 e^6+5 e^9\right ) x^3+\left (94+350 e^3+435 e^6+200 e^9+25 e^{12}\right ) x^2+10 \left (24+109 e^3+175 e^6+115 e^9+25 e^{12}\right ) x+25 \left (3+8 e^3+5 e^6\right )^2\right )}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {5 \left (1+20 e^3+40 e^6+25 e^9\right )}{\left (2+5 e^3+5 e^6\right ) \left (x+5 e^3+5\right )^2}+\frac {5 \left (3+8 e^3+5 e^6\right )}{\left (2+5 e^3+5 e^6\right ) \left (\left (1+e^3\right ) x+5 e^3+3\right )^2}-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5 \left (1+20 e^3+40 e^6+25 e^9\right )}{\left (2+5 e^3+5 e^6\right ) \left (x+5 \left (1+e^3\right )\right )}-\frac {5 \left (3+5 e^3\right )}{\left (2+5 e^3+5 e^6\right ) \left (\left (1+e^3\right ) x+5 e^3+3\right )}-\log (x)\) |
Int[(-225 - 30*x - 4*x^2 - 6*x^3 - x^4 + E^12*(-625 - 250*x - 25*x^2) + E^ 9*(-2000 - 525*x + 50*x^2 + 15*x^3) + E^3*(-1200 - 215*x + 40*x^2 + 4*x^3 - 2*x^4) + E^6*(-2350 - 500*x + 115*x^2 + 25*x^3 - x^4))/(225*x + 240*x^2 + 94*x^3 + 16*x^4 + x^5 + E^12*(625*x + 250*x^2 + 25*x^3) + E^9*(2000*x + 1150*x^2 + 200*x^3 + 10*x^4) + E^6*(2350*x + 1750*x^2 + 435*x^3 + 40*x^4 + x^5) + E^3*(1200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5)),x]
(-5*(1 + 20*E^3 + 40*E^6 + 25*E^9))/((2 + 5*E^3 + 5*E^6)*(5*(1 + E^3) + x) ) - (5*(3 + 5*E^3))/((2 + 5*E^3 + 5*E^6)*(3 + 5*E^3 + (1 + E^3)*x)) - Log[ x]
3.20.98.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(31)=62\).
Time = 0.45 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.86
method | result | size |
risch | \(\frac {\left (-5 \,{\mathrm e}^{6}-8 \,{\mathrm e}^{3}-2\right ) x -25 \,{\mathrm e}^{6}-30 \,{\mathrm e}^{3}-9}{x \,{\mathrm e}^{6}+\frac {x^{2} {\mathrm e}^{3}}{5}+5 \,{\mathrm e}^{6}+3 x \,{\mathrm e}^{3}+\frac {x^{2}}{5}+8 \,{\mathrm e}^{3}+\frac {8 x}{5}+3}-\ln \left (x \right )\) | \(65\) |
norman | \(\frac {-\frac {\left (50 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{9}+65 \,{\mathrm e}^{6}+10\right ) x}{{\mathrm e}^{3}+1}-125 \,{\mathrm e}^{6}-45-150 \,{\mathrm e}^{3}}{x^{2} {\mathrm e}^{3}+15 x \,{\mathrm e}^{3}+5 x \,{\mathrm e}^{6}+x^{2}+40 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{6}+8 x +15}-\ln \left (x \right )\) | \(84\) |
parallelrisch | \(-\frac {45+{\mathrm e}^{6} \ln \left (x \right ) x^{2}+20 x \,{\mathrm e}^{6} \ln \left (x \right )+5 x \,{\mathrm e}^{9} \ln \left (x \right )+10 x +65 \,{\mathrm e}^{6} \ln \left (x \right )+65 x \,{\mathrm e}^{6}+25 x \,{\mathrm e}^{9}+50 x \,{\mathrm e}^{3}+8 x \ln \left (x \right )+55 \ln \left (x \right ) {\mathrm e}^{3}+23 x \,{\mathrm e}^{3} \ln \left (x \right )+125 \,{\mathrm e}^{9}+275 \,{\mathrm e}^{6}+15 \ln \left (x \right )+195 \,{\mathrm e}^{3}+x^{2} \ln \left (x \right )+2 \,{\mathrm e}^{3} \ln \left (x \right ) x^{2}+25 \,{\mathrm e}^{9} \ln \left (x \right )}{\left ({\mathrm e}^{3}+1\right ) \left (x^{2} {\mathrm e}^{3}+15 x \,{\mathrm e}^{3}+5 x \,{\mathrm e}^{6}+x^{2}+40 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{6}+8 x +15\right )}\) | \(168\) |
default | \(-\ln \left (x \right )+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) \textit {\_Z}^{4}+\left (46 \,{\mathrm e}^{3}+10 \,{\mathrm e}^{9}+40 \,{\mathrm e}^{6}+16\right ) \textit {\_Z}^{3}+\left (350 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{12}+200 \,{\mathrm e}^{9}+435 \,{\mathrm e}^{6}+94\right ) \textit {\_Z}^{2}+\left (1090 \,{\mathrm e}^{3}+250 \,{\mathrm e}^{12}+1150 \,{\mathrm e}^{9}+1750 \,{\mathrm e}^{6}+240\right ) \textit {\_Z} +625 \,{\mathrm e}^{12}+2000 \,{\mathrm e}^{9}+2350 \,{\mathrm e}^{6}+1200 \,{\mathrm e}^{3}+225\right )}{\sum }\frac {\left (42+\left (2+10 \,{\mathrm e}^{3}+5 \,{\mathrm e}^{9}+13 \,{\mathrm e}^{6}\right ) \textit {\_R}^{2}+2 \left (9+39 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{9}+55 \,{\mathrm e}^{6}\right ) \textit {\_R} +175 \,{\mathrm e}^{3}+125 \,{\mathrm e}^{9}+250 \,{\mathrm e}^{6}\right ) \ln \left (x -\textit {\_R} \right )}{120+25 \textit {\_R} \,{\mathrm e}^{12}+15 \textit {\_R}^{2} {\mathrm e}^{9}+2 \textit {\_R}^{3} {\mathrm e}^{6}+125 \,{\mathrm e}^{12}+200 \textit {\_R} \,{\mathrm e}^{9}+60 \textit {\_R}^{2} {\mathrm e}^{6}+4 \textit {\_R}^{3} {\mathrm e}^{3}+575 \,{\mathrm e}^{9}+435 \textit {\_R} \,{\mathrm e}^{6}+69 \textit {\_R}^{2} {\mathrm e}^{3}+2 \textit {\_R}^{3}+875 \,{\mathrm e}^{6}+350 \textit {\_R} \,{\mathrm e}^{3}+24 \textit {\_R}^{2}+545 \,{\mathrm e}^{3}+94 \textit {\_R}}\right )}{2}\) | \(246\) |
int(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^3+(-x^ 4+25*x^3+115*x^2-500*x-2350)*exp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-1200)*exp (3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10*x^4+200 *x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)*exp(3) ^2+(2*x^5+46*x^4+350*x^3+1090*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+240*x^2 +225*x),x,method=_RETURNVERBOSE)
((-5*exp(6)-8*exp(3)-2)*x-25*exp(6)-30*exp(3)-9)/(x*exp(6)+1/5*x^2*exp(3)+ 5*exp(6)+3*x*exp(3)+1/5*x^2+8*exp(3)+8/5*x+3)-ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (32) = 64\).
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.29 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=-\frac {25 \, {\left (x + 5\right )} e^{6} + 10 \, {\left (4 \, x + 15\right )} e^{3} + {\left (x^{2} + 5 \, {\left (x + 5\right )} e^{6} + {\left (x^{2} + 15 \, x + 40\right )} e^{3} + 8 \, x + 15\right )} \log \left (x\right ) + 10 \, x + 45}{x^{2} + 5 \, {\left (x + 5\right )} e^{6} + {\left (x^{2} + 15 \, x + 40\right )} e^{3} + 8 \, x + 15} \]
integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^ 3+(-x^4+25*x^3+115*x^2-500*x-2350)*exp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-120 0)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10*x ^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)* exp(3)^2+(2*x^5+46*x^4+350*x^3+1090*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+2 40*x^2+225*x),x, algorithm=\
-(25*(x + 5)*e^6 + 10*(4*x + 15)*e^3 + (x^2 + 5*(x + 5)*e^6 + (x^2 + 15*x + 40)*e^3 + 8*x + 15)*log(x) + 10*x + 45)/(x^2 + 5*(x + 5)*e^6 + (x^2 + 15 *x + 40)*e^3 + 8*x + 15)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
Time = 3.63 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.80 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=- \frac {x \left (10 + 40 e^{3} + 25 e^{6}\right ) + 45 + 150 e^{3} + 125 e^{6}}{x^{2} \cdot \left (1 + e^{3}\right ) + x \left (8 + 15 e^{3} + 5 e^{6}\right ) + 15 + 40 e^{3} + 25 e^{6}} - \log {\left (x \right )} \]
integrate(((-25*x**2-250*x-625)*exp(3)**4+(15*x**3+50*x**2-525*x-2000)*exp (3)**3+(-x**4+25*x**3+115*x**2-500*x-2350)*exp(3)**2+(-2*x**4+4*x**3+40*x* *2-215*x-1200)*exp(3)-x**4-6*x**3-4*x**2-30*x-225)/((25*x**3+250*x**2+625* x)*exp(3)**4+(10*x**4+200*x**3+1150*x**2+2000*x)*exp(3)**3+(x**5+40*x**4+4 35*x**3+1750*x**2+2350*x)*exp(3)**2+(2*x**5+46*x**4+350*x**3+1090*x**2+120 0*x)*exp(3)+x**5+16*x**4+94*x**3+240*x**2+225*x),x)
-(x*(10 + 40*exp(3) + 25*exp(6)) + 45 + 150*exp(3) + 125*exp(6))/(x**2*(1 + exp(3)) + x*(8 + 15*exp(3) + 5*exp(6)) + 15 + 40*exp(3) + 25*exp(6)) - l og(x)
Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.74 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=-\frac {5 \, {\left (x {\left (5 \, e^{6} + 8 \, e^{3} + 2\right )} + 25 \, e^{6} + 30 \, e^{3} + 9\right )}}{x^{2} {\left (e^{3} + 1\right )} + x {\left (5 \, e^{6} + 15 \, e^{3} + 8\right )} + 25 \, e^{6} + 40 \, e^{3} + 15} - \log \left (x\right ) \]
integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^ 3+(-x^4+25*x^3+115*x^2-500*x-2350)*exp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-120 0)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10*x ^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)* exp(3)^2+(2*x^5+46*x^4+350*x^3+1090*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+2 40*x^2+225*x),x, algorithm=\
-5*(x*(5*e^6 + 8*e^3 + 2) + 25*e^6 + 30*e^3 + 9)/(x^2*(e^3 + 1) + x*(5*e^6 + 15*e^3 + 8) + 25*e^6 + 40*e^3 + 15) - log(x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (32) = 64\).
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.86 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=-\frac {5 \, {\left (5 \, x e^{6} + 8 \, x e^{3} + 2 \, x + 25 \, e^{6} + 30 \, e^{3} + 9\right )}}{x^{2} e^{3} + x^{2} + 5 \, x e^{6} + 15 \, x e^{3} + 8 \, x + 25 \, e^{6} + 40 \, e^{3} + 15} - \log \left ({\left | x \right |}\right ) \]
integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^ 3+(-x^4+25*x^3+115*x^2-500*x-2350)*exp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-120 0)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10*x ^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)* exp(3)^2+(2*x^5+46*x^4+350*x^3+1090*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+2 40*x^2+225*x),x, algorithm=\
-5*(5*x*e^6 + 8*x*e^3 + 2*x + 25*e^6 + 30*e^3 + 9)/(x^2*e^3 + x^2 + 5*x*e^ 6 + 15*x*e^3 + 8*x + 25*e^6 + 40*e^3 + 15) - log(abs(x))
Time = 12.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.74 \[ \int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} \left (-625-250 x-25 x^2\right )+e^9 \left (-2000-525 x+50 x^2+15 x^3\right )+e^3 \left (-1200-215 x+40 x^2+4 x^3-2 x^4\right )+e^6 \left (-2350-500 x+115 x^2+25 x^3-x^4\right )}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} \left (625 x+250 x^2+25 x^3\right )+e^9 \left (2000 x+1150 x^2+200 x^3+10 x^4\right )+e^6 \left (2350 x+1750 x^2+435 x^3+40 x^4+x^5\right )+e^3 \left (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5\right )} \, dx=-\ln \left (x\right )-\frac {150\,{\mathrm {e}}^3+125\,{\mathrm {e}}^6+x\,\left (40\,{\mathrm {e}}^3+25\,{\mathrm {e}}^6+10\right )+45}{\left ({\mathrm {e}}^3+1\right )\,x^2+\left (15\,{\mathrm {e}}^3+5\,{\mathrm {e}}^6+8\right )\,x+40\,{\mathrm {e}}^3+25\,{\mathrm {e}}^6+15} \]
int(-(30*x + exp(12)*(250*x + 25*x^2 + 625) + exp(6)*(500*x - 115*x^2 - 25 *x^3 + x^4 + 2350) + exp(9)*(525*x - 50*x^2 - 15*x^3 + 2000) + exp(3)*(215 *x - 40*x^2 - 4*x^3 + 2*x^4 + 1200) + 4*x^2 + 6*x^3 + x^4 + 225)/(225*x + exp(12)*(625*x + 250*x^2 + 25*x^3) + exp(6)*(2350*x + 1750*x^2 + 435*x^3 + 40*x^4 + x^5) + exp(9)*(2000*x + 1150*x^2 + 200*x^3 + 10*x^4) + exp(3)*(1 200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5) + 240*x^2 + 94*x^3 + 16*x^4 + x^5),x)