Integrand size = 68, antiderivative size = 31 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=5 e^2 \log \left (-2+\frac {x}{x-\frac {5-\frac {x^2}{(-5+x)^2}}{x}}\right ) \] Output:
5*exp(2)*ln(x/(x-1/x*(5-x^2/(-5+x)^2))-2)
Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=10 e^2 \left (-\frac {1}{2} \log \left (125-50 x-21 x^2+10 x^3-x^4\right )+\frac {1}{2} \log \left (250-100 x-17 x^2+10 x^3-x^4\right )\right ) \] Input:
Integrate[(E^2*(31250*x - 25000*x^2 + 7500*x^3 - 950*x^4 + 40*x^5))/(31250 - 25000*x - 2375*x^2 + 6700*x^3 - 1518*x^4 - 230*x^5 + 138*x^6 - 20*x^7 + x^8),x]
Output:
10*E^2*(-1/2*Log[125 - 50*x - 21*x^2 + 10*x^3 - x^4] + Log[250 - 100*x - 1 7*x^2 + 10*x^3 - x^4]/2)
Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {27, 27, 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.
\(\displaystyle \int \frac {e^2 \left (40 x^5-950 x^4+7500 x^3-25000 x^2+31250 x\right )}{x^8-20 x^7+138 x^6-230 x^5-1518 x^4+6700 x^3-2375 x^2-25000 x+31250} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^2 \int \frac {10 \left (4 x^5-95 x^4+750 x^3-2500 x^2+3125 x\right )}{x^8-20 x^7+138 x^6-230 x^5-1518 x^4+6700 x^3-2375 x^2-25000 x+31250}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 e^2 \int \frac {4 x^5-95 x^4+750 x^3-2500 x^2+3125 x}{x^8-20 x^7+138 x^6-230 x^5-1518 x^4+6700 x^3-2375 x^2-25000 x+31250}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle 10 e^2 \int \left (\frac {-2 x^3+15 x^2-21 x-25}{x^4-10 x^3+21 x^2+50 x-125}+\frac {2 x^3-15 x^2+17 x+50}{x^4-10 x^3+17 x^2+100 x-250}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 10 e^2 \left (\frac {1}{2} \log \left (-x^4+10 x^3-17 x^2-100 x+250\right )-\frac {1}{2} \log \left (-x^4+10 x^3-21 x^2-50 x+125\right )\right )\) |
Input:
Int[(E^2*(31250*x - 25000*x^2 + 7500*x^3 - 950*x^4 + 40*x^5))/(31250 - 250 00*x - 2375*x^2 + 6700*x^3 - 1518*x^4 - 230*x^5 + 138*x^6 - 20*x^7 + x^8), x]
Output:
10*E^2*(-1/2*Log[125 - 50*x - 21*x^2 + 10*x^3 - x^4] + Log[250 - 100*x - 1 7*x^2 + 10*x^3 - x^4]/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52
method | result | size |
parallelrisch | \({\mathrm e}^{2} \left (5 \ln \left (x^{4}-10 x^{3}+17 x^{2}+100 x -250\right )-5 \ln \left (x^{4}-10 x^{3}+21 x^{2}+50 x -125\right )\right )\) | \(47\) |
default | \(10 \,{\mathrm e}^{2} \left (-\frac {\ln \left (x^{4}-10 x^{3}+21 x^{2}+50 x -125\right )}{2}+\frac {\ln \left (x^{4}-10 x^{3}+17 x^{2}+100 x -250\right )}{2}\right )\) | \(48\) |
norman | \(5 \,{\mathrm e}^{2} \ln \left (x^{4}-10 x^{3}+17 x^{2}+100 x -250\right )-5 \,{\mathrm e}^{2} \ln \left (x^{4}-10 x^{3}+21 x^{2}+50 x -125\right )\) | \(48\) |
risch | \(5 \,{\mathrm e}^{2} \ln \left (x^{4}-10 x^{3}+17 x^{2}+100 x -250\right )-5 \,{\mathrm e}^{2} \ln \left (x^{4}-10 x^{3}+21 x^{2}+50 x -125\right )\) | \(48\) |
Input:
int((40*x^5-950*x^4+7500*x^3-25000*x^2+31250*x)*exp(2)/(x^8-20*x^7+138*x^6 -230*x^5-1518*x^4+6700*x^3-2375*x^2-25000*x+31250),x,method=_RETURNVERBOSE )
Output:
exp(2)*(5*ln(x^4-10*x^3+17*x^2+100*x-250)-5*ln(x^4-10*x^3+21*x^2+50*x-125) )
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=-5 \, e^{2} \log \left (x^{4} - 10 \, x^{3} + 21 \, x^{2} + 50 \, x - 125\right ) + 5 \, e^{2} \log \left (x^{4} - 10 \, x^{3} + 17 \, x^{2} + 100 \, x - 250\right ) \] Input:
integrate((40*x^5-950*x^4+7500*x^3-25000*x^2+31250*x)*exp(2)/(x^8-20*x^7+1 38*x^6-230*x^5-1518*x^4+6700*x^3-2375*x^2-25000*x+31250),x, algorithm="fri cas")
Output:
-5*e^2*log(x^4 - 10*x^3 + 21*x^2 + 50*x - 125) + 5*e^2*log(x^4 - 10*x^3 + 17*x^2 + 100*x - 250)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=5 e^{2} \log {\left (x^{4} - 10 x^{3} + 17 x^{2} + 100 x - 250 \right )} - 5 e^{2} \log {\left (x^{4} - 10 x^{3} + 21 x^{2} + 50 x - 125 \right )} \] Input:
integrate((40*x**5-950*x**4+7500*x**3-25000*x**2+31250*x)*exp(2)/(x**8-20* x**7+138*x**6-230*x**5-1518*x**4+6700*x**3-2375*x**2-25000*x+31250),x)
Output:
5*exp(2)*log(x**4 - 10*x**3 + 17*x**2 + 100*x - 250) - 5*exp(2)*log(x**4 - 10*x**3 + 21*x**2 + 50*x - 125)
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=-5 \, {\left (\log \left (x^{4} - 10 \, x^{3} + 21 \, x^{2} + 50 \, x - 125\right ) - \log \left (x^{4} - 10 \, x^{3} + 17 \, x^{2} + 100 \, x - 250\right )\right )} e^{2} \] Input:
integrate((40*x^5-950*x^4+7500*x^3-25000*x^2+31250*x)*exp(2)/(x^8-20*x^7+1 38*x^6-230*x^5-1518*x^4+6700*x^3-2375*x^2-25000*x+31250),x, algorithm="max ima")
Output:
-5*(log(x^4 - 10*x^3 + 21*x^2 + 50*x - 125) - log(x^4 - 10*x^3 + 17*x^2 + 100*x - 250))*e^2
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=-5 \, {\left (\log \left ({\left | x^{4} - 10 \, x^{3} + 21 \, x^{2} + 50 \, x - 125 \right |}\right ) - \log \left ({\left | x^{4} - 10 \, x^{3} + 17 \, x^{2} + 100 \, x - 250 \right |}\right )\right )} e^{2} \] Input:
integrate((40*x^5-950*x^4+7500*x^3-25000*x^2+31250*x)*exp(2)/(x^8-20*x^7+1 38*x^6-230*x^5-1518*x^4+6700*x^3-2375*x^2-25000*x+31250),x, algorithm="gia c")
Output:
-5*(log(abs(x^4 - 10*x^3 + 21*x^2 + 50*x - 125)) - log(abs(x^4 - 10*x^3 + 17*x^2 + 100*x - 250)))*e^2
Time = 1.96 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=-10\,{\mathrm {e}}^2\,\mathrm {atanh}\left (\frac {15254\,x^4-152540\,x^3+332970\,x^2+604750\,x-1511875}{21572\,x^4-215720\,x^3+440376\,x^2+1236550\,x-3091375}\right ) \] Input:
int(-(exp(2)*(31250*x - 25000*x^2 + 7500*x^3 - 950*x^4 + 40*x^5))/(25000*x + 2375*x^2 - 6700*x^3 + 1518*x^4 + 230*x^5 - 138*x^6 + 20*x^7 - x^8 - 312 50),x)
Output:
-10*exp(2)*atanh((604750*x + 332970*x^2 - 152540*x^3 + 15254*x^4 - 1511875 )/(1236550*x + 440376*x^2 - 215720*x^3 + 21572*x^4 - 3091375))
\[ \int \frac {e^2 \left (31250 x-25000 x^2+7500 x^3-950 x^4+40 x^5\right )}{31250-25000 x-2375 x^2+6700 x^3-1518 x^4-230 x^5+138 x^6-20 x^7+x^8} \, dx=10 e^{2} \left (4 \left (\int \frac {x^{5}}{x^{8}-20 x^{7}+138 x^{6}-230 x^{5}-1518 x^{4}+6700 x^{3}-2375 x^{2}-25000 x +31250}d x \right )-95 \left (\int \frac {x^{4}}{x^{8}-20 x^{7}+138 x^{6}-230 x^{5}-1518 x^{4}+6700 x^{3}-2375 x^{2}-25000 x +31250}d x \right )+750 \left (\int \frac {x^{3}}{x^{8}-20 x^{7}+138 x^{6}-230 x^{5}-1518 x^{4}+6700 x^{3}-2375 x^{2}-25000 x +31250}d x \right )-2500 \left (\int \frac {x^{2}}{x^{8}-20 x^{7}+138 x^{6}-230 x^{5}-1518 x^{4}+6700 x^{3}-2375 x^{2}-25000 x +31250}d x \right )+3125 \left (\int \frac {x}{x^{8}-20 x^{7}+138 x^{6}-230 x^{5}-1518 x^{4}+6700 x^{3}-2375 x^{2}-25000 x +31250}d x \right )\right ) \] Input:
int((40*x^5-950*x^4+7500*x^3-25000*x^2+31250*x)*exp(2)/(x^8-20*x^7+138*x^6 -230*x^5-1518*x^4+6700*x^3-2375*x^2-25000*x+31250),x)
Output:
10*e**2*(4*int(x**5/(x**8 - 20*x**7 + 138*x**6 - 230*x**5 - 1518*x**4 + 67 00*x**3 - 2375*x**2 - 25000*x + 31250),x) - 95*int(x**4/(x**8 - 20*x**7 + 138*x**6 - 230*x**5 - 1518*x**4 + 6700*x**3 - 2375*x**2 - 25000*x + 31250) ,x) + 750*int(x**3/(x**8 - 20*x**7 + 138*x**6 - 230*x**5 - 1518*x**4 + 670 0*x**3 - 2375*x**2 - 25000*x + 31250),x) - 2500*int(x**2/(x**8 - 20*x**7 + 138*x**6 - 230*x**5 - 1518*x**4 + 6700*x**3 - 2375*x**2 - 25000*x + 31250 ),x) + 3125*int(x/(x**8 - 20*x**7 + 138*x**6 - 230*x**5 - 1518*x**4 + 6700 *x**3 - 2375*x**2 - 25000*x + 31250),x))