Integrand size = 151, antiderivative size = 31 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=\frac {4}{-e^3-x+\log \left (5 \left (1+(2+x)^2\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )} \]
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=-\frac {4}{e^3+x-\log \left (-5 \left (5+4 x+x^2\right ) \left (-2+\log \left (\frac {25}{3}\right )\right )\right )} \]
Integrate[(4 + 8*x + 4*x^2)/(5*x^2 + 4*x^3 + x^4 + E^6*(5 + 4*x + x^2) + E ^3*(10*x + 8*x^2 + 2*x^3) + (-10*x - 8*x^2 - 2*x^3 + E^3*(-10 - 8*x - 2*x^ 2))*Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]] + (5 + 4*x + x ^2)*Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]]^2),x]
Time = 0.60 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2006, 7239, 27, 7237}
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.
\(\displaystyle \int \frac {4 x^2+8 x+4}{x^4+4 x^3+5 x^2+e^6 \left (x^2+4 x+5\right )+\left (x^2+4 x+5\right ) \log ^2\left (10 x^2-\left (5 x^2+20 x+25\right ) \log \left (\frac {25}{3}\right )+40 x+50\right )+e^3 \left (2 x^3+8 x^2+10 x\right )+\left (-2 x^3-8 x^2+e^3 \left (-2 x^2-8 x-10\right )-10 x\right ) \log \left (10 x^2-\left (5 x^2+20 x+25\right ) \log \left (\frac {25}{3}\right )+40 x+50\right )} \, dx\) |
\(\Big \downarrow \) 2006 |
\(\displaystyle \int \frac {(2 x+2)^2}{x^4+4 x^3+5 x^2+e^6 \left (x^2+4 x+5\right )+\left (x^2+4 x+5\right ) \log ^2\left (10 x^2-\left (5 x^2+20 x+25\right ) \log \left (\frac {25}{3}\right )+40 x+50\right )+e^3 \left (2 x^3+8 x^2+10 x\right )+\left (-2 x^3-8 x^2+e^3 \left (-2 x^2-8 x-10\right )-10 x\right ) \log \left (10 x^2-\left (5 x^2+20 x+25\right ) \log \left (\frac {25}{3}\right )+40 x+50\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 (x+1)^2}{\left (x^2+4 x+5\right ) \left (-\log \left (-5 \left (x^2+4 x+5\right ) \left (\log \left (\frac {25}{3}\right )-2\right )\right )+x+e^3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {(x+1)^2}{\left (x^2+4 x+5\right ) \left (x-\log \left (5 \left (x^2+4 x+5\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )+e^3\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {4}{-\log \left (5 \left (x^2+4 x+5\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )+x+e^3}\) |
Int[(4 + 8*x + 4*x^2)/(5*x^2 + 4*x^3 + x^4 + E^6*(5 + 4*x + x^2) + E^3*(10 *x + 8*x^2 + 2*x^3) + (-10*x - 8*x^2 - 2*x^3 + E^3*(-10 - 8*x - 2*x^2))*Lo g[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]] + (5 + 4*x + x^2)*Lo g[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]]^2),x]
3.22.56.3.1 Defintions of rubi rules used
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_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^Expon[Px , x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; PolyQ[Px, x] && GtQ[Expon[P x, x], 1] && NeQ[Coeff[Px, x, 0], 0] && !MatchQ[Px, (a_.)*(v_)^Expon[Px, x ] /; FreeQ[a, x] && LinearQ[v, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13
method | result | size |
norman | \(-\frac {4}{{\mathrm e}^{3}+x -\ln \left (\left (5 x^{2}+20 x +25\right ) \ln \left (\frac {3}{25}\right )+10 x^{2}+40 x +50\right )}\) | \(35\) |
parallelrisch | \(-\frac {4}{{\mathrm e}^{3}+x -\ln \left (\left (5 x^{2}+20 x +25\right ) \ln \left (\frac {3}{25}\right )+10 x^{2}+40 x +50\right )}\) | \(35\) |
risch | \(-\frac {4}{{\mathrm e}^{3}+x -\ln \left (\left (5 x^{2}+20 x +25\right ) \left (\ln \left (3\right )-2 \ln \left (5\right )\right )+10 x^{2}+40 x +50\right )}\) | \(40\) |
int((4*x^2+8*x+4)/((x^2+4*x+5)*ln((5*x^2+20*x+25)*ln(3/25)+10*x^2+40*x+50) ^2+((-2*x^2-8*x-10)*exp(3)-2*x^3-8*x^2-10*x)*ln((5*x^2+20*x+25)*ln(3/25)+1 0*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+x^4+4*x^3+5* x^2),x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=-\frac {4}{x + e^{3} - \log \left (10 \, x^{2} + 5 \, {\left (x^{2} + 4 \, x + 5\right )} \log \left (\frac {3}{25}\right ) + 40 \, x + 50\right )} \]
integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+ 40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*l og(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+x^ 4+4*x^3+5*x^2),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=\frac {4}{- x + \log {\left (10 x^{2} + 40 x + \left (5 x^{2} + 20 x + 25\right ) \log {\left (\frac {3}{25} \right )} + 50 \right )} - e^{3}} \]
integrate((4*x**2+8*x+4)/((x**2+4*x+5)*ln((5*x**2+20*x+25)*ln(3/25)+10*x** 2+40*x+50)**2+((-2*x**2-8*x-10)*exp(3)-2*x**3-8*x**2-10*x)*ln((5*x**2+20*x +25)*ln(3/25)+10*x**2+40*x+50)+(x**2+4*x+5)*exp(3)**2+(2*x**3+8*x**2+10*x) *exp(3)+x**4+4*x**3+5*x**2),x)
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=\frac {4}{i \, \pi - x - e^{3} + \log \left (5\right ) + \log \left (x^{2} + 4 \, x + 5\right ) + \log \left (2 \, \log \left (5\right ) - \log \left (3\right ) - 2\right )} \]
integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+ 40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*l og(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+x^ 4+4*x^3+5*x^2),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=-\frac {4}{x + e^{3} - \log \left (5\right ) - \log \left (x^{2} \log \left (\frac {3}{25}\right ) + 2 \, x^{2} + 4 \, x \log \left (\frac {3}{25}\right ) + 8 \, x + 5 \, \log \left (\frac {3}{25}\right ) + 10\right )} \]
integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+ 40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*l og(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+x^ 4+4*x^3+5*x^2),x, algorithm=\
Time = 10.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx=-\frac {4}{x-\ln \left (40\,x+\ln \left (\frac {3}{25}\right )\,\left (5\,x^2+20\,x+25\right )+10\,x^2+50\right )+{\mathrm {e}}^3} \]