Integrand size = 58, antiderivative size = 25 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (\frac {1}{2} \left (1+e^5\right ) x \left (2-\frac {5}{\log (2)}\right )\right )}{(4+x)^2} \] Output:
ln(1/2*(exp(5)+1)*(2-5/ln(2))*x)/(4+x)^2
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{(4+x)^2} \] Input:
Integrate[(4 + x - 2*x*Log[(-5*x - 5*E^5*x + (2*x + 2*E^5*x)*Log[2])/(2*Lo g[2])])/(64*x + 48*x^2 + 12*x^3 + x^4),x]
Output:
Log[((1 + E^5)*x*(-5 + Log[4]))/Log[4]]/(4 + x)^2
Time = 0.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2026, 2007, 7239, 7293, 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.
\(\displaystyle \int \frac {x-2 x \log \left (\frac {-5 e^5 x-5 x+\left (2 e^5 x+2 x\right ) \log (2)}{2 \log (2)}\right )+4}{x^4+12 x^3+48 x^2+64 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x-2 x \log \left (\frac {-5 e^5 x-5 x+\left (2 e^5 x+2 x\right ) \log (2)}{2 \log (2)}\right )+4}{x \left (x^3+12 x^2+48 x+64\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {x-2 x \log \left (\frac {-5 e^5 x-5 x+\left (2 e^5 x+2 x\right ) \log (2)}{2 \log (2)}\right )+4}{x (x+4)^3}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x-2 x \log \left (\frac {\left (1+e^5\right ) x (\log (4)-5)}{\log (4)}\right )+4}{x (x+4)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{x (x+4)^2}-\frac {2 \log \left (\frac {\left (1+e^5\right ) x (\log (4)-5)}{\log (4)}\right )}{(x+4)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(x+4)^2}\) |
Input:
Int[(4 + x - 2*x*Log[(-5*x - 5*E^5*x + (2*x + 2*E^5*x)*Log[2])/(2*Log[2])] )/(64*x + 48*x^2 + 12*x^3 + x^4),x]
Output:
Log[-(((1 + E^5)*x*(5 - Log[4]))/Log[4])]/(4 + x)^2
Int[(u_.)*(Px_)^(p_), 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)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.60 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40
method | result | size |
norman | \(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )}{\left (4+x \right )^{2}}\) | \(35\) |
parallelrisch | \(\frac {\ln \left (\frac {x \left (2 \,{\mathrm e}^{5} \ln \left (2\right )-5 \,{\mathrm e}^{5}+2 \ln \left (2\right )-5\right )}{2 \ln \left (2\right )}\right )}{x^{2}+8 x +16}\) | \(36\) |
risch | \(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )}{x^{2}+8 x +16}\) | \(40\) |
orering | \(-\frac {\left (4+x \right ) \left (4+5 x \right ) \left (-2 x \ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )+4+x \right )}{4 \left (2+x \right ) \left (x^{4}+12 x^{3}+48 x^{2}+64 x \right )}-\frac {x \left (4+x \right )^{2} \left (\frac {-2 \ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )-\frac {2 x \left (\left (2 \,{\mathrm e}^{5}+2\right ) \ln \left (2\right )-5 \,{\mathrm e}^{5}-5\right )}{\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}+1}{x^{4}+12 x^{3}+48 x^{2}+64 x}-\frac {\left (-2 x \ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )+4+x \right ) \left (4 x^{3}+36 x^{2}+96 x +64\right )}{\left (x^{4}+12 x^{3}+48 x^{2}+64 x \right )^{2}}\right )}{4 \left (2+x \right )}\) | \(247\) |
parts | \(\text {Expression too large to display}\) | \(6143\) |
derivativedivides | \(\text {Expression too large to display}\) | \(6305\) |
default | \(\text {Expression too large to display}\) | \(6305\) |
Input:
int((-2*x*ln(1/2*((2*x*exp(5)+2*x)*ln(2)-5*x*exp(5)-5*x)/ln(2))+4+x)/(x^4+ 12*x^3+48*x^2+64*x),x,method=_RETURNVERBOSE)
Output:
ln(1/2*((2*x*exp(5)+2*x)*ln(2)-5*x*exp(5)-5*x)/ln(2))/(4+x)^2
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (-\frac {5 \, x e^{5} - 2 \, {\left (x e^{5} + x\right )} \log \left (2\right ) + 5 \, x}{2 \, \log \left (2\right )}\right )}{x^{2} + 8 \, x + 16} \] Input:
integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4 +x)/(x^4+12*x^3+48*x^2+64*x),x, algorithm="fricas")
Output:
log(-1/2*(5*x*e^5 - 2*(x*e^5 + x)*log(2) + 5*x)/log(2))/(x^2 + 8*x + 16)
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log {\left (\frac {- \frac {5 x e^{5}}{2} - \frac {5 x}{2} + \frac {\left (2 x + 2 x e^{5}\right ) \log {\left (2 \right )}}{2}}{\log {\left (2 \right )}} \right )}}{x^{2} + 8 x + 16} \] Input:
integrate((-2*x*ln(1/2*((2*x*exp(5)+2*x)*ln(2)-5*x*exp(5)-5*x)/ln(2))+4+x) /(x**4+12*x**3+48*x**2+64*x),x)
Output:
log((-5*x*exp(5)/2 - 5*x/2 + (2*x + 2*x*exp(5))*log(2)/2)/log(2))/(x**2 + 8*x + 16)
Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (22) = 44\).
Time = 0.03 (sec) , antiderivative size = 176, normalized size of antiderivative = 7.04 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=-\frac {1}{16} \, {\left (\frac {\log \left (2\right ) \log \left (x + 4\right )}{2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5} - \frac {\log \left (2\right ) \log \left (x\right )}{2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5} - \frac {4 \, \log \left (2\right )}{{\left (2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5\right )} x + 8 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 20 \, e^{5} - 20}\right )} {\left (\frac {5 \, e^{5}}{\log \left (2\right )} + \frac {5}{\log \left (2\right )} - 2 \, e^{5} - 2\right )} + \frac {x + 6}{4 \, {\left (x^{2} + 8 \, x + 16\right )}} + \frac {\log \left (x e^{5} + x - \frac {5 \, x e^{5}}{2 \, \log \left (2\right )} - \frac {5 \, x}{2 \, \log \left (2\right )}\right )}{x^{2} + 8 \, x + 16} - \frac {1}{2 \, {\left (x^{2} + 8 \, x + 16\right )}} - \frac {1}{16} \, \log \left (x + 4\right ) + \frac {1}{16} \, \log \left (x\right ) \] Input:
integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4 +x)/(x^4+12*x^3+48*x^2+64*x),x, algorithm="maxima")
Output:
-1/16*(log(2)*log(x + 4)/(2*(e^5 + 1)*log(2) - 5*e^5 - 5) - log(2)*log(x)/ (2*(e^5 + 1)*log(2) - 5*e^5 - 5) - 4*log(2)/((2*(e^5 + 1)*log(2) - 5*e^5 - 5)*x + 8*(e^5 + 1)*log(2) - 20*e^5 - 20))*(5*e^5/log(2) + 5/log(2) - 2*e^ 5 - 2) + 1/4*(x + 6)/(x^2 + 8*x + 16) + log(x*e^5 + x - 5/2*x*e^5/log(2) - 5/2*x/log(2))/(x^2 + 8*x + 16) - 1/2/(x^2 + 8*x + 16) - 1/16*log(x + 4) + 1/16*log(x)
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=-\frac {\log \left (2\right ) - \log \left (2 \, x e^{5} \log \left (2\right ) - 5 \, x e^{5} + 2 \, x \log \left (2\right ) - 5 \, x\right ) + \log \left (\log \left (2\right )\right )}{x^{2} + 8 \, x + 16} \] Input:
integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4 +x)/(x^4+12*x^3+48*x^2+64*x),x, algorithm="giac")
Output:
-(log(2) - log(2*x*e^5*log(2) - 5*x*e^5 + 2*x*log(2) - 5*x) + log(log(2))) /(x^2 + 8*x + 16)
Time = 3.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {x^2\,\left (\ln \left (\frac {\ln \left (2\right )\,\left (2\,x+2\,x\,{\mathrm {e}}^5\right )}{2}-\frac {5\,x\,{\mathrm {e}}^5}{2}-\frac {5\,x}{2}\right )-\ln \left (\ln \left (2\right )\right )\right )}{x^4+8\,x^3+16\,x^2} \] Input:
int((x - 2*x*log(-((5*x)/2 + (5*x*exp(5))/2 - (log(2)*(2*x + 2*x*exp(5)))/ 2)/log(2)) + 4)/(64*x + 48*x^2 + 12*x^3 + x^4),x)
Output:
(x^2*(log((log(2)*(2*x + 2*x*exp(5)))/2 - (5*x*exp(5))/2 - (5*x)/2) - log( log(2))))/(16*x^2 + 8*x^3 + x^4)
Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.28 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {-2 \,\mathrm {log}\left (\frac {2 \,\mathrm {log}\left (2\right ) e^{5} x +2 \,\mathrm {log}\left (2\right ) x -5 e^{5} x -5 x}{2 \,\mathrm {log}\left (2\right )}\right ) x^{2}-16 \,\mathrm {log}\left (\frac {2 \,\mathrm {log}\left (2\right ) e^{5} x +2 \,\mathrm {log}\left (2\right ) x -5 e^{5} x -5 x}{2 \,\mathrm {log}\left (2\right )}\right ) x +2 \,\mathrm {log}\left (x \right ) x^{2}+16 \,\mathrm {log}\left (x \right ) x +32 \,\mathrm {log}\left (x \right )-x^{2}-8 x -16}{32 x^{2}+256 x +512} \] Input:
int((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4+x)/(x ^4+12*x^3+48*x^2+64*x),x)
Output:
( - 2*log((2*log(2)*e**5*x + 2*log(2)*x - 5*e**5*x - 5*x)/(2*log(2)))*x**2 - 16*log((2*log(2)*e**5*x + 2*log(2)*x - 5*e**5*x - 5*x)/(2*log(2)))*x + 2*log(x)*x**2 + 16*log(x)*x + 32*log(x) - x**2 - 8*x - 16)/(32*(x**2 + 8*x + 16))