Integrand size = 60, antiderivative size = 28 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {1-\log (5)-e (4-\log (x))-\frac {\log (x)}{x^2}}{16+x} \] Output:
(1-ln(5)-ln(x)/x^2-exp(1)*(-ln(x)+4))/(x+16)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {-x^2 (-1+4 e+\log (5))+\left (-1+e x^2\right ) \log (x)}{x^2 (16+x)} \] Input:
Integrate[(-16 - x - x^3 + E*(16*x^2 + 5*x^3) + x^3*Log[5] + (32 + 3*x - E *x^3)*Log[x])/(256*x^3 + 32*x^4 + x^5),x]
Output:
(-(x^2*(-1 + 4*E + Log[5])) + (-1 + E*x^2)*Log[x])/(x^2*(16 + x))
Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(28)=56\).
Time = 0.63 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.54, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 2026, 2007, 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.
\(\displaystyle \int \frac {-x^3+x^3 \log (5)+\left (-e x^3+3 x+32\right ) \log (x)+e \left (5 x^3+16 x^2\right )-x-16}{x^5+32 x^4+256 x^3} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^3 (\log (5)-1)+\left (-e x^3+3 x+32\right ) \log (x)+e \left (5 x^3+16 x^2\right )-x-16}{x^5+32 x^4+256 x^3}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^3 (\log (5)-1)+\left (-e x^3+3 x+32\right ) \log (x)+e \left (5 x^3+16 x^2\right )-x-16}{x^3 \left (x^2+32 x+256\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {x^3 (\log (5)-1)+\left (-e x^3+3 x+32\right ) \log (x)+e \left (5 x^3+16 x^2\right )-x-16}{x^3 (x+16)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {16}{x^3 (x+16)^2}-\frac {\left (e x^3-3 x-32\right ) \log (x)}{x^3 (x+16)^2}-\frac {1}{x^2 (x+16)^2}+\frac {e (5 x+16)}{x (x+16)^2}+\frac {\log (5)-1}{(x+16)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (x)}{16 x^2}-\frac {4 e}{x+16}+\frac {\log (x)}{256 x}+\frac {(1-256 e) x \log (x)}{4096 (x+16)}+\frac {1}{16} e \log (x)-\frac {\log (x)}{4096}-\frac {1}{16} e \log (x+16)-\frac {(1-256 e) \log (x+16)}{4096}+\frac {\log (x+16)}{4096}+\frac {1-\log (5)}{x+16}\) |
Input:
Int[(-16 - x - x^3 + E*(16*x^2 + 5*x^3) + x^3*Log[5] + (32 + 3*x - E*x^3)* Log[x])/(256*x^3 + 32*x^4 + x^5),x]
Output:
(-4*E)/(16 + x) + (1 - Log[5])/(16 + x) - Log[x]/4096 + (E*Log[x])/16 - Lo g[x]/(16*x^2) + Log[x]/(256*x) + ((1 - 256*E)*x*Log[x])/(4096*(16 + x)) + Log[16 + x]/4096 - ((1 - 256*E)*Log[16 + x])/4096 - (E*Log[16 + x])/16
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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])
Time = 0.50 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
norman | \(\frac {\left (1-4 \,{\mathrm e}-\ln \left (5\right )\right ) x^{2}+x^{2} {\mathrm e} \ln \left (x \right )-\ln \left (x \right )}{x^{2} \left (x +16\right )}\) | \(37\) |
parallelrisch | \(-\frac {-x^{2} {\mathrm e} \ln \left (x \right )+x^{2} \ln \left (5\right )+4 x^{2} {\mathrm e}-x^{2}+\ln \left (x \right )}{x^{2} \left (x +16\right )}\) | \(41\) |
risch | \(\frac {\left (x^{2} {\mathrm e}-1\right ) \ln \left (x \right )}{x^{2} \left (x +16\right )}+\frac {1}{x +16}-\frac {4 \,{\mathrm e}}{x +16}-\frac {\ln \left (5\right )}{x +16}\) | \(44\) |
parts | \(\left (-\frac {1}{4096}+\frac {{\mathrm e}}{16}\right ) \ln \left (x \right )-\frac {-1+4 \,{\mathrm e}+\ln \left (5\right )}{x +16}+\left (-\frac {{\mathrm e}}{16}+\frac {1}{4096}\right ) \ln \left (x +16\right )-\frac {\left (-1+256 \,{\mathrm e}\right ) \left (-\frac {\ln \left (x +16\right )}{16}+\frac {x \ln \left (x \right )}{16 x +256}\right )}{256}+\frac {\ln \left (x \right )}{256 x}-\frac {\ln \left (x \right )}{16 x^{2}}\) | \(76\) |
default | \(-\frac {\ln \left (5\right )}{x +16}+\left (-\frac {1}{4096}+\frac {{\mathrm e}}{16}\right ) \ln \left (x \right )-\frac {4 \,{\mathrm e}-1}{x +16}+\left (-\frac {{\mathrm e}}{16}+\frac {1}{4096}\right ) \ln \left (x +16\right )-\frac {\left (-1+256 \,{\mathrm e}\right ) \left (-\frac {\ln \left (x +16\right )}{16}+\frac {x \ln \left (x \right )}{16 x +256}\right )}{256}+\frac {\ln \left (x \right )}{256 x}-\frac {\ln \left (x \right )}{16 x^{2}}\) | \(83\) |
Input:
int(((-x^3*exp(1)+3*x+32)*ln(x)+x^3*ln(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16)/ (x^5+32*x^4+256*x^3),x,method=_RETURNVERBOSE)
Output:
((1-4*exp(1)-ln(5))*x^2+x^2*exp(1)*ln(x)-ln(x))/x^2/(x+16)
Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=-\frac {4 \, x^{2} e + x^{2} \log \left (5\right ) - x^{2} - {\left (x^{2} e - 1\right )} \log \left (x\right )}{x^{3} + 16 \, x^{2}} \] Input:
integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^ 3-x-16)/(x^5+32*x^4+256*x^3),x, algorithm="fricas")
Output:
-(4*x^2*e + x^2*log(5) - x^2 - (x^2*e - 1)*log(x))/(x^3 + 16*x^2)
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {\left (e x^{2} - 1\right ) \log {\left (x \right )}}{x^{3} + 16 x^{2}} - \frac {-1 + \log {\left (5 \right )} + 4 e}{x + 16} \] Input:
integrate(((-x**3*exp(1)+3*x+32)*ln(x)+x**3*ln(5)+(5*x**3+16*x**2)*exp(1)- x**3-x-16)/(x**5+32*x**4+256*x**3),x)
Output:
(E*x**2 - 1)*log(x)/(x**3 + 16*x**2) - (-1 + log(5) + 4*E)/(x + 16)
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (26) = 52\).
Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 5.25 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {1}{16} \, {\left (\frac {16}{x + 16} - \log \left (x + 16\right ) + \log \left (x\right )\right )} e + \frac {1}{4096} \, {\left (256 \, e - 1\right )} \log \left (x + 16\right ) + \frac {16 \, x^{2} - {\left (x^{3} {\left (256 \, e - 1\right )} - 16 \, x^{2} + 4096\right )} \log \left (x\right ) + 128 \, x - 2048}{4096 \, {\left (x^{3} + 16 \, x^{2}\right )}} - \frac {3 \, x^{2} + 24 \, x - 128}{256 \, {\left (x^{3} + 16 \, x^{2}\right )}} + \frac {x + 8}{128 \, {\left (x^{2} + 16 \, x\right )}} - \frac {5 \, e}{x + 16} - \frac {\log \left (5\right )}{x + 16} + \frac {1}{x + 16} + \frac {1}{4096} \, \log \left (x + 16\right ) - \frac {1}{4096} \, \log \left (x\right ) \] Input:
integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^ 3-x-16)/(x^5+32*x^4+256*x^3),x, algorithm="maxima")
Output:
1/16*(16/(x + 16) - log(x + 16) + log(x))*e + 1/4096*(256*e - 1)*log(x + 1 6) + 1/4096*(16*x^2 - (x^3*(256*e - 1) - 16*x^2 + 4096)*log(x) + 128*x - 2 048)/(x^3 + 16*x^2) - 1/256*(3*x^2 + 24*x - 128)/(x^3 + 16*x^2) + 1/128*(x + 8)/(x^2 + 16*x) - 5*e/(x + 16) - log(5)/(x + 16) + 1/(x + 16) + 1/4096* log(x + 16) - 1/4096*log(x)
Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {x^{2} e \log \left (x\right ) - 4 \, x^{2} e - x^{2} \log \left (5\right ) + x^{2} - \log \left (x\right )}{x^{3} + 16 \, x^{2}} \] Input:
integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^ 3-x-16)/(x^5+32*x^4+256*x^3),x, algorithm="giac")
Output:
(x^2*e*log(x) - 4*x^2*e - x^2*log(5) + x^2 - log(x))/(x^3 + 16*x^2)
Time = 2.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {x^4\,\left (\frac {\mathrm {e}}{4}+\frac {\ln \left (5\right )}{16}-\frac {1}{16}\right )-x\,\ln \left (x\right )+x^3\,\mathrm {e}\,\ln \left (x\right )}{x^4+16\,x^3} \] Input:
int(-(x - log(x)*(3*x - x^3*exp(1) + 32) - exp(1)*(16*x^2 + 5*x^3) - x^3*l og(5) + x^3 + 16)/(256*x^3 + 32*x^4 + x^5),x)
Output:
(x^4*(exp(1)/4 + log(5)/16 - 1/16) - x*log(x) + x^3*exp(1)*log(x))/(16*x^3 + x^4)
Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-16-x-x^3+e \left (16 x^2+5 x^3\right )+x^3 \log (5)+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx=\frac {16 \,\mathrm {log}\left (x \right ) e \,x^{2}-16 \,\mathrm {log}\left (x \right )+\mathrm {log}\left (5\right ) x^{3}+4 e \,x^{3}-x^{3}}{16 x^{2} \left (x +16\right )} \] Input:
int(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16 )/(x^5+32*x^4+256*x^3),x)
Output:
(16*log(x)*e*x**2 - 16*log(x) + log(5)*x**3 + 4*e*x**3 - x**3)/(16*x**2*(x + 16))