Integrand size = 127, antiderivative size = 34 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=-\frac {x \left (-x+x^2\right )}{2 (4-x) \left (4+x+\frac {5}{x-\log (x)}\right )} \]
Time = 0.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {-20-11 x+x^4-\left (-16+x^3\right ) \log (x)}{2 (-4+x) \left (5+4 x+x^2-(4+x) \log (x)\right )} \]
Integrate[(-20*x + 85*x^2 - 63*x^3 - 33*x^4 + x^6 + (-40*x + x^2 + 86*x^3 - 2*x^5)*Log[x] + (32*x - 48*x^2 + x^4)*Log[x]^2)/(800 + 880*x + 242*x^2 - 80*x^3 - 44*x^4 + 2*x^6 + (-1280 - 704*x + 80*x^2 + 108*x^3 - 4*x^5)*Log[ x] + (512 - 64*x^2 + 2*x^4)*Log[x]^2),x]
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.
\(\displaystyle \int \frac {x^6-33 x^4-63 x^3+85 x^2+\left (x^4-48 x^2+32 x\right ) \log ^2(x)+\left (-2 x^5+86 x^3+x^2-40 x\right ) \log (x)-20 x}{2 x^6-44 x^4-80 x^3+242 x^2+\left (2 x^4-64 x^2+512\right ) \log ^2(x)+\left (-4 x^5+108 x^3+80 x^2-704 x-1280\right ) \log (x)+880 x+800} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x \left (x^5-33 x^3+\left (x^3-48 x+32\right ) \log ^2(x)-63 x^2+\left (-2 x^4+86 x^2+x-40\right ) \log (x)+85 x-20\right )}{2 (4-x)^2 \left (x^2+4 x-(x+4) \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {x \left (-x^5+33 x^3+63 x^2-85 x-\left (x^3-48 x+32\right ) \log ^2(x)+\left (2 x^4-86 x^2-x+40\right ) \log (x)+20\right )}{(4-x)^2 \left (x^2+4 x-(x+4) \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x \left (-x^5+33 x^3+63 x^2-85 x-\left (x^3-48 x+32\right ) \log ^2(x)+\left (2 x^4-86 x^2-x+40\right ) \log (x)+20\right )}{(4-x)^2 \left (x^2+4 x-(x+4) \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {5 x \left (5 x^2-52 x+32\right )}{\left (x^2-16\right )^2 \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )}-\frac {x \left (x^3-48 x+32\right )}{\left (x^2-16\right )^2}-\frac {5 x \left (x^4+6 x^3-4 x^2-19 x+16\right )}{(x-4) (x+4)^2 \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (20 \int \frac {1}{\left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx+\frac {645}{4} \int \frac {1}{(x-4) \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx+10 \int \frac {x}{\left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx+5 \int \frac {x^2}{\left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx-250 \int \frac {1}{(x+4)^2 \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx+\frac {575}{4} \int \frac {1}{(x+4) \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )^2}dx+30 \int \frac {1}{(x-4)^2 \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )}dx+\frac {15}{4} \int \frac {1}{(x-4) \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )}dx+100 \int \frac {1}{(x+4)^2 \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )}dx-\frac {115}{4} \int \frac {1}{(x+4) \left (x^2-\log (x) x+4 x-4 \log (x)+5\right )}dx-\frac {(16-x) x}{16-x^2}+x\right )\) |
Int[(-20*x + 85*x^2 - 63*x^3 - 33*x^4 + x^6 + (-40*x + x^2 + 86*x^3 - 2*x^ 5)*Log[x] + (32*x - 48*x^2 + x^4)*Log[x]^2)/(800 + 880*x + 242*x^2 - 80*x^ 3 - 44*x^4 + 2*x^6 + (-1280 - 704*x + 80*x^2 + 108*x^3 - 4*x^5)*Log[x] + ( 512 - 64*x^2 + 2*x^4)*Log[x]^2),x]
3.9.56.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.83 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26
method | result | size |
parallelrisch | \(\frac {x^{4}-x^{3} \ln \left (x \right )-20-11 x +16 \ln \left (x \right )}{2 x^{3}-2 x^{2} \ln \left (x \right )-22 x +32 \ln \left (x \right )-40}\) | \(43\) |
norman | \(\frac {8 \ln \left (x \right )-\frac {11 x}{2}+\frac {x^{4}}{2}-\frac {x^{3} \ln \left (x \right )}{2}-10}{x^{3}-x^{2} \ln \left (x \right )-11 x +16 \ln \left (x \right )-20}\) | \(44\) |
default | \(\frac {-16 \ln \left (x \right )+11 x +x^{3} \ln \left (x \right )-x^{4}+20}{2 \left (x -4\right ) \left (x \ln \left (x \right )-x^{2}+4 \ln \left (x \right )-4 x -5\right )}\) | \(48\) |
risch | \(\frac {x^{3}-16}{2 x^{2}-32}-\frac {5 \left (-1+x \right ) x^{2}}{2 \left (4+x \right ) \left (x -4\right ) \left (x^{2}-x \ln \left (x \right )+4 x -4 \ln \left (x \right )+5\right )}\) | \(53\) |
int(((x^4-48*x^2+32*x)*ln(x)^2+(-2*x^5+86*x^3+x^2-40*x)*ln(x)+x^6-33*x^4-6 3*x^3+85*x^2-20*x)/((2*x^4-64*x^2+512)*ln(x)^2+(-4*x^5+108*x^3+80*x^2-704* x-1280)*ln(x)+2*x^6-44*x^4-80*x^3+242*x^2+880*x+800),x,method=_RETURNVERBO SE)
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x^{4} - {\left (x^{3} - 16\right )} \log \left (x\right ) - 11 \, x - 20}{2 \, {\left (x^{3} - {\left (x^{2} - 16\right )} \log \left (x\right ) - 11 \, x - 20\right )}} \]
integrate(((x^4-48*x^2+32*x)*log(x)^2+(-2*x^5+86*x^3+x^2-40*x)*log(x)+x^6- 33*x^4-63*x^3+85*x^2-20*x)/((2*x^4-64*x^2+512)*log(x)^2+(-4*x^5+108*x^3+80 *x^2-704*x-1280)*log(x)+2*x^6-44*x^4-80*x^3+242*x^2+880*x+800),x, algorith m=\
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (22) = 44\).
Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x}{2} + \frac {8 x - 8}{x^{2} - 16} + \frac {5 x^{3} - 5 x^{2}}{- 2 x^{4} - 8 x^{3} + 22 x^{2} + 128 x + \left (2 x^{3} + 8 x^{2} - 32 x - 128\right ) \log {\left (x \right )} + 160} \]
integrate(((x**4-48*x**2+32*x)*ln(x)**2+(-2*x**5+86*x**3+x**2-40*x)*ln(x)+ x**6-33*x**4-63*x**3+85*x**2-20*x)/((2*x**4-64*x**2+512)*ln(x)**2+(-4*x**5 +108*x**3+80*x**2-704*x-1280)*ln(x)+2*x**6-44*x**4-80*x**3+242*x**2+880*x+ 800),x)
x/2 + (8*x - 8)/(x**2 - 16) + (5*x**3 - 5*x**2)/(-2*x**4 - 8*x**3 + 22*x** 2 + 128*x + (2*x**3 + 8*x**2 - 32*x - 128)*log(x) + 160)
Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x^{4} - {\left (x^{3} - 16\right )} \log \left (x\right ) - 11 \, x - 20}{2 \, {\left (x^{3} - {\left (x^{2} - 16\right )} \log \left (x\right ) - 11 \, x - 20\right )}} \]
integrate(((x^4-48*x^2+32*x)*log(x)^2+(-2*x^5+86*x^3+x^2-40*x)*log(x)+x^6- 33*x^4-63*x^3+85*x^2-20*x)/((2*x^4-64*x^2+512)*log(x)^2+(-4*x^5+108*x^3+80 *x^2-704*x-1280)*log(x)+2*x^6-44*x^4-80*x^3+242*x^2+880*x+800),x, algorith m=\
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.06 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {1}{2} \, x - \frac {5 \, {\left (x^{3} - x^{2}\right )}}{2 \, {\left (x^{4} - x^{3} \log \left (x\right ) + 4 \, x^{3} - 4 \, x^{2} \log \left (x\right ) - 11 \, x^{2} + 16 \, x \log \left (x\right ) - 64 \, x + 64 \, \log \left (x\right ) - 80\right )}} + \frac {8 \, {\left (x - 1\right )}}{x^{2} - 16} \]
integrate(((x^4-48*x^2+32*x)*log(x)^2+(-2*x^5+86*x^3+x^2-40*x)*log(x)+x^6- 33*x^4-63*x^3+85*x^2-20*x)/((2*x^4-64*x^2+512)*log(x)^2+(-4*x^5+108*x^3+80 *x^2-704*x-1280)*log(x)+2*x^6-44*x^4-80*x^3+242*x^2+880*x+800),x, algorith m=\
1/2*x - 5/2*(x^3 - x^2)/(x^4 - x^3*log(x) + 4*x^3 - 4*x^2*log(x) - 11*x^2 + 16*x*log(x) - 64*x + 64*log(x) - 80) + 8*(x - 1)/(x^2 - 16)
Time = 12.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=-\frac {11\,x-16\,\ln \left (x\right )+x^3\,\ln \left (x\right )-x^4+20}{2\,\left (x-4\right )\,\left (4\,x-4\,\ln \left (x\right )-x\,\ln \left (x\right )+x^2+5\right )} \]
int(-(20*x + log(x)*(40*x - x^2 - 86*x^3 + 2*x^5) - 85*x^2 + 63*x^3 + 33*x ^4 - x^6 - log(x)^2*(32*x - 48*x^2 + x^4))/(880*x - log(x)*(704*x - 80*x^2 - 108*x^3 + 4*x^5 + 1280) + log(x)^2*(2*x^4 - 64*x^2 + 512) + 242*x^2 - 8 0*x^3 - 44*x^4 + 2*x^6 + 800),x)