\(\int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+(-40 x+x^2+86 x^3-2 x^5) \log (x)+(32 x-48 x^2+x^4) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+(-1280-704 x+80 x^2+108 x^3-4 x^5) \log (x)+(512-64 x^2+2 x^4) \log ^2(x)} \, dx\) [6856]

Optimal result

Integrand size = 127, antiderivative size = 34 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=-\frac {x \left (-x+x^2\right )}{2 (4-x) \left (4+x+\frac {5}{x-\log (x)}\right )} \]

[Out]

x/(-x+4)/(-8-2*x-10/(x-ln(x)))*(x^2-x)

Rubi [F]

\[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx \]

[In]

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)*Lo
g[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]

[Out]

x/2 - ((16 - x)*x)/(2*(16 - x^2)) + 10*Defer[Int][(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])^(-2), x] + (645*Defer[
Int][1/((-4 + x)*(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])^2), x])/8 + 5*Defer[Int][x/(5 + 4*x + x^2 - 4*Log[x] -
x*Log[x])^2, x] + (5*Defer[Int][x^2/(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])^2, x])/2 - 125*Defer[Int][1/((4 + x)
^2*(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])^2), x] + (575*Defer[Int][1/((4 + x)*(5 + 4*x + x^2 - 4*Log[x] - x*Log
[x])^2), x])/8 + 15*Defer[Int][1/((-4 + x)^2*(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])), x] + (15*Defer[Int][1/((-
4 + x)*(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])), x])/8 + 50*Defer[Int][1/((4 + x)^2*(5 + 4*x + x^2 - 4*Log[x] -
x*Log[x])), x] - (115*Defer[Int][1/((4 + x)*(5 + 4*x + x^2 - 4*Log[x] - x*Log[x])), x])/8

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (-20+85 x-63 x^2-33 x^3+x^5+\left (-40+x+86 x^2-2 x^4\right ) \log (x)+\left (32-48 x+x^3\right ) \log ^2(x)\right )}{2 (4-x)^2 \left (5+4 x+x^2-(4+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {x \left (-20+85 x-63 x^2-33 x^3+x^5+\left (-40+x+86 x^2-2 x^4\right ) \log (x)+\left (32-48 x+x^3\right ) \log ^2(x)\right )}{(4-x)^2 \left (5+4 x+x^2-(4+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {x \left (32-48 x+x^3\right )}{\left (-16+x^2\right )^2}+\frac {5 x \left (16-19 x-4 x^2+6 x^3+x^4\right )}{(-4+x) (4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}-\frac {5 x \left (32-52 x+5 x^2\right )}{\left (-16+x^2\right )^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {x \left (32-48 x+x^3\right )}{\left (-16+x^2\right )^2} \, dx+\frac {5}{2} \int \frac {x \left (16-19 x-4 x^2+6 x^3+x^4\right )}{(-4+x) (4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx-\frac {5}{2} \int \frac {x \left (32-52 x+5 x^2\right )}{\left (-16+x^2\right )^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx \\ & = -\frac {(16-x) x}{2 \left (16-x^2\right )}+\frac {1}{64} \int \frac {-512+32 x^2}{-16+x^2} \, dx+\frac {5}{2} \int \left (\frac {4}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}+\frac {129}{4 (-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}+\frac {2 x}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}+\frac {x^2}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}-\frac {50}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}+\frac {115}{4 (4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2}\right ) \, dx-\frac {5}{2} \int \left (-\frac {6}{(-4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )}-\frac {3}{4 (-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )}-\frac {20}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )}+\frac {23}{4 (4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )}\right ) \, dx \\ & = -\frac {(16-x) x}{2 \left (16-x^2\right )}+\frac {\int 1 \, dx}{2}+\frac {15}{8} \int \frac {1}{(-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+\frac {5}{2} \int \frac {x^2}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+5 \int \frac {x}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+10 \int \frac {1}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx-\frac {115}{8} \int \frac {1}{(4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+15 \int \frac {1}{(-4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+50 \int \frac {1}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+\frac {575}{8} \int \frac {1}{(4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+\frac {645}{8} \int \frac {1}{(-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx-125 \int \frac {1}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx \\ & = \frac {x}{2}-\frac {(16-x) x}{2 \left (16-x^2\right )}+\frac {15}{8} \int \frac {1}{(-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+\frac {5}{2} \int \frac {x^2}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+5 \int \frac {x}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+10 \int \frac {1}{\left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx-\frac {115}{8} \int \frac {1}{(4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+15 \int \frac {1}{(-4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+50 \int \frac {1}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )} \, dx+\frac {575}{8} \int \frac {1}{(4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx+\frac {645}{8} \int \frac {1}{(-4+x) \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx-125 \int \frac {1}{(4+x)^2 \left (5+4 x+x^2-4 \log (x)-x \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {-20-11 x+x^4-\left (-16+x^3\right ) \log (x)}{2 (-4+x) \left (5+4 x+x^2-(4+x) \log (x)\right )} \]

[In]

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)*Lo
g[x] + (512 - 64*x^2 + 2*x^4)*Log[x]^2),x]

[Out]

(-20 - 11*x + x^4 - (-16 + x^3)*Log[x])/(2*(-4 + x)*(5 + 4*x + x^2 - (4 + x)*Log[x]))

Maple [A] (verified)

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	\(\frac {x^{3}-16}{2 x^{2}-32}-\frac {5 \left (-1+x \right ) x^{2}}{2 \left (4+x \right ) \left (x -4\right ) \left (x^{2}-x \ln \left (x \right )+4 x -4 \ln \left (x \right )+5\right )}\)	\(53\)

[In]

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^4-63*x^3+85*x^2-20*x)/((2*x^4-64*x^2+51
2)*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=_RETURNVER
BOSE)

[Out]

1/2*(x^4-x^3*ln(x)-20-11*x+16*ln(x))/(x^3-x^2*ln(x)-11*x+16*ln(x)-20)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x^{4} - {\left (x^{3} - 16\right )} \log \left (x\right ) - 11 \, x - 20}{2 \, {\left (x^{3} - {\left (x^{2} - 16\right )} \log \left (x\right ) - 11 \, x - 20\right )}} \]

[In]

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^4-6
4*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, algori
thm="fricas")

[Out]

1/2*(x^4 - (x^3 - 16)*log(x) - 11*x - 20)/(x^3 - (x^2 - 16)*log(x) - 11*x - 20)

Sympy [B] (verification not implemented)

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 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x}{2} + \frac {8 x - 8}{x^{2} - 16} + \frac {5 x^{3} - 5 x^{2}}{- 2 x^{4} - 8 x^{3} + 22 x^{2} + 128 x + \left (2 x^{3} + 8 x^{2} - 32 x - 128\right ) \log {\left (x \right )} + 160} \]

[In]

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)

[Out]

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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {x^{4} - {\left (x^{3} - 16\right )} \log \left (x\right ) - 11 \, x - 20}{2 \, {\left (x^{3} - {\left (x^{2} - 16\right )} \log \left (x\right ) - 11 \, x - 20\right )}} \]

[In]

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^4-6
4*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, algori
thm="maxima")

[Out]

1/2*(x^4 - (x^3 - 16)*log(x) - 11*x - 20)/(x^3 - (x^2 - 16)*log(x) - 11*x - 20)

Giac [B] (verification not implemented)

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 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=\frac {1}{2} \, x - \frac {5 \, {\left (x^{3} - x^{2}\right )}}{2 \, {\left (x^{4} - x^{3} \log \left (x\right ) + 4 \, x^{3} - 4 \, x^{2} \log \left (x\right ) - 11 \, x^{2} + 16 \, x \log \left (x\right ) - 64 \, x + 64 \, \log \left (x\right ) - 80\right )}} + \frac {8 \, {\left (x - 1\right )}}{x^{2} - 16} \]

[In]

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^4-6
4*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, algori
thm="giac")

[Out]

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) - 8
0) + 8*(x - 1)/(x^2 - 16)

Mupad [B] (verification not implemented)

Time = 12.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {-20 x+85 x^2-63 x^3-33 x^4+x^6+\left (-40 x+x^2+86 x^3-2 x^5\right ) \log (x)+\left (32 x-48 x^2+x^4\right ) \log ^2(x)}{800+880 x+242 x^2-80 x^3-44 x^4+2 x^6+\left (-1280-704 x+80 x^2+108 x^3-4 x^5\right ) \log (x)+\left (512-64 x^2+2 x^4\right ) \log ^2(x)} \, dx=-\frac {11\,x-16\,\ln \left (x\right )+x^3\,\ln \left (x\right )-x^4+20}{2\,\left (x-4\right )\,\left (4\,x-4\,\ln \left (x\right )-x\,\ln \left (x\right )+x^2+5\right )} \]

[In]

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 -
80*x^3 - 44*x^4 + 2*x^6 + 800),x)

[Out]

-(11*x - 16*log(x) + x^3*log(x) - x^4 + 20)/(2*(x - 4)*(4*x - 4*log(x) - x*log(x) + x^2 + 5))