Integrand size = 78, antiderivative size = 30 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=x \left (1+x-\left (-e^4-x+4 \log \left (\frac {2}{x}\right )-\log (x)\right )^2\right ) \] Output:
x*(1+x-(4*ln(2/x)-ln(x)-x-exp(4))^2)
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=-x \left (-1+e^8-x+2 e^4 x+x^2+16 \log ^2\left (\frac {2}{x}\right )+2 \left (e^4+x\right ) \log (x)+\log ^2(x)-8 \log \left (\frac {2}{x}\right ) \left (e^4+x+\log (x)\right )\right ) \] Input:
Integrate[1 - E^8 + E^4*(-10 - 4*x) - 8*x - 3*x^2 + (40 + 8*E^4 + 16*x)*Lo g[2/x] - 16*Log[2/x]^2 + (-10 - 2*E^4 - 4*x + 8*Log[2/x])*Log[x] - Log[x]^ 2,x]
Output:
-(x*(-1 + E^8 - x + 2*E^4*x + x^2 + 16*Log[2/x]^2 + 2*(E^4 + x)*Log[x] + L og[x]^2 - 8*Log[2/x]*(E^4 + x + Log[x])))
Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(30)=60\).
Time = 0.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.40, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {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.
\(\displaystyle \int \left (-3 x^2-8 x+e^4 (-4 x-10)-16 \log ^2\left (\frac {2}{x}\right )-\log ^2(x)+\left (16 x+8 e^4+40\right ) \log \left (\frac {2}{x}\right )+\left (-4 x+8 \log \left (\frac {2}{x}\right )-2 e^4-10\right ) \log (x)-e^8+1\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^3+x^2-2 x^2 \log (x)+\left (1-e^8\right ) x+10 \left (5+e^4\right ) x-50 x-\frac {1}{2} e^4 (2 x+5)^2-16 x \log ^2\left (\frac {2}{x}\right )-x \log ^2(x)-40 x \log \left (\frac {2}{x}\right )+8 x \log \left (\frac {2}{x}\right ) \log (x)-2 \left (5+e^4\right ) x \log (x)+10 x \log (x)+2 \left (2 x+e^4+5\right )^2 \log \left (\frac {2}{x}\right )+2 \left (5+e^4\right )^2 \log (x)\) |
Input:
Int[1 - E^8 + E^4*(-10 - 4*x) - 8*x - 3*x^2 + (40 + 8*E^4 + 16*x)*Log[2/x] - 16*Log[2/x]^2 + (-10 - 2*E^4 - 4*x + 8*Log[2/x])*Log[x] - Log[x]^2,x]
Output:
-50*x + 10*(5 + E^4)*x + (1 - E^8)*x + x^2 - x^3 - (E^4*(5 + 2*x)^2)/2 - 4 0*x*Log[2/x] + 2*(5 + E^4 + 2*x)^2*Log[2/x] - 16*x*Log[2/x]^2 + 2*(5 + E^4 )^2*Log[x] + 10*x*Log[x] - 2*(5 + E^4)*x*Log[x] - 2*x^2*Log[x] + 8*x*Log[2 /x]*Log[x] - x*Log[x]^2
Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(29)=58\).
Time = 0.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.07
method | result | size |
parallelrisch | \(-2 x^{2} {\mathrm e}^{4}-2 x \,{\mathrm e}^{4} \ln \left (x \right )+8 \,{\mathrm e}^{4} x \ln \left (\frac {2}{x}\right )-x^{3}-2 x^{2} \ln \left (x \right )+8 \ln \left (\frac {2}{x}\right ) x^{2}-x \ln \left (x \right )^{2}+8 x \ln \left (\frac {2}{x}\right ) \ln \left (x \right )-16 \ln \left (\frac {2}{x}\right )^{2} x +x^{2}+\left (1-{\mathrm e}^{8}\right ) x\) | \(92\) |
risch | \(-9 x \ln \left (x \right )^{2}-2 x \,{\mathrm e}^{4} \ln \left (x \right )+8 x \ln \left (2\right ) \ln \left (x \right )-2 x^{2} \ln \left (x \right )-16 \ln \left (\frac {2}{x}\right )^{2} x +8 \,{\mathrm e}^{4} x \ln \left (\frac {2}{x}\right )+8 \ln \left (\frac {2}{x}\right ) x^{2}-x \,{\mathrm e}^{8}-2 x^{2} {\mathrm e}^{4}-x^{3}+8 x \ln \left (x \right )+8 x \ln \left (\frac {2}{x}\right )-8 x \ln \left (2\right )+x^{2}+x\) | \(103\) |
parts | \(x +8 \,{\mathrm e}^{4} x \ln \left (\frac {2}{x}\right )-2 x \,{\mathrm e}^{4}+8 x \ln \left (\frac {2}{x}\right )+8 \ln \left (\frac {2}{x}\right ) x^{2}+x^{2}-2 x^{2} \ln \left (x \right )-2 \,{\mathrm e}^{4} \left (x \ln \left (x \right )-x \right )+8 \ln \left (2\right ) \left (x \ln \left (x \right )-x \right )-8 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right ) \left (-x \ln \left (\frac {1}{x}\right )-x \right )-8 \ln \left (\frac {1}{x}\right )^{2} x -16 x \ln \left (\frac {1}{x}\right )-8 x \ln \left (x \right )-x^{3}-x \,{\mathrm e}^{8}-x \ln \left (x \right )^{2}-16 \ln \left (\frac {2}{x}\right )^{2} x -2 x^{2} {\mathrm e}^{4}\) | \(151\) |
default | \(x +2 \,{\mathrm e}^{4} \left (-x^{2}-5 x \right )+8 \,{\mathrm e}^{4} x \ln \left (\frac {2}{x}\right )+8 x \,{\mathrm e}^{4}+8 x \ln \left (\frac {2}{x}\right )+8 \ln \left (\frac {2}{x}\right ) x^{2}+x^{2}-2 x^{2} \ln \left (x \right )-2 \,{\mathrm e}^{4} \left (x \ln \left (x \right )-x \right )+8 \ln \left (2\right ) \left (x \ln \left (x \right )-x \right )-8 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right ) \left (-x \ln \left (\frac {1}{x}\right )-x \right )-8 \ln \left (\frac {1}{x}\right )^{2} x -16 x \ln \left (\frac {1}{x}\right )-8 x \ln \left (x \right )-x^{3}-x \,{\mathrm e}^{8}-x \ln \left (x \right )^{2}-16 \ln \left (\frac {2}{x}\right )^{2} x\) | \(157\) |
orering | \(\left (-\ln \left (x \right )^{2}+\left (8 \ln \left (\frac {2}{x}\right )-2 \,{\mathrm e}^{4}-4 x -10\right ) \ln \left (x \right )-16 \ln \left (\frac {2}{x}\right )^{2}+\left (8 \,{\mathrm e}^{4}+16 x +40\right ) \ln \left (\frac {2}{x}\right )-{\mathrm e}^{8}+\left (-4 x -10\right ) {\mathrm e}^{4}-3 x^{2}-8 x +1\right ) x -\frac {x \left (6 x^{4}+60 x^{3}+145 x^{2}+625\right ) \left (-\frac {2 \ln \left (x \right )}{x}+\left (-\frac {8}{x}-4\right ) \ln \left (x \right )+\frac {8 \ln \left (\frac {2}{x}\right )-2 \,{\mathrm e}^{4}-4 x -10}{x}+\frac {32 \ln \left (\frac {2}{x}\right )}{x}+16 \ln \left (\frac {2}{x}\right )-\frac {8 \,{\mathrm e}^{4}+16 x +40}{x}-4 \,{\mathrm e}^{4}-6 x -8\right )}{2 \left (6 x^{3}+50 x^{2}+120 x +125\right )}+\frac {\left (2 x^{4}+30 x^{3}+145 x^{2}-625\right ) x^{2} \left (-\frac {34}{x^{2}}+\frac {10 \ln \left (x \right )}{x^{2}}+\frac {-\frac {16}{x}-8}{x}-\frac {8 \ln \left (\frac {2}{x}\right )-2 \,{\mathrm e}^{4}-4 x -10}{x^{2}}-\frac {32 \ln \left (\frac {2}{x}\right )}{x^{2}}-\frac {32}{x}+\frac {8 \,{\mathrm e}^{4}+16 x +40}{x^{2}}-6\right )}{12 x^{3}+100 x^{2}+240 x +250}\) | \(313\) |
Input:
int(-ln(x)^2+(8*ln(2/x)-2*exp(4)-4*x-10)*ln(x)-16*ln(2/x)^2+(8*exp(4)+16*x +40)*ln(2/x)-exp(4)^2+(-4*x-10)*exp(4)-3*x^2-8*x+1,x,method=_RETURNVERBOSE )
Output:
-2*x^2*exp(4)-2*x*exp(4)*ln(x)+8*exp(4)*x*ln(2/x)-x^3-2*x^2*ln(x)+8*ln(2/x )*x^2-x*ln(x)^2+8*x*ln(2/x)*ln(x)-16*ln(2/x)^2*x+x^2+(1-exp(4)^2)*x
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (24) = 48\).
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=-x^{3} - 2 \, x^{2} e^{4} - x \log \left (2\right )^{2} - 25 \, x \log \left (\frac {2}{x}\right )^{2} + x^{2} - x e^{8} - 2 \, {\left (x^{2} + x e^{4}\right )} \log \left (2\right ) + 10 \, {\left (x^{2} + x e^{4} + x \log \left (2\right )\right )} \log \left (\frac {2}{x}\right ) + x \] Input:
integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*e xp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm ="fricas")
Output:
-x^3 - 2*x^2*e^4 - x*log(2)^2 - 25*x*log(2/x)^2 + x^2 - x*e^8 - 2*(x^2 + x *e^4)*log(2) + 10*(x^2 + x*e^4 + x*log(2))*log(2/x) + x
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=- x^{3} + x^{2} \left (- 2 e^{4} + 1 + 8 \log {\left (2 \right )}\right ) - 25 x \log {\left (x \right )}^{2} + x \left (- e^{8} - 16 \log {\left (2 \right )}^{2} + 1 + 8 e^{4} \log {\left (2 \right )}\right ) + \left (- 10 x^{2} - 10 x e^{4} + 40 x \log {\left (2 \right )}\right ) \log {\left (x \right )} \] Input:
integrate(-ln(x)**2+(8*ln(2/x)-2*exp(4)-4*x-10)*ln(x)-16*ln(2/x)**2+(8*exp (4)+16*x+40)*ln(2/x)-exp(4)**2+(-4*x-10)*exp(4)-3*x**2-8*x+1,x)
Output:
-x**3 + x**2*(-2*exp(4) + 1 + 8*log(2)) - 25*x*log(x)**2 + x*(-exp(8) - 16 *log(2)**2 + 1 + 8*exp(4)*log(2)) + (-10*x**2 - 10*x*exp(4) + 40*x*log(2)) *log(x)
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 4.17 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=-x^{3} - 16 \, x \log \left (\frac {2}{x}\right )^{2} - {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x + x^{2} + 2 \, x {\left (e^{4} - 4 \, \log \left (2\right ) - 3\right )} + 8 \, x {\left (e^{4} + 5\right )} - x e^{8} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{4} - 2 \, {\left (x^{2} + x e^{4} - 4 \, x \log \left (\frac {2}{x}\right ) + x\right )} \log \left (x\right ) + 8 \, x \log \left (x\right ) + 8 \, {\left (x^{2} + x e^{4} + 5 \, x\right )} \log \left (\frac {2}{x}\right ) - 32 \, x \log \left (\frac {2}{x}\right ) - 31 \, x \] Input:
integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*e xp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm ="maxima")
Output:
-x^3 - 16*x*log(2/x)^2 - (log(x)^2 - 2*log(x) + 2)*x + x^2 + 2*x*(e^4 - 4* log(2) - 3) + 8*x*(e^4 + 5) - x*e^8 - 2*(x^2 + 5*x)*e^4 - 2*(x^2 + x*e^4 - 4*x*log(2/x) + x)*log(x) + 8*x*log(x) + 8*(x^2 + x*e^4 + 5*x)*log(2/x) - 32*x*log(2/x) - 31*x
Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 4.37 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=-x^{3} + 4 \, x^{2} {\left (\frac {2 \, e^{4}}{x} + \frac {10}{x} + 1\right )} - 2 \, x^{2} \log \left (x\right ) - 2 \, x e^{4} \log \left (x\right ) + 8 \, x \log \left (2\right ) \log \left (x\right ) - 9 \, x \log \left (x\right )^{2} - 16 \, x \log \left (\frac {2}{x}\right )^{2} - 3 \, x^{2} - x e^{8} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{4} + 2 \, x e^{4} - 8 \, x \log \left (2\right ) + 8 \, x \log \left (x\right ) + 8 \, {\left (x^{2} + x e^{4} + 5 \, x\right )} \log \left (\frac {2}{x}\right ) - 32 \, x \log \left (\frac {2}{x}\right ) - 39 \, x \] Input:
integrate(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*e xp(4)+16*x+40)*log(2/x)-exp(4)^2+(-4*x-10)*exp(4)-3*x^2-8*x+1,x, algorithm ="giac")
Output:
-x^3 + 4*x^2*(2*e^4/x + 10/x + 1) - 2*x^2*log(x) - 2*x*e^4*log(x) + 8*x*lo g(2)*log(x) - 9*x*log(x)^2 - 16*x*log(2/x)^2 - 3*x^2 - x*e^8 - 2*(x^2 + 5* x)*e^4 + 2*x*e^4 - 8*x*log(2) + 8*x*log(x) + 8*(x^2 + x*e^4 + 5*x)*log(2/x ) - 32*x*log(2/x) - 39*x
Time = 3.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=-x\,\left ({\mathrm {e}}^8-x+16\,{\ln \left (\frac {2}{x}\right )}^2+2\,x\,{\mathrm {e}}^4+{\ln \left (x\right )}^2-8\,{\mathrm {e}}^4\,\ln \left (\frac {2}{x}\right )-8\,x\,\ln \left (\frac {2}{x}\right )+2\,{\mathrm {e}}^4\,\ln \left (x\right )+2\,x\,\ln \left (x\right )+x^2-8\,\ln \left (\frac {2}{x}\right )\,\ln \left (x\right )-1\right ) \] Input:
int(log(2/x)*(16*x + 8*exp(4) + 40) - exp(8) - 16*log(2/x)^2 - log(x)*(4*x + 2*exp(4) - 8*log(2/x) + 10) - log(x)^2 - 3*x^2 - exp(4)*(4*x + 10) - 8* x + 1,x)
Output:
-x*(exp(8) - x + 16*log(2/x)^2 + 2*x*exp(4) + log(x)^2 - 8*exp(4)*log(2/x) - 8*x*log(2/x) + 2*exp(4)*log(x) + 2*x*log(x) + x^2 - 8*log(2/x)*log(x) - 1)
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.63 \[ \int \left (1-e^8+e^4 (-10-4 x)-8 x-3 x^2+\left (40+8 e^4+16 x\right ) \log \left (\frac {2}{x}\right )-16 \log ^2\left (\frac {2}{x}\right )+\left (-10-2 e^4-4 x+8 \log \left (\frac {2}{x}\right )\right ) \log (x)-\log ^2(x)\right ) \, dx=x \left (-16 \mathrm {log}\left (\frac {2}{x}\right )^{2}+8 \,\mathrm {log}\left (\frac {2}{x}\right ) \mathrm {log}\left (x \right )+8 \,\mathrm {log}\left (\frac {2}{x}\right ) e^{4}+8 \,\mathrm {log}\left (\frac {2}{x}\right ) x -\mathrm {log}\left (x \right )^{2}-2 \,\mathrm {log}\left (x \right ) e^{4}-2 \,\mathrm {log}\left (x \right ) x -e^{8}-2 e^{4} x -x^{2}+x +1\right ) \] Input:
int(-log(x)^2+(8*log(2/x)-2*exp(4)-4*x-10)*log(x)-16*log(2/x)^2+(8*exp(4)+ 16*x+40)*log(2/x)-exp(4)^2+(-4*x-10)*exp(4)-3*x^2-8*x+1,x)
Output:
x*( - 16*log(2/x)**2 + 8*log(2/x)*log(x) + 8*log(2/x)*e**4 + 8*log(2/x)*x - log(x)**2 - 2*log(x)*e**4 - 2*log(x)*x - e**8 - 2*e**4*x - x**2 + x + 1)