Integrand size = 180, antiderivative size = 27 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {\log \left (\frac {2}{x}\right )}{-1+e^3+2 x \left (2+\frac {\log (x)}{2+x}\right )} \]
Time = 0.69 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {(2+x) \left (-1+e^3+4 x\right )+2 (2+x) \log \left (\frac {2}{x}\right )+2 x \log (x)}{2 (2+x) \left (-1+e^3+4 x\right )+4 x \log (x)} \]
Integrate[(4 - 12*x - 15*x^2 - 4*x^3 + E^3*(-4 - 4*x - x^2) + (-20*x - 18* x^2 - 4*x^3)*Log[2/x] + (-4*x - 2*x^2 - 4*x*Log[2/x])*Log[x])/(4*x - 28*x^ 2 + 33*x^3 + 56*x^4 + 16*x^5 + E^6*(4*x + 4*x^2 + x^3) + E^3*(-8*x + 24*x^ 2 + 30*x^3 + 8*x^4) + (-8*x^2 + 28*x^3 + 16*x^4 + E^3*(8*x^2 + 4*x^3))*Log [x] + 4*x^3*Log[x]^2),x]
((2 + x)*(-1 + E^3 + 4*x) + 2*(2 + x)*Log[2/x] + 2*x*Log[x])/(2*(2 + x)*(- 1 + E^3 + 4*x) + 4*x*Log[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.
| \(\displaystyle \int \frac {-4 x^3-15 x^2+e^3 \left (-x^2-4 x-4\right )+\left (-2 x^2-4 x-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)+\left (-4 x^3-18 x^2-20 x\right ) \log \left (\frac {2}{x}\right )-12 x+4}{16 x^5+56 x^4+33 x^3+4 x^3 \log ^2(x)-28 x^2+e^6 \left (x^3+4 x^2+4 x\right )+e^3 \left (8 x^4+30 x^3+24 x^2-8 x\right )+\left (16 x^4+28 x^3-8 x^2+e^3 \left (4 x^3+8 x^2\right )\right ) \log (x)+4 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-2 x \log \left (\frac {2}{x}\right ) \left (2 x^2+9 x+2 \log (x)+10\right )-(x+2) \left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )}{x \left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 (x+2) \left (-2 x^2-x+e^3-1\right ) \log \left (\frac {2}{x}\right )}{x \left (-4 x^2-7 \left (1+\frac {e^3}{7}\right ) x-2 x \log (x)+2 \left (1-e^3\right )\right )^2}+\frac {x+2 \log \left (\frac {2}{x}\right )+2}{x \left (-4 x^2-7 \left (1+\frac {e^3}{7}\right ) x-2 x \log (x)+2 \left (1-e^3\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {1}{-4 x^2-2 \log (x) x-7 \left (1+\frac {e^3}{7}\right ) x+2 \left (1-e^3\right )}dx+2 \int \frac {1}{x \left (-4 x^2-2 \log (x) x-7 \left (1+\frac {e^3}{7}\right ) x+2 \left (1-e^3\right )\right )}dx+2 \int \frac {\log \left (\frac {2}{x}\right )}{x \left (-4 x^2-2 \log (x) x-7 \left (1+\frac {e^3}{7}\right ) x+2 \left (1-e^3\right )\right )}dx-4 \int \frac {x^2 \log \left (\frac {2}{x}\right )}{\left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )^2}dx-2 \left (3-e^3\right ) \int \frac {\log \left (\frac {2}{x}\right )}{\left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )^2}dx-4 \left (1-e^3\right ) \int \frac {\log \left (\frac {2}{x}\right )}{x \left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )^2}dx-10 \int \frac {x \log \left (\frac {2}{x}\right )}{\left ((x+2) \left (4 x+e^3-1\right )+2 x \log (x)\right )^2}dx\) |
Int[(4 - 12*x - 15*x^2 - 4*x^3 + E^3*(-4 - 4*x - x^2) + (-20*x - 18*x^2 - 4*x^3)*Log[2/x] + (-4*x - 2*x^2 - 4*x*Log[2/x])*Log[x])/(4*x - 28*x^2 + 33 *x^3 + 56*x^4 + 16*x^5 + E^6*(4*x + 4*x^2 + x^3) + E^3*(-8*x + 24*x^2 + 30 *x^3 + 8*x^4) + (-8*x^2 + 28*x^3 + 16*x^4 + E^3*(8*x^2 + 4*x^3))*Log[x] + 4*x^3*Log[x]^2),x]
3.14.6.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 144.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70
| method | result | size |
| parallelrisch | \(-\frac {-8 x \ln \left (\frac {2}{x}\right )-16 \ln \left (\frac {2}{x}\right )}{8 \left (2 x \ln \left (x \right )+x \,{\mathrm e}^{3}+4 x^{2}+2 \,{\mathrm e}^{3}+7 x -2\right )}\) | \(46\) |
| risch | \(-\frac {1}{x}+\frac {-4+x^{2} {\mathrm e}^{3}+2 x^{2} \ln \left (2\right )+4 x^{3}+4 x \,{\mathrm e}^{3}+4 x \ln \left (2\right )+15 x^{2}+4 \,{\mathrm e}^{3}+12 x}{2 x \left (2 x \ln \left (x \right )+x \,{\mathrm e}^{3}+4 x^{2}+2 \,{\mathrm e}^{3}+7 x -2\right )}\) | \(79\) |
int(((-4*x*ln(2/x)-2*x^2-4*x)*ln(x)+(-4*x^3-18*x^2-20*x)*ln(2/x)+(-x^2-4*x -4)*exp(3)-4*x^3-15*x^2-12*x+4)/(4*x^3*ln(x)^2+((4*x^3+8*x^2)*exp(3)+16*x^ 4+28*x^3-8*x^2)*ln(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+24*x^2-8*x)*e xp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {4 \, x^{2} + {\left (x + 2\right )} e^{3} + 2 \, x \log \left (2\right ) + 7 \, x + 4 \, \log \left (\frac {2}{x}\right ) - 2}{2 \, {\left (4 \, x^{2} + {\left (x + 2\right )} e^{3} + 2 \, x \log \left (2\right ) - 2 \, x \log \left (\frac {2}{x}\right ) + 7 \, x - 2\right )}} \]
integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+ (-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*ex p(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+24 *x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm=\
1/2*(4*x^2 + (x + 2)*e^3 + 2*x*log(2) + 7*x + 4*log(2/x) - 2)/(4*x^2 + (x + 2)*e^3 + 2*x*log(2) - 2*x*log(2/x) + 7*x - 2)
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (20) = 40\).
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.26 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {4 x^{3} + 2 x^{2} \log {\left (2 \right )} + 15 x^{2} + x^{2} e^{3} + 4 x \log {\left (2 \right )} + 12 x + 4 x e^{3} - 4 + 4 e^{3}}{8 x^{3} + 4 x^{2} \log {\left (x \right )} + 14 x^{2} + 2 x^{2} e^{3} - 4 x + 4 x e^{3}} - \frac {1}{x} \]
integrate(((-4*x*ln(2/x)-2*x**2-4*x)*ln(x)+(-4*x**3-18*x**2-20*x)*ln(2/x)+ (-x**2-4*x-4)*exp(3)-4*x**3-15*x**2-12*x+4)/(4*x**3*ln(x)**2+((4*x**3+8*x* *2)*exp(3)+16*x**4+28*x**3-8*x**2)*ln(x)+(x**3+4*x**2+4*x)*exp(3)**2+(8*x* *4+30*x**3+24*x**2-8*x)*exp(3)+16*x**5+56*x**4+33*x**3-28*x**2+4*x),x)
(4*x**3 + 2*x**2*log(2) + 15*x**2 + x**2*exp(3) + 4*x*log(2) + 12*x + 4*x* exp(3) - 4 + 4*exp(3))/(8*x**3 + 4*x**2*log(x) + 14*x**2 + 2*x**2*exp(3) - 4*x + 4*x*exp(3)) - 1/x
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {4 \, x^{2} + x {\left (e^{3} + 2 \, \log \left (2\right ) + 7\right )} + 2 \, e^{3} + 4 \, \log \left (2\right ) - 4 \, \log \left (x\right ) - 2}{2 \, {\left (4 \, x^{2} + x {\left (e^{3} + 7\right )} + 2 \, x \log \left (x\right ) + 2 \, e^{3} - 2\right )}} \]
integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+ (-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*ex p(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+24 *x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm=\
1/2*(4*x^2 + x*(e^3 + 2*log(2) + 7) + 2*e^3 + 4*log(2) - 4*log(x) - 2)/(4* x^2 + x*(e^3 + 7) + 2*x*log(x) + 2*e^3 - 2)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\frac {4 \, x^{2} + x e^{3} + 2 \, x \log \left (2\right ) + 7 \, x + 2 \, e^{3} + 4 \, \log \left (2\right ) - 4 \, \log \left (x\right ) - 2}{2 \, {\left (4 \, x^{2} + x e^{3} + 2 \, x \log \left (x\right ) + 7 \, x + 2 \, e^{3} - 2\right )}} \]
integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+ (-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*ex p(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+24 *x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm=\
1/2*(4*x^2 + x*e^3 + 2*x*log(2) + 7*x + 2*e^3 + 4*log(2) - 4*log(x) - 2)/( 4*x^2 + x*e^3 + 2*x*log(x) + 7*x + 2*e^3 - 2)
Timed out. \[ \int \frac {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx=\int -\frac {12\,x+\ln \left (x\right )\,\left (4\,x+4\,x\,\ln \left (\frac {2}{x}\right )+2\,x^2\right )+\ln \left (\frac {2}{x}\right )\,\left (4\,x^3+18\,x^2+20\,x\right )+{\mathrm {e}}^3\,\left (x^2+4\,x+4\right )+15\,x^2+4\,x^3-4}{4\,x+{\mathrm {e}}^6\,\left (x^3+4\,x^2+4\,x\right )+4\,x^3\,{\ln \left (x\right )}^2+{\mathrm {e}}^3\,\left (8\,x^4+30\,x^3+24\,x^2-8\,x\right )-28\,x^2+33\,x^3+56\,x^4+16\,x^5+\ln \left (x\right )\,\left ({\mathrm {e}}^3\,\left (4\,x^3+8\,x^2\right )-8\,x^2+28\,x^3+16\,x^4\right )} \,d x \]
int(-(12*x + log(x)*(4*x + 4*x*log(2/x) + 2*x^2) + log(2/x)*(20*x + 18*x^2 + 4*x^3) + exp(3)*(4*x + x^2 + 4) + 15*x^2 + 4*x^3 - 4)/(4*x + exp(6)*(4* x + 4*x^2 + x^3) + 4*x^3*log(x)^2 + exp(3)*(24*x^2 - 8*x + 30*x^3 + 8*x^4) - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + log(x)*(exp(3)*(8*x^2 + 4*x^3) - 8* x^2 + 28*x^3 + 16*x^4)),x)
int(-(12*x + log(x)*(4*x + 4*x*log(2/x) + 2*x^2) + log(2/x)*(20*x + 18*x^2 + 4*x^3) + exp(3)*(4*x + x^2 + 4) + 15*x^2 + 4*x^3 - 4)/(4*x + exp(6)*(4* x + 4*x^2 + x^3) + 4*x^3*log(x)^2 + exp(3)*(24*x^2 - 8*x + 30*x^3 + 8*x^4) - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + log(x)*(exp(3)*(8*x^2 + 4*x^3) - 8* x^2 + 28*x^3 + 16*x^4)), x)