Integrand size = 122, antiderivative size = 28 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {2 x^3 \log (x)}{\left (4+\frac {e^{5-x}}{4 x}-x\right )^2} \] Output:
2*ln(x)*x^3/(4-x+1/4*exp(5-ln(x)-x))^2
Time = 5.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 e^{2 x} x^5 \log (x)}{\left (e^5-4 e^x (-4+x) x\right )^2} \] Input:
Integrate[(512*x^2 - 128*x^3 + (1536*x^2 - 128*x^3)*Log[x] + (E^(5 - x)*(3 2*x^2 + (160*x^2 + 64*x^3)*Log[x]))/x)/(4096 + E^(15 - 3*x)/x^3 + (E^(10 - 2*x)*(48 - 12*x))/x^2 - 3072*x + 768*x^2 - 64*x^3 + (E^(5 - x)*(768 - 384 *x + 48*x^2))/x),x]
Output:
(32*E^(2*x)*x^5*Log[x])/(E^5 - 4*E^x*(-4 + x)*x)^2
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.
| \(\displaystyle \int \frac {-128 x^3+512 x^2+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (64 x^3+160 x^2\right ) \log (x)\right )}{x}}{-64 x^3+\frac {e^{15-3 x}}{x^3}+768 x^2+\frac {e^{5-x} \left (48 x^2-384 x+768\right )}{x}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+4096} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {32 e^{2 x} x^4 \left (-4 e^x (x-4) x-4 e^x (x-12) x \log (x)+e^5 (2 x+5) \log (x)+e^5\right )}{\left (e^5-4 e^x (x-4) x\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 32 \int \frac {e^{2 x} x^4 \left (4 e^x (4-x) x+4 e^x (12-x) \log (x) x+e^5 (2 x+5) \log (x)+e^5\right )}{\left (4 e^x (4-x) x+e^5\right )^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 32 \int \left (\frac {2 e^{2 x+5} \left (x^2-2 x-4\right ) \log (x) x^4}{(x-4) \left (-4 e^x x^2+16 e^x x+e^5\right )^3}+\frac {e^{2 x} (\log (x) x+x-12 \log (x)-4) x^4}{(x-4) \left (4 e^x x^2-16 e^x x-e^5\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 32 \int \left (\frac {e^{2 x} x^4 (\log (x) x+x-12 \log (x)-4)}{(x-4) \left (4 e^x x^2-16 e^x x-e^5\right )^2}-\frac {2 e^{2 x+5} x^4 \left (x^2-2 x-4\right ) \log (x)}{(x-4) \left (4 e^x x^2-16 e^x x-e^5\right )^3}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle 32 \int \left (\frac {e^{2 x} x^4 (\log (x) x+x-12 \log (x)-4)}{(x-4) \left (4 e^x x^2-16 e^x x-e^5\right )^2}-\frac {2 e^{2 x+5} x^4 \left (x^2-2 x-4\right ) \log (x)}{(x-4) \left (4 e^x x^2-16 e^x x-e^5\right )^3}\right )dx\) |
Input:
Int[(512*x^2 - 128*x^3 + (1536*x^2 - 128*x^3)*Log[x] + (E^(5 - x)*(32*x^2 + (160*x^2 + 64*x^3)*Log[x]))/x)/(4096 + E^(15 - 3*x)/x^3 + (E^(10 - 2*x)* (48 - 12*x))/x^2 - 3072*x + 768*x^2 - 64*x^3 + (E^(5 - x)*(768 - 384*x + 4 8*x^2))/x),x]
Output:
$Aborted
Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
| method | result | size |
| risch | \(\frac {32 x^{3} \ln \left (x \right )}{\left (4 x -\frac {{\mathrm e}^{5-x}}{x}-16\right )^{2}}\) | \(26\) |
| parallelrisch | \(\frac {32 x^{3} \ln \left (x \right )}{16 x^{2}-8 \,{\mathrm e}^{5-\ln \left (x \right )-x} x +\frac {{\mathrm e}^{-2 x +10}}{x^{2}}-128 x +32 \,{\mathrm e}^{5-\ln \left (x \right )-x}+256}\) | \(57\) |
Input:
int((((64*x^3+160*x^2)*ln(x)+32*x^2)*exp(5-ln(x)-x)+(-128*x^3+1536*x^2)*ln (x)-128*x^3+512*x^2)/(exp(5-ln(x)-x)^3+(-12*x+48)*exp(5-ln(x)-x)^2+(48*x^2 -384*x+768)*exp(5-ln(x)-x)-64*x^3+768*x^2-3072*x+4096),x,method=_RETURNVER BOSE)
Output:
32*x^3*ln(x)/(4*x-1/x*exp(5-x)-16)^2
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 \, x^{3} \log \left (x\right )}{16 \, x^{2} - 8 \, {\left (x - 4\right )} e^{\left (-x - \log \left (x\right ) + 5\right )} - 128 \, x + e^{\left (-2 \, x - 2 \, \log \left (x\right ) + 10\right )} + 256} \] Input:
integrate((((64*x^3+160*x^2)*log(x)+32*x^2)*exp(5-log(x)-x)+(-128*x^3+1536 *x^2)*log(x)-128*x^3+512*x^2)/(exp(5-log(x)-x)^3+(-12*x+48)*exp(5-log(x)-x )^2+(48*x^2-384*x+768)*exp(5-log(x)-x)-64*x^3+768*x^2-3072*x+4096),x, algo rithm="fricas")
Output:
32*x^3*log(x)/(16*x^2 - 8*(x - 4)*e^(-x - log(x) + 5) - 128*x + e^(-2*x - 2*log(x) + 10) + 256)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 x^{5} \log {\left (x \right )}}{16 x^{4} - 128 x^{3} + 256 x^{2} + \left (- 8 x^{2} + 32 x\right ) e^{5 - x} + e^{10 - 2 x}} \] Input:
integrate((((64*x**3+160*x**2)*ln(x)+32*x**2)*exp(5-ln(x)-x)+(-128*x**3+15 36*x**2)*ln(x)-128*x**3+512*x**2)/(exp(5-ln(x)-x)**3+(-12*x+48)*exp(5-ln(x )-x)**2+(48*x**2-384*x+768)*exp(5-ln(x)-x)-64*x**3+768*x**2-3072*x+4096),x )
Output:
32*x**5*log(x)/(16*x**4 - 128*x**3 + 256*x**2 + (-8*x**2 + 32*x)*exp(5 - x ) + exp(10 - 2*x))
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 \, x^{5} e^{\left (2 \, x\right )} \log \left (x\right )}{16 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} e^{\left (2 \, x\right )} - 8 \, {\left (x^{2} e^{5} - 4 \, x e^{5}\right )} e^{x} + e^{10}} \] Input:
integrate((((64*x^3+160*x^2)*log(x)+32*x^2)*exp(5-log(x)-x)+(-128*x^3+1536 *x^2)*log(x)-128*x^3+512*x^2)/(exp(5-log(x)-x)^3+(-12*x+48)*exp(5-log(x)-x )^2+(48*x^2-384*x+768)*exp(5-log(x)-x)-64*x^3+768*x^2-3072*x+4096),x, algo rithm="maxima")
Output:
32*x^5*e^(2*x)*log(x)/(16*(x^4 - 8*x^3 + 16*x^2)*e^(2*x) - 8*(x^2*e^5 - 4* x*e^5)*e^x + e^10)
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 \, x^{5} \log \left (x\right )}{16 \, x^{4} - 128 \, x^{3} - 8 \, x^{2} e^{\left (-x + 5\right )} + 256 \, x^{2} + 32 \, x e^{\left (-x + 5\right )} + e^{\left (-2 \, x + 10\right )}} \] Input:
integrate((((64*x^3+160*x^2)*log(x)+32*x^2)*exp(5-log(x)-x)+(-128*x^3+1536 *x^2)*log(x)-128*x^3+512*x^2)/(exp(5-log(x)-x)^3+(-12*x+48)*exp(5-log(x)-x )^2+(48*x^2-384*x+768)*exp(5-log(x)-x)-64*x^3+768*x^2-3072*x+4096),x, algo rithm="giac")
Output:
32*x^5*log(x)/(16*x^4 - 128*x^3 - 8*x^2*e^(-x + 5) + 256*x^2 + 32*x*e^(-x + 5) + e^(-2*x + 10))
Timed out. \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\int \frac {\ln \left (x\right )\,\left (1536\,x^2-128\,x^3\right )+512\,x^2-128\,x^3+{\mathrm {e}}^{5-\ln \left (x\right )-x}\,\left (\ln \left (x\right )\,\left (64\,x^3+160\,x^2\right )+32\,x^2\right )}{{\mathrm {e}}^{15-3\,\ln \left (x\right )-3\,x}-3072\,x-{\mathrm {e}}^{10-2\,\ln \left (x\right )-2\,x}\,\left (12\,x-48\right )+{\mathrm {e}}^{5-\ln \left (x\right )-x}\,\left (48\,x^2-384\,x+768\right )+768\,x^2-64\,x^3+4096} \,d x \] Input:
int((log(x)*(1536*x^2 - 128*x^3) + 512*x^2 - 128*x^3 + exp(5 - log(x) - x) *(log(x)*(160*x^2 + 64*x^3) + 32*x^2))/(exp(15 - 3*log(x) - 3*x) - 3072*x - exp(10 - 2*log(x) - 2*x)*(12*x - 48) + exp(5 - log(x) - x)*(48*x^2 - 384 *x + 768) + 768*x^2 - 64*x^3 + 4096),x)
Output:
int((log(x)*(1536*x^2 - 128*x^3) + 512*x^2 - 128*x^3 + exp(5 - log(x) - x) *(log(x)*(160*x^2 + 64*x^3) + 32*x^2))/(exp(15 - 3*log(x) - 3*x) - 3072*x - exp(10 - 2*log(x) - 2*x)*(12*x - 48) + exp(5 - log(x) - x)*(48*x^2 - 384 *x + 768) + 768*x^2 - 64*x^3 + 4096), x)
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx=\frac {32 e^{2 x} \mathrm {log}\left (x \right ) x^{5}}{16 e^{2 x} x^{4}-128 e^{2 x} x^{3}+256 e^{2 x} x^{2}-8 e^{x} e^{5} x^{2}+32 e^{x} e^{5} x +e^{10}} \] Input:
int((((64*x^3+160*x^2)*log(x)+32*x^2)*exp(5-log(x)-x)+(-128*x^3+1536*x^2)* log(x)-128*x^3+512*x^2)/(exp(5-log(x)-x)^3+(-12*x+48)*exp(5-log(x)-x)^2+(4 8*x^2-384*x+768)*exp(5-log(x)-x)-64*x^3+768*x^2-3072*x+4096),x)
Output:
(32*e**(2*x)*log(x)*x**5)/(16*e**(2*x)*x**4 - 128*e**(2*x)*x**3 + 256*e**( 2*x)*x**2 - 8*e**x*e**5*x**2 + 32*e**x*e**5*x + e**10)