Integrand size = 98, antiderivative size = 20 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {\log \left (e^x\right )}{\log ^2\left (\left (-8+e^{8-2 x}+x\right )^2\right )} \] Output:
ln(exp(x))/ln((exp(4-x)^2-8+x)^2)^2
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \] Input:
Integrate[(-4*x + 8*E^(8 - 2*x)*x + (-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)])/((-8 + E^(8 - 2*x) + x)*Lo g[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)]^3),x]
Output:
x/Log[(E^8 + E^(2*x)*(-8 + x))^2/E^(4*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.
| \(\displaystyle \int \frac {\left (x+e^{8-2 x}-8\right ) \log \left (x^2-16 x+e^{16-4 x}+e^{8-2 x} (2 x-16)+64\right )+8 e^{8-2 x} x-4 x}{\left (x+e^{8-2 x}-8\right ) \log ^3\left (x^2-16 x+e^{16-4 x}+e^{8-2 x} (2 x-16)+64\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (x+e^{8-2 x}-8\right ) \log \left (x^2-16 x+e^{16-4 x}+e^{8-2 x} (2 x-16)+64\right )-8 e^{8-2 x} x+4 x}{\left (-x-e^{8-2 x}+8\right ) \log ^3\left (e^{-4 x} \left (e^{2 x} x-8 e^{2 x}+e^8\right )^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4 e^8 x (2 x-15)}{(x-8) \left (e^{2 x} x-8 e^{2 x}+e^8\right ) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}+\frac {-4 x+x \log \left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )-8 \log \left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}{(x-8) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \int \frac {1}{\log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx-32 \int \frac {1}{(x-8) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx+4 e^8 \int \frac {1}{\left (e^{2 x} x-8 e^{2 x}+e^8\right ) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx+32 e^8 \int \frac {1}{(x-8) \left (e^{2 x} x-8 e^{2 x}+e^8\right ) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx+8 e^8 \int \frac {x}{\left (e^{2 x} x-8 e^{2 x}+e^8\right ) \log ^3\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx+\int \frac {1}{\log ^2\left (e^{-4 x} \left (e^{2 x} (x-8)+e^8\right )^2\right )}dx\) |
Input:
Int[(-4*x + 8*E^(8 - 2*x)*x + (-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)])/((-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)]^3),x]
Output:
$Aborted
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80
| method | result | size |
| parallelrisch | \(\frac {x}{\ln \left ({\mathrm e}^{-4 x +16}+\left (2 x -16\right ) {\mathrm e}^{-2 x +8}+x^{2}-16 x +64\right )^{2}}\) | \(36\) |
| risch | \(-\frac {4 x}{{\left (\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )}^{3}+4 i \ln \left ({\mathrm e}^{-2 x +8}-8+x \right )\right )}^{2}}\) | \(104\) |
Input:
int(((exp(-x+4)^2-8+x)*ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8* x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)/ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x ^2-16*x+64)^3,x,method=_RETURNVERBOSE)
Output:
x/ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^2
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log \left (x^{2} + 2 \, {\left (x - 8\right )} e^{\left (-2 \, x + 8\right )} - 16 \, x + e^{\left (-4 \, x + 16\right )} + 64\right )^{2}} \] Input:
integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x +64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(- x+4)^2+x^2-16*x+64)^3,x, algorithm="fricas")
Output:
x/log(x^2 + 2*(x - 8)*e^(-2*x + 8) - 16*x + e^(-4*x + 16) + 64)^2
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log {\left (x^{2} - 16 x + \left (2 x - 16\right ) e^{8 - 2 x} + e^{16 - 4 x} + 64 \right )}^{2}} \] Input:
integrate(((exp(-x+4)**2-8+x)*ln(exp(-x+4)**4+(2*x-16)*exp(-x+4)**2+x**2-1 6*x+64)+8*x*exp(-x+4)**2-4*x)/(exp(-x+4)**2-8+x)/ln(exp(-x+4)**4+(2*x-16)* exp(-x+4)**2+x**2-16*x+64)**3,x)
Output:
x/log(x**2 - 16*x + (2*x - 16)*exp(8 - 2*x) + exp(16 - 4*x) + 64)**2
Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).
Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{4 \, {\left (4 \, x^{2} - 4 \, x \log \left ({\left (x - 8\right )} e^{\left (2 \, x\right )} + e^{8}\right ) + \log \left ({\left (x - 8\right )} e^{\left (2 \, x\right )} + e^{8}\right )^{2}\right )}} \] Input:
integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x +64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(- x+4)^2+x^2-16*x+64)^3,x, algorithm="maxima")
Output:
1/4*x/(4*x^2 - 4*x*log((x - 8)*e^(2*x) + e^8) + log((x - 8)*e^(2*x) + e^8) ^2)
\[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\int { \frac {8 \, x e^{\left (-2 \, x + 8\right )} + {\left (x + e^{\left (-2 \, x + 8\right )} - 8\right )} \log \left (x^{2} + 2 \, {\left (x - 8\right )} e^{\left (-2 \, x + 8\right )} - 16 \, x + e^{\left (-4 \, x + 16\right )} + 64\right ) - 4 \, x}{{\left (x + e^{\left (-2 \, x + 8\right )} - 8\right )} \log \left (x^{2} + 2 \, {\left (x - 8\right )} e^{\left (-2 \, x + 8\right )} - 16 \, x + e^{\left (-4 \, x + 16\right )} + 64\right )^{3}} \,d x } \] Input:
integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x +64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(- x+4)^2+x^2-16*x+64)^3,x, algorithm="giac")
Output:
integrate((8*x*e^(-2*x + 8) + (x + e^(-2*x + 8) - 8)*log(x^2 + 2*(x - 8)*e ^(-2*x + 8) - 16*x + e^(-4*x + 16) + 64) - 4*x)/((x + e^(-2*x + 8) - 8)*lo g(x^2 + 2*(x - 8)*e^(-2*x + 8) - 16*x + e^(-4*x + 16) + 64)^3), x)
Time = 0.36 (sec) , antiderivative size = 297, normalized size of antiderivative = 14.85 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {\frac {x}{8}-\frac {7}{8}}{2\,{\mathrm {e}}^{8-2\,x}-1}-\frac {\frac {x+{\mathrm {e}}^{8-2\,x}-8}{4\,\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}-\frac {\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )\,\left (x+{\mathrm {e}}^{8-2\,x}-8\right )\,\left (28\,{\mathrm {e}}^{8-2\,x}-4\,x\,{\mathrm {e}}^{8-2\,x}+1\right )}{8\,{\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}^3}}{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )}+\frac {\frac {x^2}{4}-\frac {15\,x}{4}+\frac {225}{16}}{6\,{\mathrm {e}}^{8-2\,x}-12\,{\mathrm {e}}^{16-4\,x}+8\,{\mathrm {e}}^{24-6\,x}-1}+\frac {\frac {x^2}{4}-\frac {7\,x}{2}+\frac {195}{16}}{4\,{\mathrm {e}}^{16-4\,x}-4\,{\mathrm {e}}^{8-2\,x}+1}+\frac {x+\frac {\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )\,\left (x+{\mathrm {e}}^{8-2\,x}-8\right )}{4\,\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}}{{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )}^2} \] Input:
int((log(exp(16 - 4*x) - 16*x + exp(8 - 2*x)*(2*x - 16) + x^2 + 64)*(x + e xp(8 - 2*x) - 8) - 4*x + 8*x*exp(8 - 2*x))/(log(exp(16 - 4*x) - 16*x + exp (8 - 2*x)*(2*x - 16) + x^2 + 64)^3*(x + exp(8 - 2*x) - 8)),x)
Output:
(x/8 - 7/8)/(2*exp(8 - 2*x) - 1) - ((x + exp(8 - 2*x) - 8)/(4*(2*exp(8 - 2 *x) - 1)) - (log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x)*exp(8)*(2*x - 16) + 64)*(x + exp(8 - 2*x) - 8)*(28*exp(8 - 2*x) - 4*x*exp(8 - 2*x) + 1)) /(8*(2*exp(8 - 2*x) - 1)^3))/log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x )*exp(8)*(2*x - 16) + 64) + (x^2/4 - (15*x)/4 + 225/16)/(6*exp(8 - 2*x) - 12*exp(16 - 4*x) + 8*exp(24 - 6*x) - 1) + (x^2/4 - (7*x)/2 + 195/16)/(4*ex p(16 - 4*x) - 4*exp(8 - 2*x) + 1) + (x + (log(exp(-4*x)*exp(16) - 16*x + x ^2 + exp(-2*x)*exp(8)*(2*x - 16) + 64)*(x + exp(8 - 2*x) - 8))/(4*(2*exp(8 - 2*x) - 1)))/log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x)*exp(8)*(2*x - 16) + 64)^2
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 3.10 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\mathrm {log}\left (\frac {e^{4 x} x^{2}-16 e^{4 x} x +64 e^{4 x}+2 e^{2 x} e^{8} x -16 e^{2 x} e^{8}+e^{16}}{e^{4 x}}\right )^{2}} \] Input:
int(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8 *x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2 +x^2-16*x+64)^3,x)
Output:
x/log((e**(4*x)*x**2 - 16*e**(4*x)*x + 64*e**(4*x) + 2*e**(2*x)*e**8*x - 1 6*e**(2*x)*e**8 + e**16)/e**(4*x))**2