Integrand size = 131, antiderivative size = 25 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=2+\frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)} \]
Time = 0.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=\frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)} \]
Integrate[(-(E^(E^4 + x)*x) + 2*x^2 + x*Log[x] + (-16 - x + E^(E^4 + x)*(- 16 + 15*x + x^2) + (16 + x)*Log[x])*Log[16 + x])/(64*x^2 + 4*x^3 + E^(2*E^ 4 + 2*x)*(16 + x) + E^(E^4 + x)*(-64*x - 4*x^2) + (E^(E^4 + x)*(-32 - 2*x) + 64*x + 4*x^2)*Log[x] + (16 + x)*Log[x]^2),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.
\(\displaystyle \int \frac {2 x^2+\left (e^{x+e^4} \left (x^2+15 x-16\right )-x+(x+16) \log (x)-16\right ) \log (x+16)-e^{x+e^4} x+x \log (x)}{4 x^3+64 x^2+e^{x+e^4} \left (-4 x^2-64 x\right )+\left (4 x^2+64 x+e^{x+e^4} (-2 x-32)\right ) \log (x)+e^{2 x+2 e^4} (x+16)+(x+16) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-\left (\left (e^{x+e^4}-2 x\right ) x\right )+\left (e^{x+e^4} (x-1)-1\right ) (x+16) \log (x+16)+\log (x) (x+(x+16) \log (x+16))}{(x+16) \left (-2 x+e^{x+e^4}-\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (2 x^2-2 x+x \log (x)-1\right ) \log (x+16)}{\left (-2 x+e^{x+e^4}-\log (x)\right )^2}-\frac {x^2 \log (x+16)-x+15 x \log (x+16)-16 \log (x+16)}{(x+16) \left (2 x-e^{x+e^4}+\log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {x^2 \log (x+16)}{\left (2 x-e^{x+e^4}+\log (x)\right )^2}dx-\int \frac {1}{-2 x+e^{x+e^4}-\log (x)}dx-16 \int \frac {1}{(x+16) \left (2 x-e^{x+e^4}+\log (x)\right )}dx-\int \frac {\log (x+16)}{\left (-2 x+e^{x+e^4}-\log (x)\right )^2}dx-\int \frac {\log (x+16)}{-2 x+e^{x+e^4}-\log (x)}dx-2 \int \frac {x \log (x+16)}{\left (2 x-e^{x+e^4}+\log (x)\right )^2}dx+\int \frac {x \log (x) \log (x+16)}{\left (2 x-e^{x+e^4}+\log (x)\right )^2}dx-\int \frac {x \log (x+16)}{2 x-e^{x+e^4}+\log (x)}dx\) |
Int[(-(E^(E^4 + x)*x) + 2*x^2 + x*Log[x] + (-16 - x + E^(E^4 + x)*(-16 + 1 5*x + x^2) + (16 + x)*Log[x])*Log[16 + x])/(64*x^2 + 4*x^3 + E^(2*E^4 + 2* x)*(16 + x) + E^(E^4 + x)*(-64*x - 4*x^2) + (E^(E^4 + x)*(-32 - 2*x) + 64* x + 4*x^2)*Log[x] + (16 + x)*Log[x]^2),x]
3.11.18.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x \ln \left (x +16\right )}{2 x -{\mathrm e}^{x +{\mathrm e}^{4}}+\ln \left (x \right )}\) | \(22\) |
parallelrisch | \(\frac {x \ln \left (x +16\right )}{2 x -{\mathrm e}^{x +{\mathrm e}^{4}}+\ln \left (x \right )}\) | \(22\) |
int((((x+16)*ln(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*ln(x+16)+x*ln(x)-x*ex p(x+exp(4))+2*x^2)/((x+16)*ln(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^2+64*x)*ln (x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+64*x^2),x,met hod=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=\frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \left (x\right )} \]
integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*lo g(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^ 2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+64 *x^2),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=- \frac {x \log {\left (x + 16 \right )}}{- 2 x + e^{x + e^{4}} - \log {\left (x \right )}} \]
integrate((((x+16)*ln(x)+(x**2+15*x-16)*exp(x+exp(4))-x-16)*ln(x+16)+x*ln( x)-x*exp(x+exp(4))+2*x**2)/((x+16)*ln(x)**2+((-2*x-32)*exp(x+exp(4))+4*x** 2+64*x)*ln(x)+(x+16)*exp(x+exp(4))**2+(-4*x**2-64*x)*exp(x+exp(4))+4*x**3+ 64*x**2),x)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=\frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \left (x\right )} \]
integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*lo g(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^ 2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+64 *x^2),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=\frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \left (x\right )} \]
integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*lo g(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^ 2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+64 *x^2),x, algorithm=\
Time = 9.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 4.04 \[ \int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx=-\frac {30\,x^4\,\ln \left (x+16\right )-33\,x^3\,\ln \left (x+16\right )-16\,x^2\,\ln \left (x+16\right )+2\,x^5\,\ln \left (x+16\right )+\ln \left (x\right )\,\left (16\,x^3\,\ln \left (x+16\right )+x^4\,\ln \left (x+16\right )\right )}{\left (x+16\right )\,\left (2\,x-{\mathrm {e}}^{x+{\mathrm {e}}^4}+\ln \left (x\right )\right )\,\left (x-x^2\,\ln \left (x\right )+2\,x^2-2\,x^3\right )} \]
int(-(x*exp(x + exp(4)) + log(x + 16)*(x - exp(x + exp(4))*(15*x + x^2 - 1 6) - log(x)*(x + 16) + 16) - x*log(x) - 2*x^2)/(log(x)^2*(x + 16) - exp(x + exp(4))*(64*x + 4*x^2) + exp(2*x + 2*exp(4))*(x + 16) + 64*x^2 + 4*x^3 + log(x)*(64*x - exp(x + exp(4))*(2*x + 32) + 4*x^2)),x)