Integrand size = 151, antiderivative size = 23 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4}{4+e^{e^x}+2 x+\frac {16 \log (2+x)}{x}} \]
Time = 0.83 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4 x}{x \left (4+e^{e^x}+2 x\right )+16 \log (2+x)} \]
Integrate[(-64*x - 16*x^2 - 8*x^3 + E^(E^x + x)*(-8*x^2 - 4*x^3) + (128 + 64*x)*Log[2 + x])/(32*x^2 + 48*x^3 + 24*x^4 + 4*x^5 + E^(2*E^x)*(2*x^2 + x ^3) + (256*x + 256*x^2 + 64*x^3)*Log[2 + x] + (512 + 256*x)*Log[2 + x]^2 + E^E^x*(16*x^2 + 16*x^3 + 4*x^4 + (64*x + 32*x^2)*Log[2 + x])),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.
\(\displaystyle \int \frac {-8 x^3-16 x^2+e^{x+e^x} \left (-4 x^3-8 x^2\right )-64 x+(64 x+128) \log (x+2)}{4 x^5+24 x^4+48 x^3+32 x^2+e^{2 e^x} \left (x^3+2 x^2\right )+\left (64 x^3+256 x^2+256 x\right ) \log (x+2)+e^{e^x} \left (4 x^4+16 x^3+16 x^2+\left (32 x^2+64 x\right ) \log (x+2)\right )+(256 x+512) \log ^2(x+2)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {64 (x+2) \log (x+2)-4 x \left (\left (e^{x+e^x}+2\right ) x^2+2 \left (e^{x+e^x}+2\right ) x+16\right )}{(x+2) \left (x \left (2 x+e^{e^x}+4\right )+16 \log (x+2)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 e^{x+e^x} x^2}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}-\frac {16 x^2}{(x+2) \left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}-\frac {64 x}{(x+2) \left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}+\frac {64 \log (x+2)}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}-\frac {8 x^3}{(x+2) \left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -64 \int \frac {1}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}dx-8 \int \frac {x^2}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}dx-4 \int \frac {e^{x+e^x} x^2}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}dx+128 \int \frac {1}{(x+2) \left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}dx+64 \int \frac {\log (x+2)}{\left (2 x^2+e^{e^x} x+4 x+16 \log (x+2)\right )^2}dx\) |
Int[(-64*x - 16*x^2 - 8*x^3 + E^(E^x + x)*(-8*x^2 - 4*x^3) + (128 + 64*x)* Log[2 + x])/(32*x^2 + 48*x^3 + 24*x^4 + 4*x^5 + E^(2*E^x)*(2*x^2 + x^3) + (256*x + 256*x^2 + 64*x^3)*Log[2 + x] + (512 + 256*x)*Log[2 + x]^2 + E^E^x *(16*x^2 + 16*x^3 + 4*x^4 + (64*x + 32*x^2)*Log[2 + x])),x]
3.17.37.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {4 x}{x \,{\mathrm e}^{{\mathrm e}^{x}}+2 x^{2}+16 \ln \left (2+x \right )+4 x}\) | \(26\) |
parallelrisch | \(\frac {4 x}{x \,{\mathrm e}^{{\mathrm e}^{x}}+2 x^{2}+16 \ln \left (2+x \right )+4 x}\) | \(26\) |
int(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*ln(2+x)-8*x^3-16*x^2-64* x)/((x^3+2*x^2)*exp(exp(x))^2+((32*x^2+64*x)*ln(2+x)+4*x^4+16*x^3+16*x^2)* exp(exp(x))+(256*x+512)*ln(2+x)^2+(64*x^3+256*x^2+256*x)*ln(2+x)+4*x^5+24* x^4+48*x^3+32*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4 \, x e^{x}}{x e^{\left (x + e^{x}\right )} + 2 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 16 \, e^{x} \log \left (x + 2\right )} \]
integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16* x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+ 16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+x) +4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4 x}{2 x^{2} + x e^{e^{x}} + 4 x + 16 \log {\left (x + 2 \right )}} \]
integrate(((-4*x**3-8*x**2)*exp(x)*exp(exp(x))+(64*x+128)*ln(2+x)-8*x**3-1 6*x**2-64*x)/((x**3+2*x**2)*exp(exp(x))**2+((32*x**2+64*x)*ln(2+x)+4*x**4+ 16*x**3+16*x**2)*exp(exp(x))+(256*x+512)*ln(2+x)**2+(64*x**3+256*x**2+256* x)*ln(2+x)+4*x**5+24*x**4+48*x**3+32*x**2),x)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \]
integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16* x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+ 16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+x) +4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \]
integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16* x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+ 16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+x) +4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm=\
Timed out. \[ \int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx=\int -\frac {64\,x+16\,x^2+8\,x^3-\ln \left (x+2\right )\,\left (64\,x+128\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (4\,x^3+8\,x^2\right )}{{\ln \left (x+2\right )}^2\,\left (256\,x+512\right )+\ln \left (x+2\right )\,\left (64\,x^3+256\,x^2+256\,x\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x^3+2\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \left (x+2\right )\,\left (32\,x^2+64\,x\right )+16\,x^2+16\,x^3+4\,x^4\right )+32\,x^2+48\,x^3+24\,x^4+4\,x^5} \,d x \]
int(-(64*x + 16*x^2 + 8*x^3 - log(x + 2)*(64*x + 128) + exp(exp(x))*exp(x) *(8*x^2 + 4*x^3))/(log(x + 2)^2*(256*x + 512) + log(x + 2)*(256*x + 256*x^ 2 + 64*x^3) + exp(2*exp(x))*(2*x^2 + x^3) + exp(exp(x))*(log(x + 2)*(64*x + 32*x^2) + 16*x^2 + 16*x^3 + 4*x^4) + 32*x^2 + 48*x^3 + 24*x^4 + 4*x^5),x )
int(-(64*x + 16*x^2 + 8*x^3 - log(x + 2)*(64*x + 128) + exp(exp(x))*exp(x) *(8*x^2 + 4*x^3))/(log(x + 2)^2*(256*x + 512) + log(x + 2)*(256*x + 256*x^ 2 + 64*x^3) + exp(2*exp(x))*(2*x^2 + x^3) + exp(exp(x))*(log(x + 2)*(64*x + 32*x^2) + 16*x^2 + 16*x^3 + 4*x^4) + 32*x^2 + 48*x^3 + 24*x^4 + 4*x^5), x)