Integrand size = 92, antiderivative size = 21 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \] Output:
(2+x)*exp(x^3/(ln(2+x)+2*x+4))
Time = 3.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \] Input:
Integrate[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4 *x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log[2 + x] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]
Output:
E^(x^3/(4 + 2*x + Log[2 + 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.
\(\displaystyle \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} \left (4 x^4+19 x^3+28 x^2+\left (3 x^3+6 x^2+4 x+8\right ) \log (x+2)+16 x+\log ^2(x+2)+16\right )}{4 x^2+16 x+\log ^2(x+2)+(4 x+8) \log (x+2)+16} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} \left (4 x^4+19 x^3+28 x^2+\left (3 x^3+6 x^2+4 x+8\right ) \log (x+2)+16 x+\log ^2(x+2)+16\right )}{(2 x+\log (x+2)+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {(2 x+5) x^3 e^{\frac {x^3}{2 x+\log (x+2)+4}}}{(2 x+\log (x+2)+4)^2}+e^{\frac {x^3}{2 x+\log (x+2)+4}}+\frac {3 (x+2) x^2 e^{\frac {x^3}{2 x+\log (x+2)+4}}}{2 x+\log (x+2)+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{\frac {x^3}{2 x+\log (x+2)+4}}dx-5 \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} x^3}{(2 x+\log (x+2)+4)^2}dx+3 \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} x^3}{2 x+\log (x+2)+4}dx-2 \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} x^4}{(2 x+\log (x+2)+4)^2}dx+6 \int \frac {e^{\frac {x^3}{2 x+\log (x+2)+4}} x^2}{2 x+\log (x+2)+4}dx\) |
Input:
Int[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4*x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log[2 + x] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]
Output:
$Aborted
Time = 0.96 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\left (2+x \right ) {\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}} x +2 \,{\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}}\) | \(38\) |
Input:
int((ln(2+x)^2+(3*x^3+6*x^2+4*x+8)*ln(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*ex p(x^3/(ln(2+x)+2*x+4))/(ln(2+x)^2+(4*x+8)*ln(2+x)+4*x^2+16*x+16),x,method= _RETURNVERBOSE)
Output:
(2+x)*exp(x^3/(ln(2+x)+2*x+4))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx={\left (x + 2\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \] Input:
integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16* x+16)*exp(x^3/(log(2+x)+2*x+4))/(log(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16 ),x, algorithm="fricas")
Output:
(x + 2)*e^(x^3/(2*x + log(x + 2) + 4))
Exception generated. \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((ln(2+x)**2+(3*x**3+6*x**2+4*x+8)*ln(2+x)+4*x**4+19*x**3+28*x**2 +16*x+16)*exp(x**3/(ln(2+x)+2*x+4))/(ln(2+x)**2+(4*x+8)*ln(2+x)+4*x**2+16* x+16),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
\[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=\int { \frac {{\left (4 \, x^{4} + 19 \, x^{3} + 28 \, x^{2} + {\left (3 \, x^{3} + 6 \, x^{2} + 4 \, x + 8\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )}}{4 \, x^{2} + 4 \, {\left (x + 2\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16} \,d x } \] Input:
integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16* x+16)*exp(x^3/(log(2+x)+2*x+4))/(log(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16 ),x, algorithm="maxima")
Output:
integrate((4*x^4 + 19*x^3 + 28*x^2 + (3*x^3 + 6*x^2 + 4*x + 8)*log(x + 2) + log(x + 2)^2 + 16*x + 16)*e^(x^3/(2*x + log(x + 2) + 4))/(4*x^2 + 4*(x + 2)*log(x + 2) + log(x + 2)^2 + 16*x + 16), x)
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=x e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} + 2 \, e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \] Input:
integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16* x+16)*exp(x^3/(log(2+x)+2*x+4))/(log(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16 ),x, algorithm="giac")
Output:
x*e^(x^3/(2*x + log(x + 2) + 4)) + 2*e^(x^3/(2*x + log(x + 2) + 4))
Time = 2.77 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx={\mathrm {e}}^{\frac {x^3}{2\,x+\ln \left (x+2\right )+4}}\,\left (x+2\right ) \] Input:
int((exp(x^3/(2*x + log(x + 2) + 4))*(16*x + log(x + 2)*(4*x + 6*x^2 + 3*x ^3 + 8) + log(x + 2)^2 + 28*x^2 + 19*x^3 + 4*x^4 + 16))/(16*x + log(x + 2) ^2 + 4*x^2 + log(x + 2)*(4*x + 8) + 16),x)
Output:
exp(x^3/(2*x + log(x + 2) + 4))*(x + 2)
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=e^{\frac {x^{3}}{\mathrm {log}\left (x +2\right )+2 x +4}} \left (x +2\right ) \] Input:
int((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)* exp(x^3/(log(2+x)+2*x+4))/(log(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x)
Output:
e**(x**3/(log(x + 2) + 2*x + 4))*(x + 2)