Integrand size = 85, antiderivative size = 33 \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=x^2 \left (e^{\frac {e^{-x+\frac {1}{2} (-5-x+\log (4+x))}}{x}+x}+x\right ) \]
Time = 7.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=e^{x+\frac {e^{-\frac {5}{2}-\frac {3 x}{2}} \sqrt {4+x}}{x}} x^2+x^3 \]
Integrate[(24*x^2 + 6*x^3 + E^((E^((-5 - 3*x + Log[4 + x])/2) + x^2)/x)*(1 6*x + 12*x^2 + 2*x^3 + E^((-5 - 3*x + Log[4 + x])/2)*(-8 - 13*x - 3*x^2))) /(8 + 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.
\(\displaystyle \int \frac {6 x^3+24 x^2+e^{\frac {x^2+e^{\frac {1}{2} (-3 x+\log (x+4)-5)}}{x}} \left (2 x^3+12 x^2+\left (-3 x^2-13 x-8\right ) e^{\frac {1}{2} (-3 x+\log (x+4)-5)}+16 x\right )}{2 x+8} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (x \left (e^{x+\frac {e^{-\frac {3 x}{2}-\frac {5}{2}} \sqrt {x+4}}{x}} x+3 x+2 e^{x+\frac {e^{-\frac {3 x}{2}-\frac {5}{2}} \sqrt {x+4}}{x}}\right )-\frac {e^{-\frac {x}{2}+\frac {e^{-\frac {3 x}{2}-\frac {5}{2}} \sqrt {x+4}}{x}-\frac {5}{2}} \left (3 x^2+13 x+8\right )}{2 \sqrt {x+4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \text {Subst}\left (\int e^{-\frac {x^2}{2}+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-\frac {1}{2}}dx,x,\sqrt {x+4}\right )+16 \text {Subst}\left (\int e^{x^2+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-4} xdx,x,\sqrt {x+4}\right )+11 \text {Subst}\left (\int e^{-\frac {x^2}{2}+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-\frac {1}{2}} x^2dx,x,\sqrt {x+4}\right )+2 \text {Subst}\left (\int e^{x^2+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-4} x^5dx,x,\sqrt {x+4}\right )-3 \text {Subst}\left (\int e^{-\frac {x^2}{2}+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-\frac {1}{2}} x^4dx,x,\sqrt {x+4}\right )-12 \text {Subst}\left (\int e^{x^2+\frac {e^{\frac {7}{2}-\frac {3 x^2}{2}} x}{x^2-4}-4} x^3dx,x,\sqrt {x+4}\right )+x^3\) |
Int[(24*x^2 + 6*x^3 + E^((E^((-5 - 3*x + Log[4 + x])/2) + x^2)/x)*(16*x + 12*x^2 + 2*x^3 + E^((-5 - 3*x + Log[4 + x])/2)*(-8 - 13*x - 3*x^2)))/(8 + 2*x),x]
3.2.51.3.1 Defintions of rubi rules used
Time = 1.58 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91
method | result | size |
risch | \(x^{3}+{\mathrm e}^{\frac {\sqrt {4+x}\, {\mathrm e}^{-\frac {5}{2}-\frac {3 x}{2}}+x^{2}}{x}} x^{2}\) | \(30\) |
parallelrisch | \(x^{3}+{\mathrm e}^{\frac {{\mathrm e}^{\frac {\ln \left (4+x \right )}{2}-\frac {3 x}{2}-\frac {5}{2}}+x^{2}}{x}} x^{2}\) | \(30\) |
int((((-3*x^2-13*x-8)*exp(1/2*ln(4+x)-3/2*x-5/2)+2*x^3+12*x^2+16*x)*exp((e xp(1/2*ln(4+x)-3/2*x-5/2)+x^2)/x)+6*x^3+24*x^2)/(2*x+8),x,method=_RETURNVE RBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=x^{3} + x^{2} e^{\left (\frac {x^{2} + e^{\left (-\frac {3}{2} \, x + \frac {1}{2} \, \log \left (x + 4\right ) - \frac {5}{2}\right )}}{x}\right )} \]
integrate((((-3*x^2-13*x-8)*exp(1/2*log(4+x)-3/2*x-5/2)+2*x^3+12*x^2+16*x) *exp((exp(1/2*log(4+x)-3/2*x-5/2)+x^2)/x)+6*x^3+24*x^2)/(2*x+8),x, algorit hm=\
Timed out. \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=\text {Timed out} \]
integrate((((-3*x**2-13*x-8)*exp(1/2*ln(4+x)-3/2*x-5/2)+2*x**3+12*x**2+16* x)*exp((exp(1/2*ln(4+x)-3/2*x-5/2)+x**2)/x)+6*x**3+24*x**2)/(2*x+8),x)
Time = 0.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=x^{3} + x^{2} e^{\left (x + \frac {\sqrt {x + 4} e^{\left (-\frac {3}{2} \, x - \frac {5}{2}\right )}}{x} + \frac {1}{2}\right )} \]
integrate((((-3*x^2-13*x-8)*exp(1/2*log(4+x)-3/2*x-5/2)+2*x^3+12*x^2+16*x) *exp((exp(1/2*log(4+x)-3/2*x-5/2)+x^2)/x)+6*x^3+24*x^2)/(2*x+8),x, algorit hm=\
\[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=\int { \frac {6 \, x^{3} + 24 \, x^{2} + {\left (2 \, x^{3} + 12 \, x^{2} - {\left (3 \, x^{2} + 13 \, x + 8\right )} e^{\left (-\frac {3}{2} \, x + \frac {1}{2} \, \log \left (x + 4\right ) - \frac {5}{2}\right )} + 16 \, x\right )} e^{\left (\frac {x^{2} + e^{\left (-\frac {3}{2} \, x + \frac {1}{2} \, \log \left (x + 4\right ) - \frac {5}{2}\right )}}{x}\right )}}{2 \, {\left (x + 4\right )}} \,d x } \]
integrate((((-3*x^2-13*x-8)*exp(1/2*log(4+x)-3/2*x-5/2)+2*x^3+12*x^2+16*x) *exp((exp(1/2*log(4+x)-3/2*x-5/2)+x^2)/x)+6*x^3+24*x^2)/(2*x+8),x, algorit hm=\
integrate(1/2*(6*x^3 + 24*x^2 + (2*x^3 + 12*x^2 - (3*x^2 + 13*x + 8)*e^(-3 /2*x + 1/2*log(x + 4) - 5/2) + 16*x)*e^((x^2 + e^(-3/2*x + 1/2*log(x + 4) - 5/2))/x))/(x + 4), x)
Time = 8.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {24 x^2+6 x^3+e^{\frac {e^{\frac {1}{2} (-5-3 x+\log (4+x))}+x^2}{x}} \left (16 x+12 x^2+2 x^3+e^{\frac {1}{2} (-5-3 x+\log (4+x))} \left (-8-13 x-3 x^2\right )\right )}{8+2 x} \, dx=x^3+x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {5}{2}}\,\sqrt {x+4}}{x\,{\left ({\mathrm {e}}^x\right )}^{3/2}}}\,{\mathrm {e}}^x \]