Integrand size = 67, antiderivative size = 27 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=-4+e^{3-2 (e-x)+\frac {5 x}{4}+\frac {5}{2+x}} x \] Output:
exp(3+ln(x)+5/(2+x)+13/4*x-2*exp(1))-4
Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=e^{3-2 e+\frac {13 x}{4}+\frac {5}{2+x}} x \] Input:
Integrate[(E^((44 + E*(-16 - 8*x) + 38*x + 13*x^2 + (8 + 4*x)*Log[x])/(8 + 4*x))*(16 + 48*x + 56*x^2 + 13*x^3))/(16*x + 16*x^2 + 4*x^3),x]
Output:
E^(3 - 2*E + (13*x)/4 + 5/(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.
\(\displaystyle \int \frac {\left (13 x^3+56 x^2+48 x+16\right ) \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{4 x^3+16 x^2+16 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (13 x^3+56 x^2+48 x+16\right ) \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{x \left (4 x^2+16 x+16\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (13 x^3+56 x^2+48 x+16\right ) \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{x (2 x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {13}{4} \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )-\frac {5 \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{(x+2)^2}+\frac {\exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {13}{4} \int \exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )dx+\int \frac {\exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{x}dx-5 \int \frac {\exp \left (\frac {13 x^2+38 x+e (-8 x-16)+(4 x+8) \log (x)+44}{4 x+8}\right )}{(x+2)^2}dx\) |
Input:
Int[(E^((44 + E*(-16 - 8*x) + 38*x + 13*x^2 + (8 + 4*x)*Log[x])/(8 + 4*x)) *(16 + 48*x + 56*x^2 + 13*x^3))/(16*x + 16*x^2 + 4*x^3),x]
Output:
$Aborted
Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (4 x +8\right ) \ln \left (x \right )+\left (-8 x -16\right ) {\mathrm e}+13 x^{2}+38 x +44}{4 x +8}}\) | \(35\) |
gosper | \({\mathrm e}^{\frac {4 x \ln \left (x \right )-8 x \,{\mathrm e}+13 x^{2}+8 \ln \left (x \right )-16 \,{\mathrm e}+38 x +44}{4 x +8}}\) | \(37\) |
risch | \({\mathrm e}^{-\frac {-4 x \ln \left (x \right )+8 x \,{\mathrm e}-13 x^{2}-8 \ln \left (x \right )+16 \,{\mathrm e}-38 x -44}{4 \left (2+x \right )}}\) | \(37\) |
orering | \(\frac {4 \left (2+x \right )^{2} x \,{\mathrm e}^{\frac {\left (4 x +8\right ) \ln \left (x \right )+\left (-8 x -16\right ) {\mathrm e}+13 x^{2}+38 x +44}{4 x +8}}}{4 x^{3}+16 x^{2}+16 x}\) | \(60\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {\left (4 x +8\right ) \ln \left (x \right )+\left (-8 x -16\right ) {\mathrm e}+13 x^{2}+38 x +44}{4 x +8}}+2 \,{\mathrm e}^{\frac {\left (4 x +8\right ) \ln \left (x \right )+\left (-8 x -16\right ) {\mathrm e}+13 x^{2}+38 x +44}{4 x +8}}}{2+x}\) | \(82\) |
Input:
int((13*x^3+56*x^2+48*x+16)*exp(((4*x+8)*ln(x)+(-8*x-16)*exp(1)+13*x^2+38* x+44)/(4*x+8))/(4*x^3+16*x^2+16*x),x,method=_RETURNVERBOSE)
Output:
exp(1/4/(2+x)*((4*x+8)*ln(x)+(-8*x-16)*exp(1)+13*x^2+38*x+44))
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=e^{\left (\frac {13 \, x^{2} - 8 \, {\left (x + 2\right )} e + 4 \, {\left (x + 2\right )} \log \left (x\right ) + 38 \, x + 44}{4 \, {\left (x + 2\right )}}\right )} \] Input:
integrate((13*x^3+56*x^2+48*x+16)*exp(((4*x+8)*log(x)+(-8*x-16)*exp(1)+13* x^2+38*x+44)/(4*x+8))/(4*x^3+16*x^2+16*x),x, algorithm="fricas")
Output:
e^(1/4*(13*x^2 - 8*(x + 2)*e + 4*(x + 2)*log(x) + 38*x + 44)/(x + 2))
Timed out. \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=\text {Timed out} \] Input:
integrate((13*x**3+56*x**2+48*x+16)*exp(((4*x+8)*ln(x)+(-8*x-16)*exp(1)+13 *x**2+38*x+44)/(4*x+8))/(4*x**3+16*x**2+16*x),x)
Output:
Timed out
\[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=\int { \frac {{\left (13 \, x^{3} + 56 \, x^{2} + 48 \, x + 16\right )} e^{\left (\frac {13 \, x^{2} - 8 \, {\left (x + 2\right )} e + 4 \, {\left (x + 2\right )} \log \left (x\right ) + 38 \, x + 44}{4 \, {\left (x + 2\right )}}\right )}}{4 \, {\left (x^{3} + 4 \, x^{2} + 4 \, x\right )}} \,d x } \] Input:
integrate((13*x^3+56*x^2+48*x+16)*exp(((4*x+8)*log(x)+(-8*x-16)*exp(1)+13* x^2+38*x+44)/(4*x+8))/(4*x^3+16*x^2+16*x),x, algorithm="maxima")
Output:
1/4*integrate((13*x^3 + 56*x^2 + 48*x + 16)*e^(1/4*(13*x^2 - 8*(x + 2)*e + 4*(x + 2)*log(x) + 38*x + 44)/(x + 2))/(x^3 + 4*x^2 + 4*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.37 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=e^{\left (\frac {13 \, x^{2}}{4 \, {\left (x + 2\right )}} - \frac {2 \, x e}{x + 2} + \frac {x \log \left (x\right )}{x + 2} + \frac {19 \, x}{2 \, {\left (x + 2\right )}} - \frac {4 \, e}{x + 2} + \frac {2 \, \log \left (x\right )}{x + 2} + \frac {11}{x + 2}\right )} \] Input:
integrate((13*x^3+56*x^2+48*x+16)*exp(((4*x+8)*log(x)+(-8*x-16)*exp(1)+13* x^2+38*x+44)/(4*x+8))/(4*x^3+16*x^2+16*x),x, algorithm="giac")
Output:
e^(13/4*x^2/(x + 2) - 2*x*e/(x + 2) + x*log(x)/(x + 2) + 19/2*x/(x + 2) - 4*e/(x + 2) + 2*log(x)/(x + 2) + 11/(x + 2))
Time = 4.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=x\,{\mathrm {e}}^{\frac {13\,x^2}{4\,x+8}}\,{\mathrm {e}}^{-\frac {4\,\mathrm {e}}{x+2}}\,{\mathrm {e}}^{\frac {19\,x}{2\,x+4}}\,{\mathrm {e}}^{-\frac {2\,x\,\mathrm {e}}{x+2}}\,{\mathrm {e}}^{\frac {11}{x+2}} \] Input:
int((exp((38*x + log(x)*(4*x + 8) + 13*x^2 - exp(1)*(8*x + 16) + 44)/(4*x + 8))*(48*x + 56*x^2 + 13*x^3 + 16))/(16*x + 16*x^2 + 4*x^3),x)
Output:
x*exp((13*x^2)/(4*x + 8))*exp(-(4*exp(1))/(x + 2))*exp((19*x)/(2*x + 4))*e xp(-(2*x*exp(1))/(x + 2))*exp(11/(x + 2))
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {44+e (-16-8 x)+38 x+13 x^2+(8+4 x) \log (x)}{8+4 x}} \left (16+48 x+56 x^2+13 x^3\right )}{16 x+16 x^2+4 x^3} \, dx=\frac {e^{\frac {13 x^{2}+26 x +20}{4 x +8}} e^{3} x}{e^{2 e}} \] Input:
int((13*x^3+56*x^2+48*x+16)*exp(((4*x+8)*log(x)+(-8*x-16)*exp(1)+13*x^2+38 *x+44)/(4*x+8))/(4*x^3+16*x^2+16*x),x)
Output:
(e**((13*x**2 + 26*x + 20)/(4*x + 8))*e**3*x)/e**(2*e)