Integrand size = 89, antiderivative size = 29 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x+\log \left (e^3+\frac {1}{5} e^{4 \left (3 e^2-x\right )^2} x+\log (x)\right ) \] Output:
x+ln(exp(3)+1/5*exp(2*(3*exp(2)-x)*(6*exp(2)-2*x))*x+ln(x))
Time = 0.63 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x+\log \left (5 e^3+e^{36 e^4-24 e^2 x+4 x^2} x+5 \log (x)\right ) \] Input:
Integrate[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x ^2 + 8*x^3) + 5*x*Log[x])/(5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5 *x*Log[x]),x]
Output:
x + Log[5*E^3 + E^(36*E^4 - 24*E^2*x + 4*x^2)*x + 5*Log[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.
\(\displaystyle \int \frac {e^{4 x^2-24 e^2 x+36 e^4} \left (8 x^3-24 e^2 x^2+x^2+x\right )+5 e^3 x+5 x \log (x)+5}{e^{4 x^2-24 e^2 x+36 e^4} x^2+5 e^3 x+5 x \log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 x^2+\left (1-24 e^2\right ) x+1}{x}-\frac {5 e^{24 e^2 x} \left (8 e^3 x^2+8 x^2 \log (x)-24 e^5 x-24 e^2 x \log (x)+\log (x)+e^3-1\right )}{x \left (e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 120 \int \frac {e^{24 e^2 x+5}}{e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)}dx+5 \left (1-e^3\right ) \int \frac {e^{24 e^2 x}}{x \left (e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)\right )}dx-40 \int \frac {e^{24 e^2 x+3} x}{e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)}dx+120 \int \frac {e^{24 e^2 x+2} \log (x)}{e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)}dx-5 \int \frac {e^{24 e^2 x} \log (x)}{x \left (e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)\right )}dx-40 \int \frac {e^{24 e^2 x} x \log (x)}{e^{4 x^2+36 e^4} x+5 e^{24 e^2 x+3}+5 e^{24 e^2 x} \log (x)}dx+4 x^2+\left (1-24 e^2\right ) x+\log (x)\) |
Input:
Int[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x^2 + 8 *x^3) + 5*x*Log[x])/(5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5*x*Log [x]),x]
Output:
$Aborted
Time = 0.62 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
risch | \(x +\ln \left ({\mathrm e}^{3}+\frac {{\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x}{5}+\ln \left (x \right )\right )\) | \(28\) |
norman | \(x +\ln \left ({\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x +5 \ln \left (x \right )+5 \,{\mathrm e}^{3}\right )\) | \(33\) |
parallelrisch | \(x +\ln \left ({\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x +5 \ln \left (x \right )+5 \,{\mathrm e}^{3}\right )\) | \(33\) |
Input:
int((5*x*ln(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4* x^2)+5*x*exp(3)+5)/(5*x*ln(x)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*e xp(3)),x,method=_RETURNVERBOSE)
Output:
x+ln(exp(3)+1/5*exp(36*exp(4)-24*exp(2)*x+4*x^2)*x+ln(x))
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x + \log \left (x e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, e^{3} + 5 \, \log \left (x\right )\right ) \] Input:
integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp( 2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^ 2)+5*x*exp(3)),x, algorithm="fricas")
Output:
x + log(x*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*e^3 + 5*log(x))
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x + \log {\left (x \right )} + \log {\left (e^{4 x^{2} - 24 x e^{2} + 36 e^{4}} + \frac {5 \log {\left (x \right )} + 5 e^{3}}{x} \right )} \] Input:
integrate((5*x*ln(x)+(-24*x**2*exp(2)+8*x**3+x**2+x)*exp(36*exp(2)**2-24*e xp(2)*x+4*x**2)+5*x*exp(3)+5)/(5*x*ln(x)+x**2*exp(36*exp(2)**2-24*exp(2)*x +4*x**2)+5*x*exp(3)),x)
Output:
x + log(x) + log(exp(4*x**2 - 24*x*exp(2) + 36*exp(4)) + (5*log(x) + 5*exp (3))/x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=-x {\left (24 \, e^{2} - 1\right )} + \log \left (\frac {x e^{\left (4 \, x^{2} + 36 \, e^{4}\right )} + 5 \, {\left (e^{3} + \log \left (x\right )\right )} e^{\left (24 \, x e^{2}\right )}}{5 \, {\left (e^{3} + \log \left (x\right )\right )}}\right ) + \log \left (e^{3} + \log \left (x\right )\right ) \] Input:
integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp( 2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^ 2)+5*x*exp(3)),x, algorithm="maxima")
Output:
-x*(24*e^2 - 1) + log(1/5*(x*e^(4*x^2 + 36*e^4) + 5*(e^3 + log(x))*e^(24*x *e^2))/(e^3 + log(x))) + log(e^3 + log(x))
\[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\int { \frac {5 \, x e^{3} + {\left (8 \, x^{3} - 24 \, x^{2} e^{2} + x^{2} + x\right )} e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, x \log \left (x\right ) + 5}{x^{2} e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, x e^{3} + 5 \, x \log \left (x\right )} \,d x } \] Input:
integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp( 2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^ 2)+5*x*exp(3)),x, algorithm="giac")
Output:
integrate((5*x*e^3 + (8*x^3 - 24*x^2*e^2 + x^2 + x)*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*x*log(x) + 5)/(x^2*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*x*e^3 + 5 *x*log(x)), x)
Timed out. \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\int \frac {5\,x\,{\mathrm {e}}^3+5\,x\,\ln \left (x\right )+{\mathrm {e}}^{4\,x^2-24\,{\mathrm {e}}^2\,x+36\,{\mathrm {e}}^4}\,\left (x-24\,x^2\,{\mathrm {e}}^2+x^2+8\,x^3\right )+5}{x^2\,{\mathrm {e}}^{4\,x^2-24\,{\mathrm {e}}^2\,x+36\,{\mathrm {e}}^4}+5\,x\,{\mathrm {e}}^3+5\,x\,\ln \left (x\right )} \,d x \] Input:
int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 2 4*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)),x)
Output:
int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 2 4*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)), x)
Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\mathrm {log}\left (e^{36 e^{4}+4 x^{2}} x +5 e^{24 e^{2} x} \mathrm {log}\left (x \right )+5 e^{24 e^{2} x} e^{3}\right )-24 e^{2} x +x \] Input:
int((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4 *x^2)+5*x*exp(3)+5)/(5*x*log(x)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x *exp(3)),x)
Output:
log(e**(36*e**4 + 4*x**2)*x + 5*e**(24*e**2*x)*log(x) + 5*e**(24*e**2*x)*e **3) - 24*e**2*x + x