Integrand size = 101, antiderivative size = 25 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\left (1-4 e^{e^x}\right ) x+\frac {x^2}{e^5+x}+\log (x) \] Output:
x^2/(exp(5)+x)+ln(x)+x*(1-4*exp(exp(x)))
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=2 x-4 e^{e^x} x+\frac {e^{10}}{e^5+x}+\log (x) \] Input:
Integrate[(x^2 + 2*x^3 + E^10*(1 + x) + E^5*(2*x + 4*x^2) + E^E^x*(-4*E^10 *x - 8*E^5*x^2 - 4*x^3 + E^x*(-4*E^10*x^2 - 8*E^5*x^3 - 4*x^4)))/(E^10*x + 2*E^5*x^2 + x^3),x]
Output:
2*x - 4*E^E^x*x + E^10/(E^5 + x) + 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.
\(\displaystyle \int \frac {2 x^3+x^2+e^5 \left (4 x^2+2 x\right )+e^{e^x} \left (-4 x^3-8 e^5 x^2+e^x \left (-4 x^4-8 e^5 x^3-4 e^{10} x^2\right )-4 e^{10} x\right )+e^{10} (x+1)}{x^3+2 e^5 x^2+e^{10} x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2 x^3+x^2+e^5 \left (4 x^2+2 x\right )+e^{e^x} \left (-4 x^3-8 e^5 x^2+e^x \left (-4 x^4-8 e^5 x^3-4 e^{10} x^2\right )-4 e^{10} x\right )+e^{10} (x+1)}{x \left (x^2+2 e^5 x+e^{10}\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {2 x^3+x^2+e^5 \left (4 x^2+2 x\right )+e^{e^x} \left (-4 x^3-8 e^5 x^2+e^x \left (-4 x^4-8 e^5 x^3-4 e^{10} x^2\right )-4 e^{10} x\right )+e^{10} (x+1)}{x \left (x+e^5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 e^{e^x} x^2}{\left (x+e^5\right )^2}+\frac {2 x^2}{\left (x+e^5\right )^2}-4 e^{x+e^x} x-\frac {8 e^{e^x+5} x}{\left (x+e^5\right )^2}+\frac {x}{\left (x+e^5\right )^2}+\frac {2 e^5 (2 x+1)}{\left (x+e^5\right )^2}-\frac {4 e^{e^x+10}}{\left (x+e^5\right )^2}+\frac {e^{10} (x+1)}{\left (x+e^5\right )^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int e^{x+e^x} xdx-4 \operatorname {ExpIntegralEi}\left (e^x\right )+2 x+\frac {e^5 \left (1-e^5\right )}{x+e^5}-\frac {2 e^5 \left (1-2 e^5\right )}{x+e^5}-\frac {2 e^{10}}{x+e^5}+\frac {e^5}{x+e^5}+\log (x)\) |
Input:
Int[(x^2 + 2*x^3 + E^10*(1 + x) + E^5*(2*x + 4*x^2) + E^E^x*(-4*E^10*x - 8 *E^5*x^2 - 4*x^3 + E^x*(-4*E^10*x^2 - 8*E^5*x^3 - 4*x^4)))/(E^10*x + 2*E^5 *x^2 + x^3),x]
Output:
$Aborted
Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
risch | \(2 x +\frac {{\mathrm e}^{10}}{{\mathrm e}^{5}+x}+\ln \left (x \right )-4 x \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(24\) |
norman | \(\frac {2 x^{2}-4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-4 x \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{10}}{{\mathrm e}^{5}+x}+\ln \left (x \right )\) | \(39\) |
parallelrisch | \(-\frac {4 x \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{x}}+4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+{\mathrm e}^{10}-{\mathrm e}^{5} \ln \left (x \right )-x \ln \left (x \right )-2 x^{2}}{{\mathrm e}^{5}+x}\) | \(46\) |
Input:
int((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5 )-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5 )^2+2*x^2*exp(5)+x^3),x,method=_RETURNVERBOSE)
Output:
2*x+exp(5)^2/(exp(5)+x)+ln(x)-4*x*exp(exp(x))
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\frac {2 \, x^{2} + 2 \, x e^{5} - 4 \, {\left (x^{2} + x e^{5}\right )} e^{\left (e^{x}\right )} + {\left (x + e^{5}\right )} \log \left (x\right ) + e^{10}}{x + e^{5}} \] Input:
integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2 *exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x *exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="fricas")
Output:
(2*x^2 + 2*x*e^5 - 4*(x^2 + x*e^5)*e^(e^x) + (x + e^5)*log(x) + e^10)/(x + e^5)
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=- 4 x e^{e^{x}} + 2 x + \log {\left (x \right )} + \frac {e^{10}}{x + e^{5}} \] Input:
integrate((((-4*x**2*exp(5)**2-8*x**3*exp(5)-4*x**4)*exp(x)-4*x*exp(5)**2- 8*x**2*exp(5)-4*x**3)*exp(exp(x))+(1+x)*exp(5)**2+(4*x**2+2*x)*exp(5)+2*x* *3+x**2)/(x*exp(5)**2+2*x**2*exp(5)+x**3),x)
Output:
-4*x*exp(exp(x)) + 2*x + log(x) + exp(10)/(x + exp(5))
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (23) = 46\).
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.72 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=-{\left (e^{\left (-10\right )} \log \left (x + e^{5}\right ) - e^{\left (-10\right )} \log \left (x\right ) - \frac {1}{x e^{5} + e^{10}}\right )} e^{10} + 4 \, {\left (\frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right )\right )} e^{5} - 4 \, x e^{\left (e^{x}\right )} - 4 \, e^{5} \log \left (x + e^{5}\right ) + 2 \, x - \frac {3 \, e^{10}}{x + e^{5}} - \frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right ) \] Input:
integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2 *exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x *exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="maxima")
Output:
-(e^(-10)*log(x + e^5) - e^(-10)*log(x) - 1/(x*e^5 + e^10))*e^10 + 4*(e^5/ (x + e^5) + log(x + e^5))*e^5 - 4*x*e^(e^x) - 4*e^5*log(x + e^5) + 2*x - 3 *e^10/(x + e^5) - e^5/(x + e^5) + log(x + e^5)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=-\frac {4 \, x^{2} e^{\left (x + e^{x}\right )} - 2 \, x^{2} e^{x} - x e^{x} \log \left (x\right ) + 4 \, x e^{\left (x + e^{x} + 5\right )} - 2 \, x e^{\left (x + 5\right )} - e^{\left (x + 5\right )} \log \left (x\right ) - e^{\left (x + 10\right )}}{x e^{x} + e^{\left (x + 5\right )}} \] Input:
integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2 *exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x *exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="giac")
Output:
-(4*x^2*e^(x + e^x) - 2*x^2*e^x - x*e^x*log(x) + 4*x*e^(x + e^x + 5) - 2*x *e^(x + 5) - e^(x + 5)*log(x) - e^(x + 10))/(x*e^x + e^(x + 5))
Time = 2.75 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=2\,x+\ln \left (x\right )-4\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+\frac {{\mathrm {e}}^{10}}{x+{\mathrm {e}}^5} \] Input:
int((exp(5)*(2*x + 4*x^2) - exp(exp(x))*(exp(x)*(8*x^3*exp(5) + 4*x^2*exp( 10) + 4*x^4) + 4*x*exp(10) + 8*x^2*exp(5) + 4*x^3) + exp(10)*(x + 1) + x^2 + 2*x^3)/(x*exp(10) + 2*x^2*exp(5) + x^3),x)
Output:
2*x + log(x) - 4*x*exp(exp(x)) + exp(10)/(x + exp(5))
Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\frac {-4 e^{e^{x}} e^{5} x -4 e^{e^{x}} x^{2}+\mathrm {log}\left (x \right ) e^{5}+\mathrm {log}\left (x \right ) x +e^{5} x +2 x^{2}}{e^{5}+x} \] Input:
int((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5 )-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5 )^2+2*x^2*exp(5)+x^3),x)
Output:
( - 4*e**(e**x)*e**5*x - 4*e**(e**x)*x**2 + log(x)*e**5 + log(x)*x + e**5* x + 2*x**2)/(e**5 + x)