Integrand size = 81, antiderivative size = 24 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=e^{\frac {4 x}{5 \left (x+e^{2 x} x^4\right )}}-2 x \] Output:
exp(2/5*x/(x+exp(x)^2*x^4))^2-2*x
Time = 0.52 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-\frac {2}{5} \left (-\frac {5}{2} e^{\frac {4}{5 \left (1+e^{2 x} x^3\right )}}+5 x\right ) \] Input:
Integrate[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2* x)*x^3))*(-12*x^2 - 8*x^3))/(5 + 10*E^(2*x)*x^3 + 5*E^(4*x)*x^6),x]
Output:
(-2*((-5*E^(4/(5*(1 + E^(2*x)*x^3))))/2 + 5*x))/5
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.
\(\displaystyle \int \frac {-10 e^{4 x} x^6-20 e^{2 x} x^3+e^{\frac {4}{5 e^{2 x} x^3+5}+2 x} \left (-8 x^3-12 x^2\right )-10}{5 e^{4 x} x^6+10 e^{2 x} x^3+5} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-10 e^{4 x} x^6-20 e^{2 x} x^3+e^{\frac {4}{5 e^{2 x} x^3+5}+2 x} \left (-8 x^3-12 x^2\right )-10}{5 \left (e^{2 x} x^3+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {2 \left (5 e^{4 x} x^6+10 e^{2 x} x^3+2 e^{2 x+\frac {4}{5 \left (e^{2 x} x^3+1\right )}} \left (2 x^3+3 x^2\right )+5\right )}{\left (e^{2 x} x^3+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{5} \int \frac {5 e^{4 x} x^6+10 e^{2 x} x^3+2 e^{2 x+\frac {4}{5 \left (e^{2 x} x^3+1\right )}} \left (2 x^3+3 x^2\right )+5}{\left (e^{2 x} x^3+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2}{5} \int \left (\frac {2 e^{2 x+\frac {4}{5 e^{2 x} x^3+5}} (2 x+3) x^2}{\left (e^{2 x} x^3+1\right )^2}+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2}{5} \left (4 \int \frac {e^{2 x+\frac {4}{5 e^{2 x} x^3+5}} x^3}{\left (e^{2 x} x^3+1\right )^2}dx+6 \int \frac {e^{2 x+\frac {4}{5 e^{2 x} x^3+5}} x^2}{\left (e^{2 x} x^3+1\right )^2}dx+5 x\right )\) |
Input:
Int[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2*x)*x^3 ))*(-12*x^2 - 8*x^3))/(5 + 10*E^(2*x)*x^3 + 5*E^(4*x)*x^6),x]
Output:
$Aborted
Time = 7.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \({\mathrm e}^{\frac {4}{5 \left ({\mathrm e}^{2 x} x^{3}+1\right )}}-2 x\) | \(20\) |
Input:
int(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4- 20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+10*exp(x)^2*x^3+5),x,method=_RETURNVER BOSE)
Output:
exp(4/5/(exp(2*x)*x^3+1))-2*x
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-{\left (2 \, x e^{\left (2 \, x\right )} - e^{\left (\frac {2 \, {\left (5 \, x^{4} e^{\left (2 \, x\right )} + 5 \, x + 2\right )}}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp (x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+10*exp(x)^2*x^3+5),x, algorithm= "fricas")
Output:
-(2*x*e^(2*x) - e^(2/5*(5*x^4*e^(2*x) + 5*x + 2)/(x^3*e^(2*x) + 1)))*e^(-2 *x)
Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=- 2 x + e^{\frac {4}{5 x^{3} e^{2 x} + 5}} \] Input:
integrate(((-8*x**3-12*x**2)*exp(x)**2*exp(2/(5*exp(x)**2*x**3+5))**2-10*x **6*exp(x)**4-20*exp(x)**2*x**3-10)/(5*x**6*exp(x)**4+10*exp(x)**2*x**3+5) ,x)
Output:
-2*x + exp(4/(5*x**3*exp(2*x) + 5))
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-2 \, x + e^{\left (\frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} \] Input:
integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp (x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+10*exp(x)^2*x^3+5),x, algorithm= "maxima")
Output:
-2*x + e^(4/5/(x^3*e^(2*x) + 1))
\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=\int { -\frac {2 \, {\left (5 \, x^{6} e^{\left (4 \, x\right )} + 10 \, x^{3} e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} + 5\right )}}{5 \, {\left (x^{6} e^{\left (4 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 1\right )}} \,d x } \] Input:
integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp (x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+10*exp(x)^2*x^3+5),x, algorithm= "giac")
Output:
integrate(-2/5*(5*x^6*e^(4*x) + 10*x^3*e^(2*x) + 2*(2*x^3 + 3*x^2)*e^(2*x + 4/5/(x^3*e^(2*x) + 1)) + 5)/(x^6*e^(4*x) + 2*x^3*e^(2*x) + 1), x)
Time = 2.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx={\mathrm {e}}^{\frac {4}{5\,x^3\,{\mathrm {e}}^{2\,x}+5}}-2\,x \] Input:
int(-(20*x^3*exp(2*x) + 10*x^6*exp(4*x) + exp(2*x)*exp(4/(5*x^3*exp(2*x) + 5))*(12*x^2 + 8*x^3) + 10)/(10*x^3*exp(2*x) + 5*x^6*exp(4*x) + 5),x)
Output:
exp(4/(5*x^3*exp(2*x) + 5)) - 2*x
\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-\frac {8 \left (\int \frac {e^{\frac {10 e^{2 x} x^{4}+10 x +4}{5 e^{2 x} x^{3}+5}} x^{3}}{e^{4 x} x^{6}+2 e^{2 x} x^{3}+1}d x \right )}{5}-\frac {12 \left (\int \frac {e^{\frac {10 e^{2 x} x^{4}+10 x +4}{5 e^{2 x} x^{3}+5}} x^{2}}{e^{4 x} x^{6}+2 e^{2 x} x^{3}+1}d x \right )}{5}-2 x \] Input:
int(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4- 20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+10*exp(x)^2*x^3+5),x)
Output:
(2*( - 4*int((e**((10*e**(2*x)*x**4 + 10*x + 4)/(5*e**(2*x)*x**3 + 5))*x** 3)/(e**(4*x)*x**6 + 2*e**(2*x)*x**3 + 1),x) - 6*int((e**((10*e**(2*x)*x**4 + 10*x + 4)/(5*e**(2*x)*x**3 + 5))*x**2)/(e**(4*x)*x**6 + 2*e**(2*x)*x**3 + 1),x) - 5*x))/5