Integrand size = 87, antiderivative size = 35 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=-x+\frac {\frac {e^x}{-5 e^x+\frac {x^2}{3}}+3 (1+x+\log (2))}{x} \] Output:
(exp(x)/(1/3*x^2-5*exp(x))+3+3*x+3*ln(2))/x-x
Time = 1.81 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=\frac {3 e^x \left (-14+5 x^2-15 \log (2)\right )+x^2 \left (3-x^2+\log (8)\right )}{-15 e^x x+x^3} \] Input:
Integrate[(-3*x^4 - x^6 + E^(2*x)*(-630 - 225*x^2 - 675*Log[2]) - 3*x^4*Lo g[2] + E^x*(81*x^2 + 3*x^3 + 30*x^4 + 90*x^2*Log[2]))/(225*E^(2*x)*x^2 - 3 0*E^x*x^4 + x^6),x]
Output:
(3*E^x*(-14 + 5*x^2 - 15*Log[2]) + x^2*(3 - x^2 + Log[8]))/(-15*E^x*x + x^ 3)
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.
\(\displaystyle \int \frac {-x^6-3 x^4-3 x^4 \log (2)+e^{2 x} \left (-225 x^2-630-675 \log (2)\right )+e^x \left (30 x^4+3 x^3+81 x^2+90 x^2 \log (2)\right )}{x^6-30 e^x x^4+225 e^{2 x} x^2} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^6+x^4 (-3-3 \log (2))+e^{2 x} \left (-225 x^2-630-675 \log (2)\right )+e^x \left (30 x^4+3 x^3+81 x^2+90 x^2 \log (2)\right )}{x^6-30 e^x x^4+225 e^{2 x} x^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^6+x^4 (-3-3 \log (2))+e^{2 x} \left (-225 x^2-630-675 \log (2)\right )+e^x \left (30 x^4+3 x^3+81 x^2+90 x^2 \log (2)\right )}{x^2 \left (15 e^x-x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(x-2) x^2}{5 \left (x^2-15 e^x\right )^2}-\frac {x-1}{5 \left (x^2-15 e^x\right )}+\frac {-5 x^2-14-15 \log (2)}{5 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{5} \int \frac {1}{15 e^x-x^2}dx-\frac {2}{5} \int \frac {x^2}{\left (x^2-15 e^x\right )^2}dx-\frac {1}{5} \int \frac {x}{x^2-15 e^x}dx+\frac {1}{5} \int \frac {x^3}{\left (x^2-15 e^x\right )^2}dx-x+\frac {14+15 \log (2)}{5 x}\) |
Input:
Int[(-3*x^4 - x^6 + E^(2*x)*(-630 - 225*x^2 - 675*Log[2]) - 3*x^4*Log[2] + E^x*(81*x^2 + 3*x^3 + 30*x^4 + 90*x^2*Log[2]))/(225*E^(2*x)*x^2 - 30*E^x* x^4 + x^6),x]
Output:
$Aborted
Time = 0.50 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86
method | result | size |
risch | \(-x +\frac {3 \ln \left (2\right )}{x}+\frac {14}{5 x}+\frac {x}{5 x^{2}-75 \,{\mathrm e}^{x}}\) | \(30\) |
norman | \(\frac {\left (-42-45 \ln \left (2\right )\right ) {\mathrm e}^{x}+\left (3 \ln \left (2\right )+3\right ) x^{2}-x^{4}+15 \,{\mathrm e}^{x} x^{2}}{x \left (x^{2}-15 \,{\mathrm e}^{x}\right )}\) | \(47\) |
parallelrisch | \(\frac {-x^{4}+3 x^{2} \ln \left (2\right )+15 \,{\mathrm e}^{x} x^{2}-45 \,{\mathrm e}^{x} \ln \left (2\right )+3 x^{2}-42 \,{\mathrm e}^{x}}{x \left (x^{2}-15 \,{\mathrm e}^{x}\right )}\) | \(50\) |
Input:
int(((-675*ln(2)-225*x^2-630)*exp(x)^2+(90*x^2*ln(2)+30*x^4+3*x^3+81*x^2)* exp(x)-3*x^4*ln(2)-x^6-3*x^4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6),x,metho d=_RETURNVERBOSE)
Output:
-x+3*ln(2)/x+14/5/x+1/5*x/(x^2-15*exp(x))
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=-\frac {x^{4} - 3 \, x^{2} \log \left (2\right ) - 3 \, x^{2} - 3 \, {\left (5 \, x^{2} - 15 \, \log \left (2\right ) - 14\right )} e^{x}}{x^{3} - 15 \, x e^{x}} \] Input:
integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+ 81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6 ),x, algorithm="fricas")
Output:
-(x^4 - 3*x^2*log(2) - 3*x^2 - 3*(5*x^2 - 15*log(2) - 14)*e^x)/(x^3 - 15*x *e^x)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=- x - \frac {x}{- 5 x^{2} + 75 e^{x}} - \frac {-14 - 15 \log {\left (2 \right )}}{5 x} \] Input:
integrate(((-675*ln(2)-225*x**2-630)*exp(x)**2+(90*x**2*ln(2)+30*x**4+3*x* *3+81*x**2)*exp(x)-3*x**4*ln(2)-x**6-3*x**4)/(225*exp(x)**2*x**2-30*exp(x) *x**4+x**6),x)
Output:
-x - x/(-5*x**2 + 75*exp(x)) - (-14 - 15*log(2))/(5*x)
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=-\frac {x^{4} - 3 \, x^{2} {\left (\log \left (2\right ) + 1\right )} - 3 \, {\left (5 \, x^{2} - 15 \, \log \left (2\right ) - 14\right )} e^{x}}{x^{3} - 15 \, x e^{x}} \] Input:
integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+ 81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6 ),x, algorithm="maxima")
Output:
-(x^4 - 3*x^2*(log(2) + 1) - 3*(5*x^2 - 15*log(2) - 14)*e^x)/(x^3 - 15*x*e ^x)
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=-\frac {x^{4} - 15 \, x^{2} e^{x} - 3 \, x^{2} \log \left (2\right ) - 3 \, x^{2} + 45 \, e^{x} \log \left (2\right ) + 42 \, e^{x}}{x^{3} - 15 \, x e^{x}} \] Input:
integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+ 81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6 ),x, algorithm="giac")
Output:
-(x^4 - 15*x^2*e^x - 3*x^2*log(2) - 3*x^2 + 45*e^x*log(2) + 42*e^x)/(x^3 - 15*x*e^x)
Time = 2.79 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=-x-\frac {x^2\,\left (\ln \left (8\right )+3\right )-{\mathrm {e}}^x\,\left (45\,\ln \left (2\right )+42\right )}{15\,x\,{\mathrm {e}}^x-x^3} \] Input:
int(-(3*x^4*log(2) + 3*x^4 + x^6 - exp(x)*(90*x^2*log(2) + 81*x^2 + 3*x^3 + 30*x^4) + exp(2*x)*(675*log(2) + 225*x^2 + 630))/(225*x^2*exp(2*x) - 30* x^4*exp(x) + x^6),x)
Output:
- x - (x^2*(log(8) + 3) - exp(x)*(45*log(2) + 42))/(15*x*exp(x) - x^3)
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51 \[ \int \frac {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx=\frac {45 e^{x} \mathrm {log}\left (2\right )-15 e^{x} x^{2}+42 e^{x}-3 \,\mathrm {log}\left (2\right ) x^{2}+x^{4}-3 x^{2}}{x \left (15 e^{x}-x^{2}\right )} \] Input:
int(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+81*x^2 )*exp(x)-3*x^4*log(2)-x^6-3*x^4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6),x)
Output:
(45*e**x*log(2) - 15*e**x*x**2 + 42*e**x - 3*log(2)*x**2 + x**4 - 3*x**2)/ (x*(15*e**x - x**2))