Integrand size = 121, antiderivative size = 24 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=e^{4 (2-2 x) \left (-1+\frac {1}{x^2 \left (e^{2 x}+x\right )}\right )} \]
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=e^{-8+8 x-\frac {8 (-1+x)}{x^2 \left (e^{2 x}+x\right )}} \]
Integrate[(E^((8 - 8*x - 8*x^3 + 8*x^4 + E^(2*x)*(-8*x^2 + 8*x^3))/(E^(2*x )*x^2 + x^3))*(-24*x + 16*x^2 + 8*E^(4*x)*x^3 + 8*x^5 + E^(2*x)*(-16 - 8*x + 16*x^2 + 16*x^4)))/(E^(4*x)*x^3 + 2*E^(2*x)*x^4 + x^5),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.
\(\displaystyle \int \frac {\left (8 x^5+8 e^{4 x} x^3+16 x^2+e^{2 x} \left (16 x^4+16 x^2-8 x-16\right )-24 x\right ) \exp \left (\frac {8 x^4-8 x^3+e^{2 x} \left (8 x^3-8 x^2\right )-8 x+8}{x^3+e^{2 x} x^2}\right )}{x^5+2 e^{2 x} x^4+e^{4 x} x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (8 x^5+8 e^{4 x} x^3+16 x^2+e^{2 x} \left (16 x^4+16 x^2-8 x-16\right )-24 x\right ) \exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^3 \left (x+e^{2 x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {8 \left (2 x^2-3 x+1\right ) \exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^2 \left (x+e^{2 x}\right )^2}+8 \exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )+\frac {8 \left (2 x^2-x-2\right ) \exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^3 \left (x+e^{2 x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \int \exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )dx-16 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{\left (x+e^{2 x}\right )^2}dx-8 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^2 \left (x+e^{2 x}\right )^2}dx+24 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x \left (x+e^{2 x}\right )^2}dx-16 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^3 \left (x+e^{2 x}\right )}dx-8 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x^2 \left (x+e^{2 x}\right )}dx+16 \int \frac {\exp \left (\frac {8 (x-1) \left (x^3+e^{2 x} x^2-1\right )}{x^3+e^{2 x} x^2}\right )}{x \left (x+e^{2 x}\right )}dx\) |
Int[(E^((8 - 8*x - 8*x^3 + 8*x^4 + E^(2*x)*(-8*x^2 + 8*x^3))/(E^(2*x)*x^2 + x^3))*(-24*x + 16*x^2 + 8*E^(4*x)*x^3 + 8*x^5 + E^(2*x)*(-16 - 8*x + 16* x^2 + 16*x^4)))/(E^(4*x)*x^3 + 2*E^(2*x)*x^4 + x^5),x]
3.1.50.3.1 Defintions of rubi rules used
Time = 3.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
risch | \({\mathrm e}^{\frac {8 \left (-1+x \right ) \left ({\mathrm e}^{2 x} x^{2}+x^{3}-1\right )}{x^{2} \left ({\mathrm e}^{2 x}+x \right )}}\) | \(31\) |
parallelrisch | \({\mathrm e}^{\frac {\left (8 x^{3}-8 x^{2}\right ) {\mathrm e}^{2 x}+8 x^{4}-8 x^{3}-8 x +8}{x^{2} \left ({\mathrm e}^{2 x}+x \right )}}\) | \(45\) |
int((8*x^3*exp(x)^4+(16*x^4+16*x^2-8*x-16)*exp(x)^2+8*x^5+16*x^2-24*x)*exp (((8*x^3-8*x^2)*exp(x)^2+8*x^4-8*x^3-8*x+8)/(exp(x)^2*x^2+x^3))/(x^3*exp(x )^4+2*exp(x)^2*x^4+x^5),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=e^{\left (\frac {8 \, {\left (x^{4} - x^{3} + {\left (x^{3} - x^{2}\right )} e^{\left (2 \, x\right )} - x + 1\right )}}{x^{3} + x^{2} e^{\left (2 \, x\right )}}\right )} \]
integrate((8*x^3*exp(x)^4+(16*x^4+16*x^2-8*x-16)*exp(x)^2+8*x^5+16*x^2-24* x)*exp(((8*x^3-8*x^2)*exp(x)^2+8*x^4-8*x^3-8*x+8)/(exp(x)^2*x^2+x^3))/(x^3 *exp(x)^4+2*exp(x)^2*x^4+x^5),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=e^{\frac {8 x^{4} - 8 x^{3} - 8 x + \left (8 x^{3} - 8 x^{2}\right ) e^{2 x} + 8}{x^{3} + x^{2} e^{2 x}}} \]
integrate((8*x**3*exp(x)**4+(16*x**4+16*x**2-8*x-16)*exp(x)**2+8*x**5+16*x **2-24*x)*exp(((8*x**3-8*x**2)*exp(x)**2+8*x**4-8*x**3-8*x+8)/(exp(x)**2*x **2+x**3))/(x**3*exp(x)**4+2*exp(x)**2*x**4+x**5),x)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (20) = 40\).
Time = 0.44 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.62 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=e^{\left (8 \, x - \frac {8 \, e^{\left (-2 \, x\right )}}{x} - \frac {8 \, e^{\left (-4 \, x\right )}}{x} + \frac {8}{x e^{\left (4 \, x\right )} + e^{\left (6 \, x\right )}} + \frac {8}{x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}} + \frac {8 \, e^{\left (-2 \, x\right )}}{x^{2}} - 8\right )} \]
integrate((8*x^3*exp(x)^4+(16*x^4+16*x^2-8*x-16)*exp(x)^2+8*x^5+16*x^2-24* x)*exp(((8*x^3-8*x^2)*exp(x)^2+8*x^4-8*x^3-8*x+8)/(exp(x)^2*x^2+x^3))/(x^3 *exp(x)^4+2*exp(x)^2*x^4+x^5),x, algorithm=\
e^(8*x - 8*e^(-2*x)/x - 8*e^(-4*x)/x + 8/(x*e^(4*x) + e^(6*x)) + 8/(x*e^(2 *x) + e^(4*x)) + 8*e^(-2*x)/x^2 - 8)
Exception generated. \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx=\text {Exception raised: TypeError} \]
integrate((8*x^3*exp(x)^4+(16*x^4+16*x^2-8*x-16)*exp(x)^2+8*x^5+16*x^2-24* x)*exp(((8*x^3-8*x^2)*exp(x)^2+8*x^4-8*x^3-8*x+8)/(exp(x)^2*x^2+x^3))/(x^3 *exp(x)^4+2*exp(x)^2*x^4+x^5),x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{512,[1,17]%%%}+%%%{-2048,[1,16]%%%}+%%%{2816,[1,15]%%%}+%% %{-6752,[
Time = 11.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75 \[ \int \frac {e^{\frac {8-8 x-8 x^3+8 x^4+e^{2 x} \left (-8 x^2+8 x^3\right )}{e^{2 x} x^2+x^3}} \left (-24 x+16 x^2+8 e^{4 x} x^3+8 x^5+e^{2 x} \left (-16-8 x+16 x^2+16 x^4\right )\right )}{e^{4 x} x^3+2 e^{2 x} x^4+x^5} \, dx={\mathrm {e}}^{\frac {8\,x^2}{x+{\mathrm {e}}^{2\,x}}}\,{\mathrm {e}}^{-\frac {8}{x\,{\mathrm {e}}^{2\,x}+x^2}}\,{\mathrm {e}}^{-\frac {8\,{\mathrm {e}}^{2\,x}}{x+{\mathrm {e}}^{2\,x}}}\,{\mathrm {e}}^{-\frac {8\,x}{x+{\mathrm {e}}^{2\,x}}}\,{\mathrm {e}}^{\frac {8}{x^2\,{\mathrm {e}}^{2\,x}+x^3}}\,{\mathrm {e}}^{\frac {8\,x\,{\mathrm {e}}^{2\,x}}{x+{\mathrm {e}}^{2\,x}}} \]
int((exp(-(8*x + exp(2*x)*(8*x^2 - 8*x^3) + 8*x^3 - 8*x^4 - 8)/(x^2*exp(2* x) + x^3))*(8*x^3*exp(4*x) - exp(2*x)*(8*x - 16*x^2 - 16*x^4 + 16) - 24*x + 16*x^2 + 8*x^5))/(2*x^4*exp(2*x) + x^3*exp(4*x) + x^5),x)