Integrand size = 104, antiderivative size = 25 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\frac {x}{4+e^{5/4}-\frac {x^2}{4+\frac {4}{x}}} \] Output:
x/(4-x^2/(4+4/x)+exp(5/4))
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\frac {4 x (1+x)}{16+16 x-x^3+4 e^{5/4} (1+x)} \] Input:
Integrate[(64 + 128*x + 64*x^2 + 8*x^3 + 4*x^4 + E^(5/4)*(16 + 32*x + 16*x ^2))/(256 + 512*x + 256*x^2 - 32*x^3 - 32*x^4 + x^6 + E^(5/2)*(16 + 32*x + 16*x^2) + E^(5/4)*(128 + 256*x + 128*x^2 - 8*x^3 - 8*x^4)),x]
Output:
(4*x*(1 + x))/(16 + 16*x - x^3 + 4*E^(5/4)*(1 + 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.
| \(\displaystyle \int \frac {4 x^4+8 x^3+64 x^2+e^{5/4} \left (16 x^2+32 x+16\right )+128 x+64}{x^6-32 x^4-32 x^3+256 x^2+e^{5/2} \left (16 x^2+32 x+16\right )+e^{5/4} \left (-8 x^4-8 x^3+128 x^2+256 x+128\right )+512 x+256} \, dx\) |
| \(\Big \downarrow \) 2462 |
| \(\displaystyle \int \left (\frac {4 (-x-2)}{-x^3+4 \left (4+e^{5/4}\right ) x+4 \left (4+e^{5/4}\right )}+\frac {16 \left (4+e^{5/4}\right ) \left (2 x^2+5 x+3\right )}{\left (-x^3+4 \left (4+e^{5/4}\right ) x+4 \left (4+e^{5/4}\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {4 (-x-2)}{-x^3+4 \left (4+e^{5/4}\right ) x+4 \left (4+e^{5/4}\right )}+\frac {16 \left (4+e^{5/4}\right ) \left (2 x^2+5 x+3\right )}{\left (-x^3+4 \left (4+e^{5/4}\right ) x+4 \left (4+e^{5/4}\right )\right )^2}\right )dx\) |
Input:
Int[(64 + 128*x + 64*x^2 + 8*x^3 + 4*x^4 + E^(5/4)*(16 + 32*x + 16*x^2))/( 256 + 512*x + 256*x^2 - 32*x^3 - 32*x^4 + x^6 + E^(5/2)*(16 + 32*x + 16*x^ 2) + E^(5/4)*(128 + 256*x + 128*x^2 - 8*x^3 - 8*x^4)),x]
Output:
$Aborted
Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(\frac {x^{2}+x}{-\frac {x^{3}}{4}+{\mathrm e}^{\frac {5}{4}} x +{\mathrm e}^{\frac {5}{4}}+4 x +4}\) | \(25\) |
| gosper | \(\frac {4 \left (1+x \right ) x}{-x^{3}+4 \,{\mathrm e}^{\frac {5}{4}} x +4 \,{\mathrm e}^{\frac {5}{4}}+16 x +16}\) | \(28\) |
| norman | \(\frac {4 x^{2}+4 x}{-x^{3}+4 \,{\mathrm e}^{\frac {5}{4}} x +4 \,{\mathrm e}^{\frac {5}{4}}+16 x +16}\) | \(32\) |
| parallelrisch | \(-\frac {-4 x^{2}-4 x}{-x^{3}+4 \,{\mathrm e}^{\frac {5}{4}} x +4 \,{\mathrm e}^{\frac {5}{4}}+16 x +16}\) | \(33\) |
| default | \(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}+\left (-8 \,{\mathrm e}^{\frac {5}{4}}-32\right ) \textit {\_Z}^{4}+\left (-8 \,{\mathrm e}^{\frac {5}{4}}-32\right ) \textit {\_Z}^{3}+\left (256+16 \,{\mathrm e}^{\frac {5}{2}}+128 \,{\mathrm e}^{\frac {5}{4}}\right ) \textit {\_Z}^{2}+\left (32 \,{\mathrm e}^{\frac {5}{2}}+256 \,{\mathrm e}^{\frac {5}{4}}+512\right ) \textit {\_Z} +256+16 \,{\mathrm e}^{\frac {5}{2}}+128 \,{\mathrm e}^{\frac {5}{4}}\right )}{\sum }\frac {\left (16+\textit {\_R}^{4}+2 \textit {\_R}^{3}+4 \left (4+{\mathrm e}^{\frac {5}{4}}\right ) \textit {\_R}^{2}+8 \left (4+{\mathrm e}^{\frac {5}{4}}\right ) \textit {\_R} +4 \,{\mathrm e}^{\frac {5}{4}}\right ) \ln \left (x -\textit {\_R} \right )}{256+3 \textit {\_R}^{5}-16 \,{\mathrm e}^{\frac {5}{4}} \textit {\_R}^{3}+16 \,{\mathrm e}^{\frac {5}{2}} \textit {\_R} -12 \,{\mathrm e}^{\frac {5}{4}} \textit {\_R}^{2}-64 \textit {\_R}^{3}+16 \,{\mathrm e}^{\frac {5}{2}}+128 \,{\mathrm e}^{\frac {5}{4}} \textit {\_R} -48 \textit {\_R}^{2}+128 \,{\mathrm e}^{\frac {5}{4}}+256 \textit {\_R}}\right )\) | \(157\) |
Input:
int(((16*x^2+32*x+16)*exp(5/4)+4*x^4+8*x^3+64*x^2+128*x+64)/((16*x^2+32*x+ 16)*exp(5/4)^2+(-8*x^4-8*x^3+128*x^2+256*x+128)*exp(5/4)+x^6-32*x^4-32*x^3 +256*x^2+512*x+256),x,method=_RETURNVERBOSE)
Output:
(x^2+x)/(-1/4*x^3+exp(5/4)*x+exp(5/4)+4*x+4)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=-\frac {4 \, {\left (x^{2} + x\right )}}{x^{3} - 4 \, {\left (x + 1\right )} e^{\frac {5}{4}} - 16 \, x - 16} \] Input:
integrate(((16*x^2+32*x+16)*exp(5/4)+4*x^4+8*x^3+64*x^2+128*x+64)/((16*x^2 +32*x+16)*exp(5/4)^2+(-8*x^4-8*x^3+128*x^2+256*x+128)*exp(5/4)+x^6-32*x^4- 32*x^3+256*x^2+512*x+256),x, algorithm="fricas")
Output:
-4*(x^2 + x)/(x^3 - 4*(x + 1)*e^(5/4) - 16*x - 16)
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (15) = 30\).
Time = 2.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 5.36 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\frac {x^{2} \left (- 660 e^{\frac {5}{2}} - 2208 e^{\frac {5}{4}} - 64 e^{\frac {15}{4}} - 2368\right ) + x \left (- 660 e^{\frac {5}{2}} - 2208 e^{\frac {5}{4}} - 64 e^{\frac {15}{4}} - 2368\right )}{x^{3} \cdot \left (592 + 16 e^{\frac {15}{4}} + 552 e^{\frac {5}{4}} + 165 e^{\frac {5}{2}}\right ) + x \left (- 4848 e^{\frac {5}{2}} - 11200 e^{\frac {5}{4}} - 916 e^{\frac {15}{4}} - 64 e^{5} - 9472\right ) - 4848 e^{\frac {5}{2}} - 11200 e^{\frac {5}{4}} - 916 e^{\frac {15}{4}} - 64 e^{5} - 9472} \] Input:
integrate(((16*x**2+32*x+16)*exp(5/4)+4*x**4+8*x**3+64*x**2+128*x+64)/((16 *x**2+32*x+16)*exp(5/4)**2+(-8*x**4-8*x**3+128*x**2+256*x+128)*exp(5/4)+x* *6-32*x**4-32*x**3+256*x**2+512*x+256),x)
Output:
(x**2*(-660*exp(5/2) - 2208*exp(5/4) - 64*exp(15/4) - 2368) + x*(-660*exp( 5/2) - 2208*exp(5/4) - 64*exp(15/4) - 2368))/(x**3*(592 + 16*exp(15/4) + 5 52*exp(5/4) + 165*exp(5/2)) + x*(-4848*exp(5/2) - 11200*exp(5/4) - 916*exp (15/4) - 64*exp(5) - 9472) - 4848*exp(5/2) - 11200*exp(5/4) - 916*exp(15/4 ) - 64*exp(5) - 9472)
\[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\int { \frac {4 \, {\left (x^{4} + 2 \, x^{3} + 16 \, x^{2} + 4 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\frac {5}{4}} + 32 \, x + 16\right )}}{x^{6} - 32 \, x^{4} - 32 \, x^{3} + 256 \, x^{2} + 16 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\frac {5}{2}} - 8 \, {\left (x^{4} + x^{3} - 16 \, x^{2} - 32 \, x - 16\right )} e^{\frac {5}{4}} + 512 \, x + 256} \,d x } \] Input:
integrate(((16*x^2+32*x+16)*exp(5/4)+4*x^4+8*x^3+64*x^2+128*x+64)/((16*x^2 +32*x+16)*exp(5/4)^2+(-8*x^4-8*x^3+128*x^2+256*x+128)*exp(5/4)+x^6-32*x^4- 32*x^3+256*x^2+512*x+256),x, algorithm="maxima")
Output:
4*integrate((x^4 + 2*x^3 + 16*x^2 + 4*(x^2 + 2*x + 1)*e^(5/4) + 32*x + 16) /(x^6 - 32*x^4 - 32*x^3 + 256*x^2 + 16*(x^2 + 2*x + 1)*e^(5/2) - 8*(x^4 + x^3 - 16*x^2 - 32*x - 16)*e^(5/4) + 512*x + 256), x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=-\frac {4 \, {\left (x^{2} + x\right )}}{x^{3} - 4 \, x e^{\frac {5}{4}} - 16 \, x - 4 \, e^{\frac {5}{4}} - 16} \] Input:
integrate(((16*x^2+32*x+16)*exp(5/4)+4*x^4+8*x^3+64*x^2+128*x+64)/((16*x^2 +32*x+16)*exp(5/4)^2+(-8*x^4-8*x^3+128*x^2+256*x+128)*exp(5/4)+x^6-32*x^4- 32*x^3+256*x^2+512*x+256),x, algorithm="giac")
Output:
-4*(x^2 + x)/(x^3 - 4*x*e^(5/4) - 16*x - 4*e^(5/4) - 16)
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\frac {4\,x\,\left (x+1\right )}{-x^3+\left (4\,{\mathrm {e}}^{5/4}+16\right )\,x+4\,{\mathrm {e}}^{5/4}+16} \] Input:
int((128*x + exp(5/4)*(32*x + 16*x^2 + 16) + 64*x^2 + 8*x^3 + 4*x^4 + 64)/ (512*x + exp(5/2)*(32*x + 16*x^2 + 16) + exp(5/4)*(256*x + 128*x^2 - 8*x^3 - 8*x^4 + 128) + 256*x^2 - 32*x^3 - 32*x^4 + x^6 + 256),x)
Output:
(4*x*(x + 1))/(4*exp(5/4) - x^3 + x*(4*exp(5/4) + 16) + 16)
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.84 \[ \int \frac {64+128 x+64 x^2+8 x^3+4 x^4+e^{5/4} \left (16+32 x+16 x^2\right )}{256+512 x+256 x^2-32 x^3-32 x^4+x^6+e^{5/2} \left (16+32 x+16 x^2\right )+e^{5/4} \left (128+256 x+128 x^2-8 x^3-8 x^4\right )} \, dx=\frac {-4 e^{\frac {5}{4}} x^{2}+4 e^{\frac {5}{4}}-x^{3}-16 x^{2}+16}{e^{\frac {5}{4}} x^{3}-32 e^{\frac {5}{4}} x -32 e^{\frac {5}{4}}-4 \sqrt {e}\, e^{2} x -4 \sqrt {e}\, e^{2}+4 x^{3}-64 x -64} \] Input:
int(((16*x^2+32*x+16)*exp(5/4)+4*x^4+8*x^3+64*x^2+128*x+64)/((16*x^2+32*x+ 16)*exp(5/4)^2+(-8*x^4-8*x^3+128*x^2+256*x+128)*exp(5/4)+x^6-32*x^4-32*x^3 +256*x^2+512*x+256),x)
Output:
( - 4*e**(1/4)*e*x**2 + 4*e**(1/4)*e - x**3 - 16*x**2 + 16)/(e**(1/4)*e*x* *3 - 32*e**(1/4)*e*x - 32*e**(1/4)*e - 4*sqrt(e)*e**2*x - 4*sqrt(e)*e**2 + 4*x**3 - 64*x - 64)