Integrand size = 88, antiderivative size = 25 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=\frac {5}{9-x-x^2+e^{-6-2 x} x^2} \]
Time = 4.82 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=-\frac {5 e^{6+2 x}}{-x^2+e^{6+2 x} \left (-9+x+x^2\right )} \]
Integrate[(E^(12 + 4*x)*(5 + 10*x) + E^(6 + 2*x)*(-10*x + 10*x^2))/(x^4 + E^(6 + 2*x)*(18*x^2 - 2*x^3 - 2*x^4) + E^(12 + 4*x)*(81 - 18*x - 17*x^2 + 2*x^3 + x^4)),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.
\(\displaystyle \int \frac {e^{2 x+6} \left (10 x^2-10 x\right )+e^{4 x+12} (10 x+5)}{x^4+e^{2 x+6} \left (-2 x^4-2 x^3+18 x^2\right )+e^{4 x+12} \left (x^4+2 x^3-17 x^2-18 x+81\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 e^{2 x+6} \left (2 (x-1) x+e^{2 x+6} (2 x+1)\right )}{\left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {e^{2 x+6} \left (2 (1-x) x-e^{2 x+6} (2 x+1)\right )}{\left (x^2+e^{2 x+6} \left (-x^2-x+9\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {e^{2 x+6} \left (2 (1-x) x-e^{2 x+6} (2 x+1)\right )}{\left (x^2+e^{2 x+6} \left (-x^2-x+9\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (-\frac {e^{2 x+6} (2 x+1)}{\left (x^2+x-9\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )}-\frac {e^{2 x+6} x \left (2 x^3+2 x^2-19 x+18\right )}{\left (x^2+x-9\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (\frac {2 \int \frac {e^{2 x+6}}{\left (-2 x+\sqrt {37}-1\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )}dx}{\sqrt {37}}-\frac {2}{37} \left (37-\sqrt {37}\right ) \int \frac {e^{2 x+6}}{\left (2 x-\sqrt {37}+1\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )}dx-\frac {2}{37} \left (37+\sqrt {37}\right ) \int \frac {e^{2 x+6}}{\left (2 x+\sqrt {37}+1\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )}dx+\frac {2 \int \frac {e^{2 x+6}}{\left (2 x+\sqrt {37}+1\right ) \left (e^{2 x+6} x^2-x^2+e^{2 x+6} x-9 e^{2 x+6}\right )}dx}{\sqrt {37}}+\int \frac {e^{2 x+6}}{\left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx-\frac {18 \int \frac {e^{2 x+6}}{\left (-2 x+\sqrt {37}-1\right ) \left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx}{\sqrt {37}}-2 \int \frac {e^{2 x+6} x^2}{\left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx-\frac {19}{37} \left (37-\sqrt {37}\right ) \int \frac {e^{2 x+6}}{\left (2 x-\sqrt {37}+1\right ) \left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx-\frac {19}{37} \left (37+\sqrt {37}\right ) \int \frac {e^{2 x+6}}{\left (2 x+\sqrt {37}+1\right ) \left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx-\frac {18 \int \frac {e^{2 x+6}}{\left (2 x+\sqrt {37}+1\right ) \left (x^2-e^{2 x+6} \left (x^2+x-9\right )\right )^2}dx}{\sqrt {37}}\right )\) |
Int[(E^(12 + 4*x)*(5 + 10*x) + E^(6 + 2*x)*(-10*x + 10*x^2))/(x^4 + E^(6 + 2*x)*(18*x^2 - 2*x^3 - 2*x^4) + E^(12 + 4*x)*(81 - 18*x - 17*x^2 + 2*x^3 + x^4)),x]
3.17.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04
method | result | size |
norman | \(-\frac {5 \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{x^{2} {\mathrm e}^{6} {\mathrm e}^{2 x}+{\mathrm e}^{6} {\mathrm e}^{2 x} x -9 \,{\mathrm e}^{2 x} {\mathrm e}^{6}-x^{2}}\) | \(51\) |
parallelrisch | \(-\frac {5 \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{x^{2} {\mathrm e}^{6} {\mathrm e}^{2 x}+{\mathrm e}^{6} {\mathrm e}^{2 x} x -9 \,{\mathrm e}^{2 x} {\mathrm e}^{6}-x^{2}}\) | \(51\) |
risch | \(-\frac {5}{x^{2}+x -9}-\frac {5 x^{2}}{\left (x^{2}+x -9\right ) \left ({\mathrm e}^{2 x +6} x^{2}+x \,{\mathrm e}^{2 x +6}-9 \,{\mathrm e}^{2 x +6}-x^{2}\right )}\) | \(59\) |
int(((10*x+5)*exp(3)^4*exp(x)^4+(10*x^2-10*x)*exp(3)^2*exp(x)^2)/((x^4+2*x ^3-17*x^2-18*x+81)*exp(3)^4*exp(x)^4+(-2*x^4-2*x^3+18*x^2)*exp(3)^2*exp(x) ^2+x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=\frac {5 \, e^{\left (2 \, x + 6\right )}}{x^{2} - {\left (x^{2} + x - 9\right )} e^{\left (2 \, x + 6\right )}} \]
integrate(((10*x+5)*exp(3)^4*exp(x)^4+(10*x^2-10*x)*exp(3)^2*exp(x)^2)/((x ^4+2*x^3-17*x^2-18*x+81)*exp(3)^4*exp(x)^4+(-2*x^4-2*x^3+18*x^2)*exp(3)^2* exp(x)^2+x^4),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (19) = 38\).
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.64 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=- \frac {5 x^{2}}{- x^{4} - x^{3} + 9 x^{2} + \left (x^{4} e^{6} + 2 x^{3} e^{6} - 17 x^{2} e^{6} - 18 x e^{6} + 81 e^{6}\right ) e^{2 x}} - \frac {5}{x^{2} + x - 9} \]
integrate(((10*x+5)*exp(3)**4*exp(x)**4+(10*x**2-10*x)*exp(3)**2*exp(x)**2 )/((x**4+2*x**3-17*x**2-18*x+81)*exp(3)**4*exp(x)**4+(-2*x**4-2*x**3+18*x* *2)*exp(3)**2*exp(x)**2+x**4),x)
-5*x**2/(-x**4 - x**3 + 9*x**2 + (x**4*exp(6) + 2*x**3*exp(6) - 17*x**2*ex p(6) - 18*x*exp(6) + 81*exp(6))*exp(2*x)) - 5/(x**2 + x - 9)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=\frac {5 \, e^{\left (2 \, x + 6\right )}}{x^{2} - {\left (x^{2} e^{6} + x e^{6} - 9 \, e^{6}\right )} e^{\left (2 \, x\right )}} \]
integrate(((10*x+5)*exp(3)^4*exp(x)^4+(10*x^2-10*x)*exp(3)^2*exp(x)^2)/((x ^4+2*x^3-17*x^2-18*x+81)*exp(3)^4*exp(x)^4+(-2*x^4-2*x^3+18*x^2)*exp(3)^2* exp(x)^2+x^4),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=-\frac {5 \, e^{\left (2 \, x + 6\right )}}{x^{2} e^{\left (2 \, x + 6\right )} - x^{2} + x e^{\left (2 \, x + 6\right )} - 9 \, e^{\left (2 \, x + 6\right )}} \]
integrate(((10*x+5)*exp(3)^4*exp(x)^4+(10*x^2-10*x)*exp(3)^2*exp(x)^2)/((x ^4+2*x^3-17*x^2-18*x+81)*exp(3)^4*exp(x)^4+(-2*x^4-2*x^3+18*x^2)*exp(3)^2* exp(x)^2+x^4),x, algorithm=\
Time = 8.65 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {e^{12+4 x} (5+10 x)+e^{6+2 x} \left (-10 x+10 x^2\right )}{x^4+e^{6+2 x} \left (18 x^2-2 x^3-2 x^4\right )+e^{12+4 x} \left (81-18 x-17 x^2+2 x^3+x^4\right )} \, dx=\frac {5\,x\,\left ({\mathrm {e}}^{2\,x+6}-x+x\,{\mathrm {e}}^{2\,x+6}\right )}{9\,\left (9\,{\mathrm {e}}^{2\,x+6}-x\,{\mathrm {e}}^{2\,x+6}-x^2\,{\mathrm {e}}^{2\,x+6}+x^2\right )} \]
int((exp(4*x)*exp(12)*(10*x + 5) - exp(2*x)*exp(6)*(10*x - 10*x^2))/(x^4 - exp(2*x)*exp(6)*(2*x^3 - 18*x^2 + 2*x^4) + exp(4*x)*exp(12)*(2*x^3 - 17*x ^2 - 18*x + x^4 + 81)),x)