Integrand size = 84, antiderivative size = 27 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=-\frac {x^3}{48 \left (2-e^{2-2 x}\right ) (4+5 x)} \]
Time = 12.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {-24 e^2 (4+5 x)+e^{2 x} \left (192+240 x-125 x^3\right )}{6000 \left (-e^2+2 e^{2 x}\right ) (4+5 x)} \]
Integrate[(-12*x^2 - 10*x^3 + E^(2 - 2*x)*(6*x^2 + 9*x^3 + 5*x^4))/(1536 + 3840*x + 2400*x^2 + E^(2 - 2*x)*(-1536 - 3840*x - 2400*x^2) + E^(4 - 4*x) *(384 + 960*x + 600*x^2)),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.
\(\displaystyle \int \frac {-10 x^3-12 x^2+e^{2-2 x} \left (5 x^4+9 x^3+6 x^2\right )}{2400 x^2+e^{2-2 x} \left (-2400 x^2-3840 x-1536\right )+e^{4-4 x} \left (600 x^2+960 x+384\right )+3840 x+1536} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{4 x} \left (-10 x^3-12 x^2+e^{2-2 x} \left (5 x^4+9 x^3+6 x^2\right )\right )}{24 \left (e^2-2 e^{2 x}\right )^2 (5 x+4)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \int -\frac {e^{4 x} \left (10 x^3+12 x^2-e^{2-2 x} \left (5 x^4+9 x^3+6 x^2\right )\right )}{\left (e^2-2 e^{2 x}\right )^2 (5 x+4)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{24} \int \frac {e^{4 x} \left (10 x^3+12 x^2-e^{2-2 x} \left (5 x^4+9 x^3+6 x^2\right )\right )}{\left (e^2-2 e^{2 x}\right )^2 (5 x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{24} \int \left (-\frac {2 e^{4 x} x^3}{\left (-e^2+2 e^{2 x}\right )^2 (5 x+4)}-\frac {e^{2 x-2} \left (5 x^2+9 x+6\right ) x^2}{(5 x+4)^2}-\frac {2 e^{4 x-2} \left (5 x^2+9 x+6\right ) x^2}{\left (e^2-2 e^{2 x}\right ) (5 x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{24} \left (-\frac {2}{5} \int \frac {e^{4 x-2} x^2}{-e^2+2 e^{2 x}}dx-\frac {2}{25} \int \frac {e^{4 x-2} x}{-e^2+2 e^{2 x}}dx-\frac {64}{25} \int \frac {e^{4 x-2}}{\left (-e^2+2 e^{2 x}\right ) (5 x+4)^2}dx-\frac {32}{125} e^4 \int \frac {1}{\left (e^2-2 e^{2 x}\right )^2 (5 x+4)}dx+\frac {64}{125} e^2 \int \frac {1}{\left (e^2-2 e^{2 x}\right ) (5 x+4)}dx+\frac {128}{125} \int \frac {e^{4 x-2}}{\left (-e^2+2 e^{2 x}\right ) (5 x+4)}dx+\frac {1}{20} x \operatorname {PolyLog}\left (2,2 e^{2 x-2}\right )+\frac {1}{200} \operatorname {PolyLog}\left (2,2 e^{2 x-2}\right )-\frac {1}{40} \operatorname {PolyLog}\left (3,2 e^{2 x-2}\right )+\frac {1}{10} e^{2 x-2} x^2+\frac {e^2 x^2}{20 \left (e^2-2 e^{2 x}\right )}-\frac {x^2}{20}+\frac {1}{20} x^2 \log \left (1-2 e^{2 x-2}\right )-\frac {2}{25} e^{2 x-2} x-\frac {e^2 x}{25 \left (e^2-2 e^{2 x}\right )}+\frac {x}{25}+\frac {1}{25} e^{2 x-2}+\frac {4 e^2}{125 \left (e^2-2 e^{2 x}\right )}-\frac {32 e^{2 x-2}}{125 (5 x+4)}+\frac {1}{100} x \log \left (1-2 e^{2 x-2}\right )-\frac {32}{625} \log (5 x+4)\right )\) |
Int[(-12*x^2 - 10*x^3 + E^(2 - 2*x)*(6*x^2 + 9*x^3 + 5*x^4))/(1536 + 3840* x + 2400*x^2 + E^(2 - 2*x)*(-1536 - 3840*x - 2400*x^2) + E^(4 - 4*x)*(384 + 960*x + 600*x^2)),x]
3.7.22.3.1 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]]
Time = 1.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
norman | \(\frac {x^{3}}{48 \left ({\mathrm e}^{2-2 x}-2\right ) \left (4+5 x \right )}\) | \(23\) |
risch | \(\frac {x^{3}}{48 \left ({\mathrm e}^{2-2 x}-2\right ) \left (4+5 x \right )}\) | \(23\) |
parallelrisch | \(\frac {x^{3}}{48 \left ({\mathrm e}^{2-2 x}-2\right ) \left (4+5 x \right )}\) | \(23\) |
int(((5*x^4+9*x^3+6*x^2)*exp(2-2*x)-10*x^3-12*x^2)/((600*x^2+960*x+384)*ex p(2-2*x)^2+(-2400*x^2-3840*x-1536)*exp(2-2*x)+2400*x^2+3840*x+1536),x,meth od=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {x^{3}}{48 \, {\left ({\left (5 \, x + 4\right )} e^{\left (-2 \, x + 2\right )} - 10 \, x - 8\right )}} \]
integrate(((5*x^4+9*x^3+6*x^2)*exp(2-2*x)-10*x^3-12*x^2)/((600*x^2+960*x+3 84)*exp(2-2*x)^2+(-2400*x^2-3840*x-1536)*exp(2-2*x)+2400*x^2+3840*x+1536), x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {25 x^{3}}{- 12000 x + \left (6000 x + 4800\right ) e^{2 - 2 x} - 9600} \]
integrate(((5*x**4+9*x**3+6*x**2)*exp(2-2*x)-10*x**3-12*x**2)/((600*x**2+9 60*x+384)*exp(2-2*x)**2+(-2400*x**2-3840*x-1536)*exp(2-2*x)+2400*x**2+3840 *x+1536),x)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {40 \, x e^{2} + {\left (125 \, x^{3} - 80 \, x - 64\right )} e^{\left (2 \, x\right )} + 32 \, e^{2}}{6000 \, {\left (5 \, x e^{2} - 2 \, {\left (5 \, x + 4\right )} e^{\left (2 \, x\right )} + 4 \, e^{2}\right )}} \]
integrate(((5*x^4+9*x^3+6*x^2)*exp(2-2*x)-10*x^3-12*x^2)/((600*x^2+960*x+3 84)*exp(2-2*x)^2+(-2400*x^2-3840*x-1536)*exp(2-2*x)+2400*x^2+3840*x+1536), x, algorithm=\
1/6000*(40*x*e^2 + (125*x^3 - 80*x - 64)*e^(2*x) + 32*e^2)/(5*x*e^2 - 2*(5 *x + 4)*e^(2*x) + 4*e^2)
Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {x^{3}}{48 \, {\left (5 \, x e^{\left (-2 \, x + 2\right )} - 10 \, x + 4 \, e^{\left (-2 \, x + 2\right )} - 8\right )}} \]
integrate(((5*x^4+9*x^3+6*x^2)*exp(2-2*x)-10*x^3-12*x^2)/((600*x^2+960*x+3 84)*exp(2-2*x)^2+(-2400*x^2-3840*x-1536)*exp(2-2*x)+2400*x^2+3840*x+1536), x, algorithm=\
Time = 9.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\right )} \, dx=\frac {5\,x^4+4\,x^3}{48\,{\left (5\,x+4\right )}^2\,\left ({\mathrm {e}}^{2-2\,x}-2\right )} \]