Integrand size = 128, antiderivative size = 28 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=-x+\frac {e^x}{3 x+\frac {1}{4} e^{-2 x} (2+x)^2} \]
Time = 4.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=\frac {e^x}{3 x}-x-\frac {e^x (2+x)^2}{3 x \left (4+4 x+12 e^{2 x} x+x^2\right )} \]
Integrate[(-16 - 32*x - 24*x^2 - 144*E^(4*x)*x^2 - 8*x^3 - x^4 + E^(5*x)*( -48 + 48*x) + E^(3*x)*(32 + 40*x + 12*x^2) + E^(2*x)*(-96*x - 96*x^2 - 24* x^3))/(16 + 32*x + 24*x^2 + 144*E^(4*x)*x^2 + 8*x^3 + x^4 + E^(2*x)*(96*x + 96*x^2 + 24*x^3)),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.
\(\displaystyle \int \frac {-x^4-8 x^3-144 e^{4 x} x^2-24 x^2+e^{3 x} \left (12 x^2+40 x+32\right )+e^{2 x} \left (-24 x^3-96 x^2-96 x\right )-32 x+e^{5 x} (48 x-48)-16}{x^4+8 x^3+144 e^{4 x} x^2+24 x^2+e^{2 x} \left (24 x^3+96 x^2+96 x\right )+32 x+16} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^4-8 x^3-144 e^{4 x} x^2-24 x^2+e^{3 x} \left (12 x^2+40 x+32\right )+e^{2 x} \left (-24 x^3-96 x^2-96 x\right )-32 x+e^{5 x} (48 x-48)-16}{\left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^x \left (2 x^2+3 x+2\right ) (x+2)^3}{3 x^2 \left (x^2+12 e^{2 x} x+4 x+4\right )^2}+\frac {e^x (x-1)}{3 x^2}+\frac {e^x \left (x^3+4 x^2+8 x+8\right )}{3 x^2 \left (x^2+12 e^{2 x} x+4 x+4\right )}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {64}{3} \int \frac {e^x}{\left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx-\frac {16}{3} \int \frac {e^x}{x^2 \left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx-16 \int \frac {e^x}{x \left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx-\frac {44}{3} \int \frac {e^x x}{\left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx-5 \int \frac {e^x x^2}{\left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx+\frac {4}{3} \int \frac {e^x}{x^2+12 e^{2 x} x+4 x+4}dx+\frac {8}{3} \int \frac {e^x}{x^2 \left (x^2+12 e^{2 x} x+4 x+4\right )}dx+\frac {8}{3} \int \frac {e^x}{x \left (x^2+12 e^{2 x} x+4 x+4\right )}dx+\frac {1}{3} \int \frac {e^x x}{x^2+12 e^{2 x} x+4 x+4}dx-\frac {2}{3} \int \frac {e^x x^3}{\left (x^2+12 e^{2 x} x+4 x+4\right )^2}dx-x+\frac {e^x}{3 x}\) |
Int[(-16 - 32*x - 24*x^2 - 144*E^(4*x)*x^2 - 8*x^3 - x^4 + E^(5*x)*(-48 + 48*x) + E^(3*x)*(32 + 40*x + 12*x^2) + E^(2*x)*(-96*x - 96*x^2 - 24*x^3))/ (16 + 32*x + 24*x^2 + 144*E^(4*x)*x^2 + 8*x^3 + x^4 + E^(2*x)*(96*x + 96*x ^2 + 24*x^3)),x]
3.28.5.3.1 Defintions of rubi rules used
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
risch | \(\frac {-12 \,{\mathrm e}^{2 x} x^{2}-x^{3}+4 \,{\mathrm e}^{3 x}-4 x^{2}-4 x}{12 x \,{\mathrm e}^{2 x}+x^{2}+4 x +4}\) | \(48\) |
parallelrisch | \(-\frac {144 \,{\mathrm e}^{2 x} x^{2}+12 x^{3}-48 \,{\mathrm e}^{3 x}+48 x^{2}+48 x}{12 \left (12 x \,{\mathrm e}^{2 x}+x^{2}+4 x +4\right )}\) | \(49\) |
norman | \(\frac {12 x +48 x \,{\mathrm e}^{2 x}-x^{3}+4 \,{\mathrm e}^{3 x}-12 \,{\mathrm e}^{2 x} x^{2}+16}{12 x \,{\mathrm e}^{2 x}+x^{2}+4 x +4}\) | \(51\) |
int(((48*x-48)*exp(x)^5-144*x^2*exp(x)^4+(12*x^2+40*x+32)*exp(x)^3+(-24*x^ 3-96*x^2-96*x)*exp(x)^2-x^4-8*x^3-24*x^2-32*x-16)/(144*x^2*exp(x)^4+(24*x^ 3+96*x^2+96*x)*exp(x)^2+x^4+8*x^3+24*x^2+32*x+16),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=-\frac {x^{3} + 12 \, x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} + 4 \, x - 4 \, e^{\left (3 \, x\right )}}{x^{2} + 12 \, x e^{\left (2 \, x\right )} + 4 \, x + 4} \]
integrate(((48*x-48)*exp(x)^5-144*x^2*exp(x)^4+(12*x^2+40*x+32)*exp(x)^3+( -24*x^3-96*x^2-96*x)*exp(x)^2-x^4-8*x^3-24*x^2-32*x-16)/(144*x^2*exp(x)^4+ (24*x^3+96*x^2+96*x)*exp(x)^2+x^4+8*x^3+24*x^2+32*x+16),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=- x + \frac {\left (- x^{2} - 4 x - 4\right ) e^{x}}{3 x^{3} + 36 x^{2} e^{2 x} + 12 x^{2} + 12 x} + \frac {e^{x}}{3 x} \]
integrate(((48*x-48)*exp(x)**5-144*x**2*exp(x)**4+(12*x**2+40*x+32)*exp(x) **3+(-24*x**3-96*x**2-96*x)*exp(x)**2-x**4-8*x**3-24*x**2-32*x-16)/(144*x* *2*exp(x)**4+(24*x**3+96*x**2+96*x)*exp(x)**2+x**4+8*x**3+24*x**2+32*x+16) ,x)
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=-\frac {x^{3} + 12 \, x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} + 4 \, x - 4 \, e^{\left (3 \, x\right )}}{x^{2} + 12 \, x e^{\left (2 \, x\right )} + 4 \, x + 4} \]
integrate(((48*x-48)*exp(x)^5-144*x^2*exp(x)^4+(12*x^2+40*x+32)*exp(x)^3+( -24*x^3-96*x^2-96*x)*exp(x)^2-x^4-8*x^3-24*x^2-32*x-16)/(144*x^2*exp(x)^4+ (24*x^3+96*x^2+96*x)*exp(x)^2+x^4+8*x^3+24*x^2+32*x+16),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (24) = 48\).
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.57 \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=-\frac {3 \, x^{4} + 36 \, x^{3} e^{\left (2 \, x\right )} + 12 \, x^{3} + x^{2} e^{x} + 12 \, x^{2} - 12 \, x e^{\left (3 \, x\right )} + 4 \, x e^{x} + 4 \, e^{x}}{3 \, {\left (x^{3} + 12 \, x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} + 4 \, x\right )}} \]
integrate(((48*x-48)*exp(x)^5-144*x^2*exp(x)^4+(12*x^2+40*x+32)*exp(x)^3+( -24*x^3-96*x^2-96*x)*exp(x)^2-x^4-8*x^3-24*x^2-32*x-16)/(144*x^2*exp(x)^4+ (24*x^3+96*x^2+96*x)*exp(x)^2+x^4+8*x^3+24*x^2+32*x+16),x, algorithm=\
-1/3*(3*x^4 + 36*x^3*e^(2*x) + 12*x^3 + x^2*e^x + 12*x^2 - 12*x*e^(3*x) + 4*x*e^x + 4*e^x)/(x^3 + 12*x^2*e^(2*x) + 4*x^2 + 4*x)
Timed out. \[ \int \frac {-16-32 x-24 x^2-144 e^{4 x} x^2-8 x^3-x^4+e^{5 x} (-48+48 x)+e^{3 x} \left (32+40 x+12 x^2\right )+e^{2 x} \left (-96 x-96 x^2-24 x^3\right )}{16+32 x+24 x^2+144 e^{4 x} x^2+8 x^3+x^4+e^{2 x} \left (96 x+96 x^2+24 x^3\right )} \, dx=\int -\frac {32\,x-{\mathrm {e}}^{3\,x}\,\left (12\,x^2+40\,x+32\right )+{\mathrm {e}}^{2\,x}\,\left (24\,x^3+96\,x^2+96\,x\right )+144\,x^2\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^{5\,x}\,\left (48\,x-48\right )+24\,x^2+8\,x^3+x^4+16}{32\,x+{\mathrm {e}}^{2\,x}\,\left (24\,x^3+96\,x^2+96\,x\right )+144\,x^2\,{\mathrm {e}}^{4\,x}+24\,x^2+8\,x^3+x^4+16} \,d x \]
int(-(32*x - exp(3*x)*(40*x + 12*x^2 + 32) + exp(2*x)*(96*x + 96*x^2 + 24* x^3) + 144*x^2*exp(4*x) - exp(5*x)*(48*x - 48) + 24*x^2 + 8*x^3 + x^4 + 16 )/(32*x + exp(2*x)*(96*x + 96*x^2 + 24*x^3) + 144*x^2*exp(4*x) + 24*x^2 + 8*x^3 + x^4 + 16),x)