Integrand size = 119, antiderivative size = 29 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=7-\frac {3}{5 \left (1-\frac {e^{5 x}}{x}+x^2\right )}-\log (3+x) \]
Time = 4.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=\frac {1}{5} \left (\frac {3 x}{e^{5 x}-x-x^3}-5 \log (3+x)\right ) \]
Integrate[(-5*E^(10*x) - 5*x^2 + 18*x^3 - 4*x^4 - 5*x^6 + E^(5*x)*(9 - 32* x - 15*x^2 + 10*x^3))/(15*x^2 + 5*x^3 + 30*x^4 + 10*x^5 + 15*x^6 + 5*x^7 + E^(10*x)*(15 + 5*x) + E^(5*x)*(-30*x - 10*x^2 - 30*x^3 - 10*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.
\(\displaystyle \int \frac {-5 x^6-4 x^4+18 x^3-5 x^2+e^{5 x} \left (10 x^3-15 x^2-32 x+9\right )-5 e^{10 x}}{5 x^7+15 x^6+10 x^5+30 x^4+5 x^3+15 x^2+e^{5 x} \left (-10 x^4-30 x^3-10 x^2-30 x\right )+e^{10 x} (5 x+15)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-5 x^6-4 x^4+18 x^3-5 x^2+e^{5 x} \left (10 x^3-15 x^2-32 x+9\right )-5 e^{10 x}}{5 (x+3) \left (-x^3-x+e^{5 x}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {5 x^6+4 x^4-18 x^3+5 x^2+5 e^{10 x}-e^{5 x} \left (10 x^3-15 x^2-32 x+9\right )}{(x+3) \left (-x^3-x+e^{5 x}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {5 x^6+4 x^4-18 x^3+5 x^2+5 e^{10 x}-e^{5 x} \left (10 x^3-15 x^2-32 x+9\right )}{(x+3) \left (-x^3-x+e^{5 x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (-\frac {3 (5 x-1)}{x^3+x-e^{5 x}}+\frac {3 x \left (5 x^3-3 x^2+5 x-1\right )}{\left (x^3+x-e^{5 x}\right )^2}+\frac {5}{x+3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (3 \int \frac {1}{-x^3-x+e^{5 x}}dx+3 \int \frac {x}{\left (x^3+x-e^{5 x}\right )^2}dx+9 \int \frac {x^3}{\left (x^3+x-e^{5 x}\right )^2}dx+15 \int \frac {x}{x^3+x-e^{5 x}}dx-15 \int \frac {x^4}{\left (x^3+x-e^{5 x}\right )^2}dx-15 \int \frac {x^2}{\left (x^3+x-e^{5 x}\right )^2}dx-5 \log (x+3)\right )\) |
Int[(-5*E^(10*x) - 5*x^2 + 18*x^3 - 4*x^4 - 5*x^6 + E^(5*x)*(9 - 32*x - 15 *x^2 + 10*x^3))/(15*x^2 + 5*x^3 + 30*x^4 + 10*x^5 + 15*x^6 + 5*x^7 + E^(10 *x)*(15 + 5*x) + E^(5*x)*(-30*x - 10*x^2 - 30*x^3 - 10*x^4)),x]
3.4.72.3.1 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 = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
norman | \(-\frac {3 x}{5 \left (x^{3}+x -{\mathrm e}^{5 x}\right )}-\ln \left (3+x \right )\) | \(24\) |
risch | \(-\frac {3 x}{5 \left (x^{3}+x -{\mathrm e}^{5 x}\right )}-\ln \left (3+x \right )\) | \(24\) |
parallelrisch | \(-\frac {5 \ln \left (3+x \right ) x^{3}+5 x \ln \left (3+x \right )-5 \,{\mathrm e}^{5 x} \ln \left (3+x \right )+3 x}{5 \left (x^{3}+x -{\mathrm e}^{5 x}\right )}\) | \(46\) |
int((-5*exp(5*x)^2+(10*x^3-15*x^2-32*x+9)*exp(5*x)-5*x^6-4*x^4+18*x^3-5*x^ 2)/((5*x+15)*exp(5*x)^2+(-10*x^4-30*x^3-10*x^2-30*x)*exp(5*x)+5*x^7+15*x^6 +10*x^5+30*x^4+5*x^3+15*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=-\frac {5 \, {\left (x^{3} + x - e^{\left (5 \, x\right )}\right )} \log \left (x + 3\right ) + 3 \, x}{5 \, {\left (x^{3} + x - e^{\left (5 \, x\right )}\right )}} \]
integrate((-5*exp(5*x)^2+(10*x^3-15*x^2-32*x+9)*exp(5*x)-5*x^6-4*x^4+18*x^ 3-5*x^2)/((5*x+15)*exp(5*x)^2+(-10*x^4-30*x^3-10*x^2-30*x)*exp(5*x)+5*x^7+ 15*x^6+10*x^5+30*x^4+5*x^3+15*x^2),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=\frac {3 x}{- 5 x^{3} - 5 x + 5 e^{5 x}} - \log {\left (x + 3 \right )} \]
integrate((-5*exp(5*x)**2+(10*x**3-15*x**2-32*x+9)*exp(5*x)-5*x**6-4*x**4+ 18*x**3-5*x**2)/((5*x+15)*exp(5*x)**2+(-10*x**4-30*x**3-10*x**2-30*x)*exp( 5*x)+5*x**7+15*x**6+10*x**5+30*x**4+5*x**3+15*x**2),x)
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=-\frac {3 \, x}{5 \, {\left (x^{3} + x - e^{\left (5 \, x\right )}\right )}} - \log \left (x + 3\right ) \]
integrate((-5*exp(5*x)^2+(10*x^3-15*x^2-32*x+9)*exp(5*x)-5*x^6-4*x^4+18*x^ 3-5*x^2)/((5*x+15)*exp(5*x)^2+(-10*x^4-30*x^3-10*x^2-30*x)*exp(5*x)+5*x^7+ 15*x^6+10*x^5+30*x^4+5*x^3+15*x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=-\frac {5 \, x^{3} \log \left (x + 3\right ) + 5 \, x \log \left (x + 3\right ) - 5 \, e^{\left (5 \, x\right )} \log \left (x + 3\right ) + 3 \, x}{5 \, {\left (x^{3} + x - e^{\left (5 \, x\right )}\right )}} \]
integrate((-5*exp(5*x)^2+(10*x^3-15*x^2-32*x+9)*exp(5*x)-5*x^6-4*x^4+18*x^ 3-5*x^2)/((5*x+15)*exp(5*x)^2+(-10*x^4-30*x^3-10*x^2-30*x)*exp(5*x)+5*x^7+ 15*x^6+10*x^5+30*x^4+5*x^3+15*x^2),x, algorithm=\
Timed out. \[ \int \frac {-5 e^{10 x}-5 x^2+18 x^3-4 x^4-5 x^6+e^{5 x} \left (9-32 x-15 x^2+10 x^3\right )}{15 x^2+5 x^3+30 x^4+10 x^5+15 x^6+5 x^7+e^{10 x} (15+5 x)+e^{5 x} \left (-30 x-10 x^2-30 x^3-10 x^4\right )} \, dx=\int -\frac {5\,{\mathrm {e}}^{10\,x}+{\mathrm {e}}^{5\,x}\,\left (-10\,x^3+15\,x^2+32\,x-9\right )+5\,x^2-18\,x^3+4\,x^4+5\,x^6}{{\mathrm {e}}^{10\,x}\,\left (5\,x+15\right )-{\mathrm {e}}^{5\,x}\,\left (10\,x^4+30\,x^3+10\,x^2+30\,x\right )+15\,x^2+5\,x^3+30\,x^4+10\,x^5+15\,x^6+5\,x^7} \,d x \]
int(-(5*exp(10*x) + exp(5*x)*(32*x + 15*x^2 - 10*x^3 - 9) + 5*x^2 - 18*x^3 + 4*x^4 + 5*x^6)/(exp(10*x)*(5*x + 15) - exp(5*x)*(30*x + 10*x^2 + 30*x^3 + 10*x^4) + 15*x^2 + 5*x^3 + 30*x^4 + 10*x^5 + 15*x^6 + 5*x^7),x)