Integrand size = 112, antiderivative size = 28 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=\frac {-1-e^x-5 (1+x)}{x \left (x+\log \left (\frac {3+x}{5}\right )\right )} \]
Time = 0.70 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=-\frac {6+e^x+5 x}{x \left (x+\log \left (\frac {3+x}{5}\right )\right )} \]
Integrate[(42*x + 32*x^2 + 5*x^3 + E^x*(7*x - x^2 - x^3) + (18 + 6*x + E^x *(3 - 2*x - x^2))*Log[(3 + x)/5])/(3*x^4 + x^5 + (6*x^3 + 2*x^4)*Log[(3 + x)/5] + (3*x^2 + x^3)*Log[(3 + x)/5]^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.
\(\displaystyle \int \frac {5 x^3+32 x^2+\left (e^x \left (-x^2-2 x+3\right )+6 x+18\right ) \log \left (\frac {x+3}{5}\right )+e^x \left (-x^3-x^2+7 x\right )+42 x}{x^5+3 x^4+\left (2 x^4+6 x^3\right ) \log \left (\frac {x+3}{5}\right )+\left (x^3+3 x^2\right ) \log ^2\left (\frac {x+3}{5}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-x \left (-5 x^2+e^x \left (x^2+x-7\right )-32 x-42\right )-\left (e^x (x-1)-6\right ) (x+3) \log \left (\frac {x+3}{5}\right )}{x^2 (x+3) \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {6 \log \left (\frac {x}{5}+\frac {3}{5}\right )}{x^2 \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}-\frac {e^x \left (x^3+x^2+x^2 \log \left (\frac {x+3}{5}\right )-7 x+2 x \log \left (\frac {x+3}{5}\right )-3 \log \left (\frac {x+3}{5}\right )\right )}{(x+3) x^2 \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}+\frac {5 x}{(x+3) \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}+\frac {32}{(x+3) \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}+\frac {42}{(x+3) x \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 6 \int \frac {\log \left (\frac {x}{5}+\frac {3}{5}\right )}{x^2 \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx+\int \frac {e^x}{x^2 \left (x+\log \left (\frac {x+3}{5}\right )\right )}dx+5 \int \frac {1}{\left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx+14 \int \frac {1}{x \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx+\frac {4}{3} \int \frac {e^x}{x \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx+3 \int \frac {1}{(x+3) \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx-\frac {1}{3} \int \frac {e^x}{(x+3) \left (x+\log \left (\frac {x+3}{5}\right )\right )^2}dx-\int \frac {e^x}{x \left (x+\log \left (\frac {x+3}{5}\right )\right )}dx\) |
Int[(42*x + 32*x^2 + 5*x^3 + E^x*(7*x - x^2 - x^3) + (18 + 6*x + E^x*(3 - 2*x - x^2))*Log[(3 + x)/5])/(3*x^4 + x^5 + (6*x^3 + 2*x^4)*Log[(3 + x)/5] + (3*x^2 + x^3)*Log[(3 + x)/5]^2),x]
3.27.46.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 37.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {5 x +{\mathrm e}^{x}+6}{x \left (\ln \left (\frac {3}{5}+\frac {x}{5}\right )+x \right )}\) | \(23\) |
parallelrisch | \(\frac {-6-5 x -{\mathrm e}^{x}}{\left (\ln \left (\frac {3}{5}+\frac {x}{5}\right )+x \right ) x}\) | \(24\) |
int((((-x^2-2*x+3)*exp(x)+18+6*x)*ln(3/5+1/5*x)+(-x^3-x^2+7*x)*exp(x)+5*x^ 3+32*x^2+42*x)/((x^3+3*x^2)*ln(3/5+1/5*x)^2+(2*x^4+6*x^3)*ln(3/5+1/5*x)+x^ 5+3*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=-\frac {5 \, x + e^{x} + 6}{x^{2} + x \log \left (\frac {1}{5} \, x + \frac {3}{5}\right )} \]
integrate((((-x^2-2*x+3)*exp(x)+18+6*x)*log(3/5+1/5*x)+(-x^3-x^2+7*x)*exp( x)+5*x^3+32*x^2+42*x)/((x^3+3*x^2)*log(3/5+1/5*x)^2+(2*x^4+6*x^3)*log(3/5+ 1/5*x)+x^5+3*x^4),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=\frac {- 5 x - 6}{x^{2} + x \log {\left (\frac {x}{5} + \frac {3}{5} \right )}} - \frac {e^{x}}{x^{2} + x \log {\left (\frac {x}{5} + \frac {3}{5} \right )}} \]
integrate((((-x**2-2*x+3)*exp(x)+18+6*x)*ln(3/5+1/5*x)+(-x**3-x**2+7*x)*ex p(x)+5*x**3+32*x**2+42*x)/((x**3+3*x**2)*ln(3/5+1/5*x)**2+(2*x**4+6*x**3)* ln(3/5+1/5*x)+x**5+3*x**4),x)
Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=-\frac {5 \, x + e^{x} + 6}{x^{2} - x \log \left (5\right ) + x \log \left (x + 3\right )} \]
integrate((((-x^2-2*x+3)*exp(x)+18+6*x)*log(3/5+1/5*x)+(-x^3-x^2+7*x)*exp( x)+5*x^3+32*x^2+42*x)/((x^3+3*x^2)*log(3/5+1/5*x)^2+(2*x^4+6*x^3)*log(3/5+ 1/5*x)+x^5+3*x^4),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.21 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=-\frac {5 \, {\left (x + 3\right )} e^{3} - 9 \, e^{3} + e^{\left (x + 3\right )}}{{\left (x + 3\right )}^{2} e^{3} + {\left (x + 3\right )} e^{3} \log \left (\frac {1}{5} \, x + \frac {3}{5}\right ) - 6 \, {\left (x + 3\right )} e^{3} - 3 \, e^{3} \log \left (\frac {1}{5} \, x + \frac {3}{5}\right ) + 9 \, e^{3}} \]
integrate((((-x^2-2*x+3)*exp(x)+18+6*x)*log(3/5+1/5*x)+(-x^3-x^2+7*x)*exp( x)+5*x^3+32*x^2+42*x)/((x^3+3*x^2)*log(3/5+1/5*x)^2+(2*x^4+6*x^3)*log(3/5+ 1/5*x)+x^5+3*x^4),x, algorithm=\
-(5*(x + 3)*e^3 - 9*e^3 + e^(x + 3))/((x + 3)^2*e^3 + (x + 3)*e^3*log(1/5* x + 3/5) - 6*(x + 3)*e^3 - 3*e^3*log(1/5*x + 3/5) + 9*e^3)
Time = 9.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {42 x+32 x^2+5 x^3+e^x \left (7 x-x^2-x^3\right )+\left (18+6 x+e^x \left (3-2 x-x^2\right )\right ) \log \left (\frac {3+x}{5}\right )}{3 x^4+x^5+\left (6 x^3+2 x^4\right ) \log \left (\frac {3+x}{5}\right )+\left (3 x^2+x^3\right ) \log ^2\left (\frac {3+x}{5}\right )} \, dx=-\frac {5\,x+{\mathrm {e}}^x+6}{x\,\left (x+\ln \left (\frac {x}{5}+\frac {3}{5}\right )\right )} \]