Integrand size = 208, antiderivative size = 35 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=x+\frac {x \log (\log (3))}{x-\frac {x^2}{\log \left (\frac {2+\frac {5}{x}+x}{5+e^3}\right )}} \]
\[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx \]
Integrate[(5*x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^ 2)/(5*x + E^3*x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)]^2 + (5 - x^2 + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)])*Log[Log[3] ])/(5*x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)]^2),x]
Integrate[(5*x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^ 2)/(5*x + E^3*x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)]^2 + (5 - x^2 + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)])*Log[Log[3] ])/(5*x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*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.
| \(\displaystyle \int \frac {x^4+2 x^3+5 x^2+\left (x^2+2 x+5\right ) \log ^2\left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )+\log (\log (3)) \left (-x^2+\left (x^2+2 x+5\right ) \log \left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )+5\right )+\left (-2 x^3-4 x^2-10 x\right ) \log \left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )}{x^4+2 x^3+5 x^2+\left (x^2+2 x+5\right ) \log ^2\left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )+\left (-2 x^3-4 x^2-10 x\right ) \log \left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {x^4+2 x^3+5 x^2+\left (x^2+2 x+5\right ) \log ^2\left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )+\log (\log (3)) \left (-x^2+\left (x^2+2 x+5\right ) \log \left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )+5\right )+\left (-2 x^3-4 x^2-10 x\right ) \log \left (\frac {x^2+2 x+5}{e^3 x+5 x}\right )}{\left (x^2+2 x+5\right ) \left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\log ^2\left (\frac {x^2+2 x+5}{\left (5+e^3\right ) x}\right )}{\left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}+\frac {5 x^2 \left (1-\frac {1}{5} \log (\log (3))\right )}{\left (x^2+2 x+5\right ) \left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}+\frac {\log \left (\frac {x^2+2 x+5}{\left (5+e^3\right ) x}\right ) (\log (\log (3))-2 x)}{\left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}+\frac {5 \log (\log (3))}{\left (x^2+2 x+5\right ) \left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}+\frac {x^4}{\left (x^2+2 x+5\right ) \left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}+\frac {2 x^3}{\left (x^2+2 x+5\right ) \left (x-\log \left (x+\frac {5}{x}+2\right )+\log \left (5+e^3\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {\log ^2\left (\frac {x^2+2 x+5}{\left (5+e^3\right ) x}\right )}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\int \frac {x^2}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\log (\log (3)) \int \frac {\log \left (\frac {x^2+2 x+5}{\left (5+e^3\right ) x}\right )}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx-2 \int \frac {x \log \left (\frac {x^2+2 x+5}{\left (5+e^3\right ) x}\right )}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+(5-\log (\log (3))) \int \frac {1}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx-5 \int \frac {1}{\left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\frac {5}{2} i \log (\log (3)) \int \frac {1}{(-2 x-(2-4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx-\frac {5}{2} i (5-\log (\log (3))) \int \frac {1}{(-2 x-(2-4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\frac {25}{2} i \int \frac {1}{(-2 x-(2-4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx-(2+i) (5-\log (\log (3))) \int \frac {1}{(2 x+(2-4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+(10+5 i) \int \frac {1}{(2 x+(2-4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\frac {5}{2} i \log (\log (3)) \int \frac {1}{(2 x+(2+4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx-\left (2+\frac {3 i}{2}\right ) (5-\log (\log (3))) \int \frac {1}{(2 x+(2+4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx+\left (10+\frac {15 i}{2}\right ) \int \frac {1}{(2 x+(2+4 i)) \left (x-\log \left (x+2+\frac {5}{x}\right )+\log \left (5+e^3\right )\right )^2}dx\) |
Int[(5*x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^2)/(5* x + E^3*x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)]^2 + (5 - x^2 + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)])*Log[Log[3]])/(5* x^2 + 2*x^3 + x^4 + (-10*x - 4*x^2 - 2*x^3)*Log[(5 + 2*x + x^2)/(5*x + E^3 *x)] + (5 + 2*x + x^2)*Log[(5 + 2*x + x^2)/(5*x + E^3*x)]^2),x]
3.3.57.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 1.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(x -\frac {x \ln \left (\ln \left (3\right )\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}\) | \(35\) |
| default | \(x +\frac {-\ln \left (\ln \left (3\right )\right ) \ln \left (\frac {x^{2}+2 x +5}{x}\right )+\ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{3}+5\right )}{\ln \left ({\mathrm e}^{3}+5\right )+x -\ln \left (\frac {x^{2}+2 x +5}{x}\right )}\) | \(56\) |
| norman | \(\frac {x^{2}-\ln \left (\ln \left (3\right )\right ) \ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )-x \ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \,{\mathrm e}^{3}+5 x}\right )}\) | \(80\) |
| parallelrisch | \(-\frac {\ln \left (\ln \left (3\right )\right ) x -x^{2}+\ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right ) x +4 x -4 \ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right )}{x -\ln \left (\frac {x^{2}+2 x +5}{x \left ({\mathrm e}^{3}+5\right )}\right )}\) | \(84\) |
int((((x^2+2*x+5)*ln((x^2+2*x+5)/(x*exp(3)+5*x))-x^2+5)*ln(ln(3))+(x^2+2*x +5)*ln((x^2+2*x+5)/(x*exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*ln((x^2+2*x+5)/(x *exp(3)+5*x))+x^4+2*x^3+5*x^2)/((x^2+2*x+5)*ln((x^2+2*x+5)/(x*exp(3)+5*x)) ^2+(-2*x^3-4*x^2-10*x)*ln((x^2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x^3+5*x^2),x,m ethod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} - x \log \left (\frac {x^{2} + 2 \, x + 5}{x e^{3} + 5 \, x}\right ) - x \log \left (\log \left (3\right )\right )}{x - \log \left (\frac {x^{2} + 2 \, x + 5}{x e^{3} + 5 \, x}\right )} \]
integrate((((x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))-x^2+5)*log(log(3)) +(x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^ 2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x^3+5*x^2)/((x^2+2*x+5)*log((x^2+2*x+5)/(x* exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x ^3+5*x^2),x, algorithm=\
(x^2 - x*log((x^2 + 2*x + 5)/(x*e^3 + 5*x)) - x*log(log(3)))/(x - log((x^2 + 2*x + 5)/(x*e^3 + 5*x)))
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=x + \frac {x \log {\left (\log {\left (3 \right )} \right )}}{- x + \log {\left (\frac {x^{2} + 2 x + 5}{5 x + x e^{3}} \right )}} \]
integrate((((x**2+2*x+5)*ln((x**2+2*x+5)/(x*exp(3)+5*x))-x**2+5)*ln(ln(3)) +(x**2+2*x+5)*ln((x**2+2*x+5)/(x*exp(3)+5*x))**2+(-2*x**3-4*x**2-10*x)*ln( (x**2+2*x+5)/(x*exp(3)+5*x))+x**4+2*x**3+5*x**2)/((x**2+2*x+5)*ln((x**2+2* x+5)/(x*exp(3)+5*x))**2+(-2*x**3-4*x**2-10*x)*ln((x**2+2*x+5)/(x*exp(3)+5* x))+x**4+2*x**3+5*x**2),x)
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.60 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} + x {\left (\log \left (e^{3} + 5\right ) - \log \left (\log \left (3\right )\right )\right )} - x \log \left (x^{2} + 2 \, x + 5\right ) + x \log \left (x\right )}{x - \log \left (x^{2} + 2 \, x + 5\right ) + \log \left (x\right ) + \log \left (e^{3} + 5\right )} \]
integrate((((x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))-x^2+5)*log(log(3)) +(x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^ 2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x^3+5*x^2)/((x^2+2*x+5)*log((x^2+2*x+5)/(x* exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x ^3+5*x^2),x, algorithm=\
(x^2 + x*(log(e^3 + 5) - log(log(3))) - x*log(x^2 + 2*x + 5) + x*log(x))/( x - log(x^2 + 2*x + 5) + log(x) + log(e^3 + 5))
Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=\frac {x^{2} - x \log \left (x^{2} + 2 \, x + 5\right ) + x \log \left (x e^{3} + 5 \, x\right ) - x \log \left (\log \left (3\right )\right )}{x - \log \left (x^{2} + 2 \, x + 5\right ) + \log \left (x e^{3} + 5 \, x\right )} \]
integrate((((x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))-x^2+5)*log(log(3)) +(x^2+2*x+5)*log((x^2+2*x+5)/(x*exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^ 2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x^3+5*x^2)/((x^2+2*x+5)*log((x^2+2*x+5)/(x* exp(3)+5*x))^2+(-2*x^3-4*x^2-10*x)*log((x^2+2*x+5)/(x*exp(3)+5*x))+x^4+2*x ^3+5*x^2),x, algorithm=\
(x^2 - x*log(x^2 + 2*x + 5) + x*log(x*e^3 + 5*x) - x*log(log(3)))/(x - log (x^2 + 2*x + 5) + log(x*e^3 + 5*x))
Time = 13.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.29 \[ \int \frac {5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5-x^2+\left (5+2 x+x^2\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )\right ) \log (\log (3))}{5 x^2+2 x^3+x^4+\left (-10 x-4 x^2-2 x^3\right ) \log \left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )+\left (5+2 x+x^2\right ) \log ^2\left (\frac {5+2 x+x^2}{5 x+e^3 x}\right )} \, dx=-\frac {x\,\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )-x^2+\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )\,\ln \left (\ln \left (3\right )\right )}{x-\ln \left (\frac {x^2+2\,x+5}{5\,x+x\,{\mathrm {e}}^3}\right )} \]
int((log((2*x + x^2 + 5)/(5*x + x*exp(3)))^2*(2*x + x^2 + 5) - log((2*x + x^2 + 5)/(5*x + x*exp(3)))*(10*x + 4*x^2 + 2*x^3) + 5*x^2 + 2*x^3 + x^4 + log(log(3))*(log((2*x + x^2 + 5)/(5*x + x*exp(3)))*(2*x + x^2 + 5) - x^2 + 5))/(log((2*x + x^2 + 5)/(5*x + x*exp(3)))^2*(2*x + x^2 + 5) - log((2*x + x^2 + 5)/(5*x + x*exp(3)))*(10*x + 4*x^2 + 2*x^3) + 5*x^2 + 2*x^3 + x^4), x)