Integrand size = 128, antiderivative size = 23 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=2 x \left (5-\frac {3 x}{\log \left (\frac {1}{e^2 x^2 (3+x)^4}\right )}\right ) \]
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=-2 \left (-5 x+\frac {3 x^2}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )}\right ) \]
Integrate[(-36*x - 36*x^2 + (-36*x - 12*x^2)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))] + (30 + 10*x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54 *x^4 + 12*x^5 + x^6))]^2)/((3 + x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))]^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.
\(\displaystyle \int \frac {-36 x^2+(10 x+30) \log ^2\left (\frac {1}{e^2 \left (x^6+12 x^5+54 x^4+108 x^3+81 x^2\right )}\right )+\left (-12 x^2-36 x\right ) \log \left (\frac {1}{e^2 \left (x^6+12 x^5+54 x^4+108 x^3+81 x^2\right )}\right )-36 x}{(x+3) \log ^2\left (\frac {1}{e^2 \left (x^6+12 x^5+54 x^4+108 x^3+81 x^2\right )}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (-6 x^2+5 (x+3) \log ^2\left (\frac {1}{x^2 (x+3)^4}\right )-\left (6 x^2+38 x+60\right ) \log \left (\frac {1}{x^2 (x+3)^4}\right )+38 x+60\right )}{(x+3) \left (2-\log \left (\frac {1}{x^2 (x+3)^4}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {-6 x^2+38 x+5 (x+3) \log ^2\left (\frac {1}{x^2 (x+3)^4}\right )-2 \left (3 x^2+19 x+30\right ) \log \left (\frac {1}{x^2 (x+3)^4}\right )+60}{(x+3) \left (2-\log \left (\frac {1}{x^2 (x+3)^4}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (-\frac {6 x}{\log \left (\frac {1}{x^2 (x+3)^4}\right )-2}-\frac {18 (x+1) x}{(x+3) \left (\log \left (\frac {1}{x^2 (x+3)^4}\right )-2\right )^2}+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (36 \int \frac {1}{\left (\log \left (\frac {1}{x^2 (x+3)^4}\right )-2\right )^2}dx-18 \int \frac {x}{\left (\log \left (\frac {1}{x^2 (x+3)^4}\right )-2\right )^2}dx-108 \int \frac {1}{(x+3) \left (\log \left (\frac {1}{x^2 (x+3)^4}\right )-2\right )^2}dx-6 \int \frac {x}{\log \left (\frac {1}{x^2 (x+3)^4}\right )-2}dx+5 x\right )\) |
Int[(-36*x - 36*x^2 + (-36*x - 12*x^2)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x ^4 + 12*x^5 + x^6))] + (30 + 10*x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))]^2)/((3 + x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^ 5 + x^6))]^2),x]
3.14.20.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.85 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83
method | result | size |
risch | \(10 x -\frac {6 x^{2}}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\) | \(42\) |
norman | \(\frac {-6 x^{2}+10 x \ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\) | \(77\) |
parallelrisch | \(\frac {-24 x^{2}+40 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right ) x -240 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right )}{4 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right )}\) | \(103\) |
int(((10*x+30)*ln(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(-12*x^2- 36*x)*ln(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x)/(3+x)/l n(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=-\frac {2 \, {\left (3 \, x^{2} - 5 \, x \log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )\right )}}{\log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )} \]
integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(- 12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x) /(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm=\
-2*(3*x^2 - 5*x*log(e^(-2)/(x^6 + 12*x^5 + 54*x^4 + 108*x^3 + 81*x^2)))/lo g(e^(-2)/(x^6 + 12*x^5 + 54*x^4 + 108*x^3 + 81*x^2))
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=- \frac {6 x^{2}}{\log {\left (\frac {1}{\left (x^{6} + 12 x^{5} + 54 x^{4} + 108 x^{3} + 81 x^{2}\right ) e^{2}} \right )}} + 10 x \]
integrate(((10*x+30)*ln(1/(x**6+12*x**5+54*x**4+108*x**3+81*x**2)/exp(2))* *2+(-12*x**2-36*x)*ln(1/(x**6+12*x**5+54*x**4+108*x**3+81*x**2)/exp(2))-36 *x**2-36*x)/(3+x)/ln(1/(x**6+12*x**5+54*x**4+108*x**3+81*x**2)/exp(2))**2, x)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=\frac {3 \, x^{2} + 20 \, x \log \left (x + 3\right ) + 10 \, x \log \left (x\right ) + 10 \, x}{2 \, \log \left (x + 3\right ) + \log \left (x\right ) + 1} \]
integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(- 12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x) /(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=10 \, x + \frac {6 \, x^{2}}{\log \left (x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}\right ) + 2} \]
integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(- 12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x) /(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm=\
Time = 8.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.30 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=14\,x+\frac {4}{x+1}+2\,x^2-\frac {6\,x^2+\frac {2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )\,\left (x+3\right )}{x+1}}{\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )} \]
int(-(36*x + log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))*(36*x + 12*x^2) - log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))^2*(10 *x + 30) + 36*x^2)/(log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6) )^2*(x + 3)),x)