Integrand size = 112, antiderivative size = 26 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=\frac {x (1+x)}{-2 x+\frac {e^{-11 x/10}}{x \log (x)}} \end {dmath*}
Time = 2.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=\frac {1}{10} \left (-5 x-\frac {5 (1+x)}{-1+2 e^{11 x/10} x^2 \log (x)}\right ) \end {dmath*}
Integrate[(-20*E^((11*x + 10*Log[3*x])/5)*x^2*Log[x]^2 + E^((11*x + 10*Log [3*x])/10)*(30 + 30*x + (60 + 123*x + 33*x^2)*Log[x]))/(90 - 120*E^((11*x + 10*Log[3*x])/10)*x*Log[x] + 40*E^((11*x + 10*Log[3*x])/5)*x^2*Log[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.
\(\displaystyle \int \frac {e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (\left (33 x^2+123 x+60\right ) \log (x)+30 x+30\right )-20 x^2 e^{\frac {1}{5} (11 x+10 \log (3 x))} \log ^2(x)}{40 x^2 e^{\frac {1}{5} (11 x+10 \log (3 x))} \log ^2(x)-120 x e^{\frac {1}{10} (11 x+10 \log (3 x))} \log (x)+90} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{11 x/10} x \left (-20 e^{11 x/10} x^3 \log ^2(x)+\left (11 x^2+41 x+20\right ) \log (x)+10 (x+1)\right )}{10 \left (1-2 e^{11 x/10} x^2 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int \frac {e^{11 x/10} x \left (-20 e^{11 x/10} \log ^2(x) x^3+10 (x+1)+\left (11 x^2+41 x+20\right ) \log (x)\right )}{\left (1-2 e^{11 x/10} x^2 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{10} \int \left (\frac {e^{11 x/10} x (x+1) (11 x \log (x)+20 \log (x)+10)}{\left (2 e^{11 x/10} x^2 \log (x)-1\right )^2}-\frac {10 e^{11 x/10} x^2 \log (x)}{2 e^{11 x/10} x^2 \log (x)-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (10 \int \frac {e^{11 x/10} x^2}{\left (2 e^{11 x/10} x^2 \log (x)-1\right )^2}dx+20 \int \frac {e^{11 x/10} x^2 \log (x)}{\left (2 e^{11 x/10} x^2 \log (x)-1\right )^2}dx-5 \int \frac {1}{2 e^{11 x/10} x^2 \log (x)-1}dx+11 \int \frac {e^{11 x/10} x^3 \log (x)}{\left (2 e^{11 x/10} x^2 \log (x)-1\right )^2}dx-\frac {10}{2-\frac {e^{-11 x/10}}{x^2 \log (x)}}-5 x\right )\) |
Int[(-20*E^((11*x + 10*Log[3*x])/5)*x^2*Log[x]^2 + E^((11*x + 10*Log[3*x]) /10)*(30 + 30*x + (60 + 123*x + 33*x^2)*Log[x]))/(90 - 120*E^((11*x + 10*L og[3*x])/10)*x*Log[x] + 40*E^((11*x + 10*Log[3*x])/5)*x^2*Log[x]^2),x]
3.6.39.3.1 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 = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {x}{2}-\frac {3 \left (1+x \right )}{2 \left (6 \ln \left (x \right ) {\mathrm e}^{\frac {11 x}{10}} x^{2}-3\right )}\) | \(25\) |
parallelrisch | \(-\frac {360+240 x^{2} \ln \left (x \right ) {\mathrm e}^{\ln \left (3 x \right )+\frac {11 x}{10}}}{240 \left (2 x \ln \left (x \right ) {\mathrm e}^{\ln \left (3 x \right )+\frac {11 x}{10}}-3\right )}\) | \(39\) |
int((-20*x^2*ln(x)^2*exp(ln(3*x)+11/10*x)^2+((33*x^2+123*x+60)*ln(x)+30*x+ 30)*exp(ln(3*x)+11/10*x))/(40*x^2*ln(x)^2*exp(ln(3*x)+11/10*x)^2-120*x*ln( x)*exp(ln(3*x)+11/10*x)+90),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=-\frac {2 \, x^{2} e^{\left (\frac {11}{10} \, x + \log \left (3\right ) + \log \left (x\right )\right )} \log \left (x\right ) + 3}{2 \, {\left (2 \, x e^{\left (\frac {11}{10} \, x + \log \left (3\right ) + \log \left (x\right )\right )} \log \left (x\right ) - 3\right )}} \end {dmath*}
integrate((-20*x^2*log(x)^2*exp(log(3*x)+11/10*x)^2+((33*x^2+123*x+60)*log (x)+30*x+30)*exp(log(3*x)+11/10*x))/(40*x^2*log(x)^2*exp(log(3*x)+11/10*x) ^2-120*x*log(x)*exp(log(3*x)+11/10*x)+90),x, algorithm=\
-1/2*(2*x^2*e^(11/10*x + log(3) + log(x))*log(x) + 3)/(2*x*e^(11/10*x + lo g(3) + log(x))*log(x) - 3)
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=- \frac {x}{2} + \frac {- x - 1}{4 x^{2} e^{\frac {11 x}{10}} \log {\left (x \right )} - 2} \end {dmath*}
integrate((-20*x**2*ln(x)**2*exp(ln(3*x)+11/10*x)**2+((33*x**2+123*x+60)*l n(x)+30*x+30)*exp(ln(3*x)+11/10*x))/(40*x**2*ln(x)**2*exp(ln(3*x)+11/10*x) **2-120*x*ln(x)*exp(ln(3*x)+11/10*x)+90),x)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=-\frac {2 \, x^{3} e^{\left (\frac {11}{10} \, x\right )} \log \left (x\right ) + 1}{2 \, {\left (2 \, x^{2} e^{\left (\frac {11}{10} \, x\right )} \log \left (x\right ) - 1\right )}} \end {dmath*}
integrate((-20*x^2*log(x)^2*exp(log(3*x)+11/10*x)^2+((33*x^2+123*x+60)*log (x)+30*x+30)*exp(log(3*x)+11/10*x))/(40*x^2*log(x)^2*exp(log(3*x)+11/10*x) ^2-120*x*log(x)*exp(log(3*x)+11/10*x)+90),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=-\frac {2 \, x^{3} e^{\left (\frac {11}{10} \, x\right )} \log \left (x\right ) + 1}{2 \, {\left (2 \, x^{2} e^{\left (\frac {11}{10} \, x\right )} \log \left (x\right ) - 1\right )}} \end {dmath*}
integrate((-20*x^2*log(x)^2*exp(log(3*x)+11/10*x)^2+((33*x^2+123*x+60)*log (x)+30*x+30)*exp(log(3*x)+11/10*x))/(40*x^2*log(x)^2*exp(log(3*x)+11/10*x) ^2-120*x*log(x)*exp(log(3*x)+11/10*x)+90),x, algorithm=\
Time = 16.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-20 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)+e^{\frac {1}{10} (11 x+10 \log (3 x))} \left (30+30 x+\left (60+123 x+33 x^2\right ) \log (x)\right )}{90-120 e^{\frac {1}{10} (11 x+10 \log (3 x))} x \log (x)+40 e^{\frac {1}{5} (11 x+10 \log (3 x))} x^2 \log ^2(x)} \, dx=-\frac {x}{2}-\frac {\frac {x}{2}+\frac {1}{2}}{2\,x^2\,{\mathrm {e}}^{\frac {11\,x}{10}}\,\ln \left (x\right )-1} \end {dmath*}