Integrand size = 97, antiderivative size = 29 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=\frac {1}{3} \left (-5-x+\log \left (-2-\frac {4 (5+5 (3+x))}{e^4+\log (x)}\right )\right ) \]
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=\frac {1}{3} \left (-x-\log \left (e^4+\log (x)\right )+\log \left (40+e^4+10 x+\log (x)\right )\right ) \]
Integrate[(-40 - 10*x - E^8*x + E^4*(-30*x - 10*x^2) + (-30*x - 2*E^4*x - 10*x^2)*Log[x] - x*Log[x]^2)/(3*E^8*x + E^4*(120*x + 30*x^2) + (120*x + 6* E^4*x + 30*x^2)*Log[x] + 3*x*Log[x]^2),x]
Time = 1.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6, 7292, 27, 25, 7292, 7293, 2009}
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.
\(\displaystyle \int \frac {e^4 \left (-10 x^2-30 x\right )+\left (-10 x^2-2 e^4 x-30 x\right ) \log (x)-e^8 x-10 x-x \log ^2(x)-40}{e^4 \left (30 x^2+120 x\right )+\left (30 x^2+6 e^4 x+120 x\right ) \log (x)+3 e^8 x+3 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^4 \left (-10 x^2-30 x\right )+\left (-10 x^2-2 e^4 x-30 x\right ) \log (x)+\left (-10-e^8\right ) x-x \log ^2(x)-40}{e^4 \left (30 x^2+120 x\right )+\left (30 x^2+6 e^4 x+120 x\right ) \log (x)+3 e^8 x+3 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^4 \left (-10 x^2-30 x\right )+\left (-10 x^2-2 e^4 x-30 x\right ) \log (x)+\left (-10-e^8\right ) x-x \log ^2(x)-40}{3 x \left (\log (x)+e^4\right ) \left (10 x+\log (x)+40 \left (1+\frac {e^4}{40}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {x \log ^2(x)+2 \left (5 x^2+e^4 x+15 x\right ) \log (x)+\left (10+e^8\right ) x+10 e^4 \left (x^2+3 x\right )+40}{x \left (\log (x)+e^4\right ) \left (10 x+\log (x)+e^4+40\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {x \log ^2(x)+2 \left (5 x^2+e^4 x+15 x\right ) \log (x)+\left (10+e^8\right ) x+10 e^4 \left (x^2+3 x\right )+40}{x \left (\log (x)+e^4\right ) \left (10 x+\log (x)+e^4+40\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{3} \int \frac {x \log ^2(x)+2 \left (5 x^2+e^4 x+15 x\right ) \log (x)+\left (10+e^8\right ) x+10 e^4 \left (x^2+3 x\right )+40}{x \left (\log (x)+e^4\right ) \left (10 x+\log (x)+40 \left (1+\frac {e^4}{40}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} \int \left (\frac {-10 x-1}{x \left (10 x+\log (x)+40 \left (1+\frac {e^4}{40}\right )\right )}+\frac {1}{x \left (\log (x)+e^4\right )}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-x-\log \left (\log (x)+e^4\right )+\log \left (10 x+\log (x)+e^4+40\right )\right )\) |
Int[(-40 - 10*x - E^8*x + E^4*(-30*x - 10*x^2) + (-30*x - 2*E^4*x - 10*x^2 )*Log[x] - x*Log[x]^2)/(3*E^8*x + E^4*(120*x + 30*x^2) + (120*x + 6*E^4*x + 30*x^2)*Log[x] + 3*x*Log[x]^2),x]
3.14.22.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.44 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {x}{3}+\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )+10 x +40\right )}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )\right )}{3}\) | \(25\) |
norman | \(-\frac {x}{3}+\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )+10 x +40\right )}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )\right )}{3}\) | \(25\) |
risch | \(-\frac {x}{3}+\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )+10 x +40\right )}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )\right )}{3}\) | \(25\) |
parallelrisch | \(-\frac {x}{3}-\frac {\ln \left ({\mathrm e}^{4}+\ln \left (x \right )\right )}{3}+\frac {\ln \left (\frac {{\mathrm e}^{4}}{10}+x +\frac {\ln \left (x \right )}{10}+4\right )}{3}\) | \(27\) |
int((-x*ln(x)^2+(-2*x*exp(4)-10*x^2-30*x)*ln(x)-x*exp(4)^2+(-10*x^2-30*x)* exp(4)-10*x-40)/(3*x*ln(x)^2+(6*x*exp(4)+30*x^2+120*x)*ln(x)+3*x*exp(4)^2+ (30*x^2+120*x)*exp(4)),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=-\frac {1}{3} \, x + \frac {1}{3} \, \log \left (10 \, x + e^{4} + \log \left (x\right ) + 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \left (x\right )\right ) \]
integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^ 2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x *exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=- \frac {x}{3} - \frac {\log {\left (\log {\left (x \right )} + e^{4} \right )}}{3} + \frac {\log {\left (10 x + \log {\left (x \right )} + 40 + e^{4} \right )}}{3} \]
integrate((-x*ln(x)**2+(-2*x*exp(4)-10*x**2-30*x)*ln(x)-x*exp(4)**2+(-10*x **2-30*x)*exp(4)-10*x-40)/(3*x*ln(x)**2+(6*x*exp(4)+30*x**2+120*x)*ln(x)+3 *x*exp(4)**2+(30*x**2+120*x)*exp(4)),x)
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=-\frac {1}{3} \, x + \frac {1}{3} \, \log \left (10 \, x + e^{4} + \log \left (x\right ) + 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \left (x\right )\right ) \]
integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^ 2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x *exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=-\frac {1}{3} \, x + \frac {1}{3} \, \log \left (-10 \, x - e^{4} - \log \left (x\right ) - 40\right ) - \frac {1}{3} \, \log \left (e^{4} + \log \left (x\right )\right ) \]
integrate((-x*log(x)^2+(-2*x*exp(4)-10*x^2-30*x)*log(x)-x*exp(4)^2+(-10*x^ 2-30*x)*exp(4)-10*x-40)/(3*x*log(x)^2+(6*x*exp(4)+30*x^2+120*x)*log(x)+3*x *exp(4)^2+(30*x^2+120*x)*exp(4)),x, algorithm=\
Timed out. \[ \int \frac {-40-10 x-e^8 x+e^4 \left (-30 x-10 x^2\right )+\left (-30 x-2 e^4 x-10 x^2\right ) \log (x)-x \log ^2(x)}{3 e^8 x+e^4 \left (120 x+30 x^2\right )+\left (120 x+6 e^4 x+30 x^2\right ) \log (x)+3 x \log ^2(x)} \, dx=\int -\frac {x\,{\ln \left (x\right )}^2+\left (30\,x+2\,x\,{\mathrm {e}}^4+10\,x^2\right )\,\ln \left (x\right )+10\,x+{\mathrm {e}}^4\,\left (10\,x^2+30\,x\right )+x\,{\mathrm {e}}^8+40}{3\,x\,{\ln \left (x\right )}^2+\left (120\,x+6\,x\,{\mathrm {e}}^4+30\,x^2\right )\,\ln \left (x\right )+{\mathrm {e}}^4\,\left (30\,x^2+120\,x\right )+3\,x\,{\mathrm {e}}^8} \,d x \]
int(-(10*x + x*log(x)^2 + exp(4)*(30*x + 10*x^2) + x*exp(8) + log(x)*(30*x + 2*x*exp(4) + 10*x^2) + 40)/(3*x*log(x)^2 + exp(4)*(120*x + 30*x^2) + 3* x*exp(8) + log(x)*(120*x + 6*x*exp(4) + 30*x^2)),x)