Integrand size = 97, antiderivative size = 25 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=\frac {e^5 \left (2+\frac {x}{x+e^x x}\right )}{2-\log (x)} \]
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {e^5 \left (3+2 e^x\right )}{\left (1+e^x\right ) (-2+\log (x))} \]
Integrate[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x ])/(4*x + 8*E^x*x + 4*E^(2*x)*x + (-4*x - 8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2*E^x*x + E^(2*x)*x)*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.
\(\displaystyle \int \frac {e^{x+5} (5-2 x)+2 e^{2 x+5}+e^{x+5} x \log (x)+3 e^5}{8 e^x x+4 e^{2 x} x+4 x+\left (2 e^x x+e^{2 x} x+x\right ) \log ^2(x)+\left (-8 e^x x-4 e^{2 x} x-4 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^5 \left (e^x (5-2 x)+2 e^{2 x}+e^x x \log (x)+3\right )}{\left (e^x+1\right )^2 x (2-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^5 \int \frac {e^x (5-2 x)+2 e^{2 x}+e^x x \log (x)+3}{\left (1+e^x\right )^2 x (2-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^5 \int \left (\frac {\log (x) x-2 x+1}{\left (1+e^x\right ) x (\log (x)-2)^2}-\frac {1}{\left (1+e^x\right )^2 (\log (x)-2)}+\frac {2}{x (\log (x)-2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^5 \left (-2 \int \frac {1}{\left (1+e^x\right ) (\log (x)-2)^2}dx+\int \frac {1}{\left (1+e^x\right ) x (\log (x)-2)^2}dx-\int \frac {1}{\left (1+e^x\right )^2 (\log (x)-2)}dx+\int \frac {\log (x)}{\left (1+e^x\right ) (\log (x)-2)^2}dx+\frac {2}{2-\log (x)}\right )\) |
Int[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x])/(4* x + 8*E^x*x + 4*E^(2*x)*x + (-4*x - 8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2 *E^x*x + E^(2*x)*x)*Log[x]^2),x]
3.14.8.3.1 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 = 0.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {\left (3+2 \,{\mathrm e}^{x}\right ) {\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (\ln \left (x \right )-2\right )}\) | \(23\) |
norman | \(\frac {-2 \,{\mathrm e}^{5} {\mathrm e}^{x}-3 \,{\mathrm e}^{5}}{\left (\ln \left (x \right )-2\right ) \left ({\mathrm e}^{x}+1\right )}\) | \(25\) |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{x}+3 \,{\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (\ln \left (x \right )-2\right )}\) | \(26\) |
int((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+3*exp(5 ))/((x*exp(x)^2+2*exp(x)*x+x)*ln(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*ln(x) +4*x*exp(x)^2+8*exp(x)*x+4*x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{10} + 2 \, e^{\left (x + 10\right )}}{{\left (e^{5} + e^{\left (x + 5\right )}\right )} \log \left (x\right ) - 2 \, e^{5} - 2 \, e^{\left (x + 5\right )}} \]
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=- \frac {e^{5}}{\left (\log {\left (x \right )} - 2\right ) e^{x} + \log {\left (x \right )} - 2} - \frac {2 e^{5}}{\log {\left (x \right )} - 2} \]
integrate((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)**2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)**2+2*exp(x)*x+x)*ln(x)**2+(-4*x*exp(x)**2-8*exp(x)*x- 4*x)*ln(x)+4*x*exp(x)**2+8*exp(x)*x+4*x),x)
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{5} + 2 \, e^{\left (x + 5\right )}}{{\left (\log \left (x\right ) - 2\right )} e^{x} + \log \left (x\right ) - 2} \]
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{10} + 2 \, e^{\left (x + 10\right )}}{e^{5} \log \left (x\right ) + e^{\left (x + 5\right )} \log \left (x\right ) - 2 \, e^{5} - 2 \, e^{\left (x + 5\right )}} \]
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm=\
Time = 9.38 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.00 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=\frac {\frac {{\mathrm {e}}^5}{2}+\frac {x\,{\mathrm {e}}^5}{2}}{{\mathrm {e}}^x+1}-\frac {\frac {{\mathrm {e}}^5\,\left (2\,{\mathrm {e}}^{2\,x}+5\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x+3\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}+\frac {x\,{\mathrm {e}}^{x+5}\,\ln \left (x\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}}{\ln \left (x\right )-2}-\frac {\frac {{\mathrm {e}}^5\,\left (x+1\right )}{2}-{\mathrm {e}}^{x+5}\,\left (\frac {x}{2}-\frac {1}{2}\right )}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1} \]
int((3*exp(5) + 2*exp(2*x)*exp(5) - exp(5)*exp(x)*(2*x - 5) + x*exp(5)*exp (x)*log(x))/(4*x + 4*x*exp(2*x) - log(x)*(4*x + 4*x*exp(2*x) + 8*x*exp(x)) + log(x)^2*(x + x*exp(2*x) + 2*x*exp(x)) + 8*x*exp(x)),x)