Integrand size = 89, antiderivative size = 26 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {5+\frac {9 x}{\log (x)}}{5+\frac {5 e^{-x}}{2}+2 x} \]
Time = 5.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 e^x (9 x+5 \log (x))}{\left (5+2 e^x (5+2 x)\right ) \log (x)} \]
Integrate[(-90*E^x + E^(2*x)*(-180 - 72*x) + (180*E^(2*x) + E^x*(90 + 90*x ))*Log[x] + (50*E^x - 40*E^(2*x))*Log[x]^2)/((25 + E^x*(100 + 40*x) + E^(2 *x)*(100 + 80*x + 16*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.
\(\displaystyle \int \frac {-90 e^x+e^{2 x} (-72 x-180)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)+\left (e^x (90 x+90)+180 e^{2 x}\right ) \log (x)}{\left (e^{2 x} \left (16 x^2+80 x+100\right )+e^x (40 x+100)+25\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 e^x \left (-90 e^x-36 e^x x-20 e^x \log ^2(x)+25 \log ^2(x)+90 e^x \log (x)+45 x \log (x)+45 \log (x)-45\right )}{\left (4 e^x x+10 e^x+5\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {e^x \left (20 e^x \log ^2(x)-25 \log ^2(x)-90 e^x \log (x)-45 x \log (x)-45 \log (x)+90 e^x+36 e^x x+45\right )}{\left (4 e^x x+10 e^x+5\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {e^x \left (20 e^x \log ^2(x)-25 \log ^2(x)-90 e^x \log (x)-45 x \log (x)-45 \log (x)+90 e^x+36 e^x x+45\right )}{\left (4 e^x x+10 e^x+5\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {e^x \left (10 \log ^2(x)-45 \log (x)+18 x+45\right )}{(2 x+5) \left (4 e^x x+10 e^x+5\right ) \log ^2(x)}-\frac {5 e^x (2 x+7) (9 x+5 \log (x))}{(2 x+5) \left (4 e^x x+10 e^x+5\right )^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (-25 \int \frac {e^x}{\left (4 e^x x+10 e^x+5\right )^2}dx-50 \int \frac {e^x}{(2 x+5) \left (4 e^x x+10 e^x+5\right )^2}dx+10 \int \frac {e^x}{(2 x+5) \left (4 e^x x+10 e^x+5\right )}dx+9 \int \frac {e^x}{\left (4 e^x x+10 e^x+5\right ) \log ^2(x)}dx-45 \int \frac {e^x}{\left (4 e^x x+10 e^x+5\right )^2 \log (x)}dx-45 \int \frac {e^x x}{\left (4 e^x x+10 e^x+5\right )^2 \log (x)}dx+225 \int \frac {e^x}{(2 x+5) \left (4 e^x x+10 e^x+5\right )^2 \log (x)}dx-45 \int \frac {e^x}{(2 x+5) \left (4 e^x x+10 e^x+5\right ) \log (x)}dx\right )\) |
Int[(-90*E^x + E^(2*x)*(-180 - 72*x) + (180*E^(2*x) + E^x*(90 + 90*x))*Log [x] + (50*E^x - 40*E^(2*x))*Log[x]^2)/((25 + E^x*(100 + 40*x) + E^(2*x)*(1 00 + 80*x + 16*x^2))*Log[x]^2),x]
3.8.84.3.1 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]]
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(-\frac {-72 \,{\mathrm e}^{x} x -40 \,{\mathrm e}^{x} \ln \left (x \right )}{4 \ln \left (x \right ) \left (4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5\right )}\) | \(32\) |
risch | \(\frac {10 \,{\mathrm e}^{x}}{4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5}+\frac {18 x \,{\mathrm e}^{x}}{\left (4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5\right ) \ln \left (x \right )}\) | \(41\) |
int(((-40*exp(x)^2+50*exp(x))*ln(x)^2+(180*exp(x)^2+(90*x+90)*exp(x))*ln(x )+(-72*x-180)*exp(x)^2-90*exp(x))/((16*x^2+80*x+100)*exp(x)^2+(40*x+100)*e xp(x)+25)/ln(x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x e^{x} + 5 \, e^{x} \log \left (x\right )\right )}}{{\left (2 \, {\left (2 \, x + 5\right )} e^{x} + 5\right )} \log \left (x\right )} \]
integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x ))*log(x)+(-72*x-180)*exp(x)^2-90*exp(x))/((16*x^2+80*x+100)*exp(x)^2+(40* x+100)*exp(x)+25)/log(x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (19) = 38\).
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {9 x}{\left (2 x + 5\right ) \log {\left (x \right )}} + \frac {- 45 x - 25 \log {\left (x \right )}}{10 x \log {\left (x \right )} + \left (8 x^{2} \log {\left (x \right )} + 40 x \log {\left (x \right )} + 50 \log {\left (x \right )}\right ) e^{x} + 25 \log {\left (x \right )}} + \frac {10}{4 x + 10} \]
integrate(((-40*exp(x)**2+50*exp(x))*ln(x)**2+(180*exp(x)**2+(90*x+90)*exp (x))*ln(x)+(-72*x-180)*exp(x)**2-90*exp(x))/((16*x**2+80*x+100)*exp(x)**2+ (40*x+100)*exp(x)+25)/ln(x)**2,x)
9*x/((2*x + 5)*log(x)) + (-45*x - 25*log(x))/(10*x*log(x) + (8*x**2*log(x) + 40*x*log(x) + 50*log(x))*exp(x) + 25*log(x)) + 10/(4*x + 10)
Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x + 5 \, \log \left (x\right )\right )} e^{x}}{2 \, {\left (2 \, x + 5\right )} e^{x} \log \left (x\right ) + 5 \, \log \left (x\right )} \]
integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x ))*log(x)+(-72*x-180)*exp(x)^2-90*exp(x))/((16*x^2+80*x+100)*exp(x)^2+(40* x+100)*exp(x)+25)/log(x)^2,x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x e^{x} + 5 \, e^{x} \log \left (x\right )\right )}}{4 \, x e^{x} \log \left (x\right ) + 10 \, e^{x} \log \left (x\right ) + 5 \, \log \left (x\right )} \]
integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x ))*log(x)+(-72*x-180)*exp(x)^2-90*exp(x))/((16*x^2+80*x+100)*exp(x)^2+(40* x+100)*exp(x)+25)/log(x)^2,x, algorithm=\
Timed out. \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\int -\frac {\left (40\,{\mathrm {e}}^{2\,x}-50\,{\mathrm {e}}^x\right )\,{\ln \left (x\right )}^2+\left (-180\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (90\,x+90\right )\right )\,\ln \left (x\right )+90\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (72\,x+180\right )}{{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+80\,x+100\right )+{\mathrm {e}}^x\,\left (40\,x+100\right )+25\right )} \,d x \]
int(-(90*exp(x) - log(x)*(180*exp(2*x) + exp(x)*(90*x + 90)) + exp(2*x)*(7 2*x + 180) + log(x)^2*(40*exp(2*x) - 50*exp(x)))/(log(x)^2*(exp(2*x)*(80*x + 16*x^2 + 100) + exp(x)*(40*x + 100) + 25)),x)