Integrand size = 83, antiderivative size = 29 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\log \left (-2+x-\left (-\frac {e^{-x}}{5}+x\right )^2-\log (4)-\log (2 x)\right ) \]
Time = 5.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 x+\log \left (1+50 e^{2 x}-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+25 e^{2 x} \log (4)+25 e^{2 x} \log (2 x)\right ) \]
Integrate[(-2*x + E^x*(-10*x + 10*x^2) + E^(2*x)*(25 - 25*x + 50*x^2))/(x - 10*E^x*x^2 + E^(2*x)*(50*x - 25*x^2 + 25*x^3 + 25*x*Log[4]) + 25*E^(2*x) *x*Log[2*x]),x]
-2*x + Log[1 + 50*E^(2*x) - 10*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 25* E^(2*x)*Log[4] + 25*E^(2*x)*Log[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.
\(\displaystyle \int \frac {e^x \left (10 x^2-10 x\right )+e^{2 x} \left (50 x^2-25 x+25\right )-2 x}{-10 e^x x^2+e^{2 x} \left (25 x^3-25 x^2+50 x+25 x \log (4)\right )+x+25 e^{2 x} x \log (2 x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2-x+1}{x \left (x^2-x+\log (8 x)+2\right )}+\frac {10 e^x x^4-2 x^3+10 e^x x^2 \log (2 x)+20 e^x x^2 (1+\log (2))-10 e^x x \log (2 x)-2 x \log (2 x)-3 x \left (1+\frac {\log (16)}{3}\right )-10 e^x x (1+\log (4))-1}{x \left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x \left (-25 e^{2 x} x^2+10 e^x x+25 e^{2 x} x-25 e^{2 x} \log (2 x)-50 e^{2 x} (1+\log (2))-1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+2 \int \frac {x^2}{\left (-25 e^{2 x} x^2+10 e^x x+25 e^{2 x} x-25 e^{2 x} \log (2 x)-50 e^{2 x} (1+\log (2))-1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+2 \int \frac {\log (2 x)}{\left (-25 e^{2 x} x^2+10 e^x x+25 e^{2 x} x-25 e^{2 x} \log (2 x)-50 e^{2 x} (1+\log (2))-1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+10 \int \frac {e^x \log (2 x)}{\left (-25 e^{2 x} x^2+10 e^x x+25 e^{2 x} x-25 e^{2 x} \log (2 x)-50 e^{2 x} (1+\log (2))-1\right ) \left (x^2-x+\log (8 x)+2\right )}dx-(3+\log (16)) \int \frac {1}{\left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}dx-10 (1+\log (4)) \int \frac {e^x}{\left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+20 (1+\log (2)) \int \frac {e^x x}{\left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+10 \int \frac {e^x x \log (2 x)}{\left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+10 \int \frac {e^x x^3}{\left (25 e^{2 x} x^2-10 e^x x-25 e^{2 x} x+25 e^{2 x} \log (2 x)+50 e^{2 x} (1+\log (2))+1\right ) \left (x^2-x+\log (8 x)+2\right )}dx+\log \left (x^2-x+\log (8 x)+2\right )\) |
Int[(-2*x + E^x*(-10*x + 10*x^2) + E^(2*x)*(25 - 25*x + 50*x^2))/(x - 10*E ^x*x^2 + E^(2*x)*(50*x - 25*x^2 + 25*x^3 + 25*x*Log[4]) + 25*E^(2*x)*x*Log [2*x]),x]
3.1.77.3.1 Defintions of rubi rules used
Time = 0.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\ln \left (\ln \left (2 x \right )+x^{2}+2 \ln \left (2\right )-\frac {2 x \,{\mathrm e}^{-x}}{5}-x +2+\frac {{\mathrm e}^{-2 x}}{25}\right )\) | \(31\) |
parallelrisch | \(\ln \left ({\mathrm e}^{2 x} x^{2}+2 \ln \left (2\right ) {\mathrm e}^{2 x}+{\mathrm e}^{2 x} \ln \left (2 x \right )-x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}-\frac {2 \,{\mathrm e}^{x} x}{5}+\frac {1}{25}\right )-2 x\) | \(51\) |
int(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x)^2*ln (2*x)+(50*x*ln(2)+25*x^3-25*x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x,method=_ RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\log \left ({\left (25 \, {\left (x^{2} - x + 2 \, \log \left (2\right ) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 25 \, e^{\left (2 \, x\right )} \log \left (2 \, x\right ) + 1\right )} e^{\left (-2 \, x\right )}\right ) \]
integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x )^2*log(2*x)+(50*x*log(2)+25*x^3-25*x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (24) = 48\).
Time = 9.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.76 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=- 2 x + \log {\left (- \frac {2 x e^{x}}{5 x^{2} - 5 x + 5 \log {\left (2 x \right )} + 10 \log {\left (2 \right )} + 10} + e^{2 x} + \frac {1}{25 x^{2} - 25 x + 25 \log {\left (2 x \right )} + 50 \log {\left (2 \right )} + 50} \right )} + \log {\left (x^{2} - x + \log {\left (2 x \right )} + 2 \log {\left (2 \right )} + 2 \right )} \]
integrate(((50*x**2-25*x+25)*exp(x)**2+(10*x**2-10*x)*exp(x)-2*x)/(25*x*ex p(x)**2*ln(2*x)+(50*x*ln(2)+25*x**3-25*x**2+50*x)*exp(x)**2-10*exp(x)*x**2 +x),x)
-2*x + log(-2*x*exp(x)/(5*x**2 - 5*x + 5*log(2*x) + 10*log(2) + 10) + exp( 2*x) + 1/(25*x**2 - 25*x + 25*log(2*x) + 50*log(2) + 50)) + log(x**2 - x + log(2*x) + 2*log(2) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (28) = 56\).
Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 \, x + \log \left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right ) + \log \left (\frac {25 \, {\left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 1}{25 \, {\left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right )}}\right ) \]
integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x )^2*log(2*x)+(50*x*log(2)+25*x^3-25*x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm=\
-2*x + log(x^2 - x + 3*log(2) + log(x) + 2) + log(1/25*(25*(x^2 - x + 3*lo g(2) + log(x) + 2)*e^(2*x) - 10*x*e^x + 1)/(x^2 - x + 3*log(2) + log(x) + 2))
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 \, x + \log \left (25 \, x^{2} e^{\left (2 \, x\right )} - 25 \, x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 75 \, e^{\left (2 \, x\right )} \log \left (2\right ) + 25 \, e^{\left (2 \, x\right )} \log \left (x\right ) + 50 \, e^{\left (2 \, x\right )} + 1\right ) \]
integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x )^2*log(2*x)+(50*x*log(2)+25*x^3-25*x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm=\
-2*x + log(25*x^2*e^(2*x) - 25*x*e^(2*x) - 10*x*e^x + 75*e^(2*x)*log(2) + 25*e^(2*x)*log(x) + 50*e^(2*x) + 1)
Timed out. \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\int -\frac {2\,x-{\mathrm {e}}^{2\,x}\,\left (50\,x^2-25\,x+25\right )+{\mathrm {e}}^x\,\left (10\,x-10\,x^2\right )}{x-10\,x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (50\,x+50\,x\,\ln \left (2\right )-25\,x^2+25\,x^3\right )+25\,x\,\ln \left (2\,x\right )\,{\mathrm {e}}^{2\,x}} \,d x \]
int(-(2*x - exp(2*x)*(50*x^2 - 25*x + 25) + exp(x)*(10*x - 10*x^2))/(x - 1 0*x^2*exp(x) + exp(2*x)*(50*x + 50*x*log(2) - 25*x^2 + 25*x^3) + 25*x*log( 2*x)*exp(2*x)),x)