Integrand size = 62, antiderivative size = 23 \[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=5+10 e^{-\frac {1}{4-e \log (x)}} (5-x) x \]
Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=-10 e^{\frac {1}{-4+e \log (x)}} (-5+x) x \]
Integrate[(E^(-4 + E*Log[x])^(-1)*(800 - 320*x + E*(-50 + 10*x) + E*(-400 + 160*x)*Log[x] + E^2*(50 - 20*x)*Log[x]^2))/(16 - 8*E*Log[x] + E^2*Log[x] ^2),x]
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {2726}
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.
\(\displaystyle \int \frac {e^{\frac {1}{e \log (x)-4}} \left (-320 x+e (10 x-50)+e^2 (50-20 x) \log ^2(x)+e (160 x-400) \log (x)+800\right )}{e^2 \log ^2(x)-8 e \log (x)+16} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {10 (5-x) x e^{\frac {1}{e \log (x)-4}} (4-e \log (x))^2}{e^2 \log ^2(x)-8 e \log (x)+16}\) |
Int[(E^(-4 + E*Log[x])^(-1)*(800 - 320*x + E*(-50 + 10*x) + E*(-400 + 160* x)*Log[x] + E^2*(50 - 20*x)*Log[x]^2))/(16 - 8*E*Log[x] + E^2*Log[x]^2),x]
3.21.89.3.1 Defintions of rubi rules used
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.54 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\left (-10 x^{2}+50 x \right ) {\mathrm e}^{\frac {1}{{\mathrm e} \ln \left (x \right )-4}}\) | \(21\) |
norman | \(\frac {\left (-200 x +40 x^{2}+50 x \,{\mathrm e} \ln \left (x \right )-10 x^{2} {\mathrm e} \ln \left (x \right )\right ) {\mathrm e}^{\frac {1}{{\mathrm e} \ln \left (x \right )-4}}}{{\mathrm e} \ln \left (x \right )-4}\) | \(50\) |
parallelrisch | \(\frac {\left (-2560 x^{2} {\mathrm e} \ln \left (x \right )+12800 x \,{\mathrm e} \ln \left (x \right )+10240 x^{2}-51200 x \right ) {\mathrm e}^{\frac {1}{{\mathrm e} \ln \left (x \right )-4}}}{256 \,{\mathrm e} \ln \left (x \right )-1024}\) | \(51\) |
int(((-20*x+50)*exp(1)^2*ln(x)^2+(160*x-400)*exp(1)*ln(x)+(10*x-50)*exp(1) -320*x+800)/(exp(1)^2*ln(x)^2-8*exp(1)*ln(x)+16)/exp(-1/(exp(1)*ln(x)-4)), x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=-10 \, {\left (x^{2} - 5 \, x\right )} e^{\left (\frac {1}{e \log \left (x\right ) - 4}\right )} \]
integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50 )*exp(1)-320*x+800)/(exp(1)^2*log(x)^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)* log(x)-4)),x, algorithm=\
Time = 2.53 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=\left (- 10 x^{2} + 50 x\right ) e^{\frac {1}{e \log {\left (x \right )} - 4}} \]
integrate(((-20*x+50)*exp(1)**2*ln(x)**2+(160*x-400)*exp(1)*ln(x)+(10*x-50 )*exp(1)-320*x+800)/(exp(1)**2*ln(x)**2-8*exp(1)*ln(x)+16)/exp(-1/(exp(1)* ln(x)-4)),x)
\[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=\int { -\frac {10 \, {\left ({\left (2 \, x - 5\right )} e^{2} \log \left (x\right )^{2} - 8 \, {\left (2 \, x - 5\right )} e \log \left (x\right ) - {\left (x - 5\right )} e + 32 \, x - 80\right )} e^{\left (\frac {1}{e \log \left (x\right ) - 4}\right )}}{e^{2} \log \left (x\right )^{2} - 8 \, e \log \left (x\right ) + 16} \,d x } \]
integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50 )*exp(1)-320*x+800)/(exp(1)^2*log(x)^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)* log(x)-4)),x, algorithm=\
320*x^2*e^(1/(e*log(x) - 4) - 1) - 10*x^2*e^(1/(e*log(x) - 4)) - 800*x*e^( 1/(e*log(x) - 4) - 1) + 50*x*e^(1/(e*log(x) - 4)) - 10*integrate(((2*x*e^2 - 5*e^2)*log(x)^2 - 8*(2*x*e - 5*e)*log(x))*e^(1/(e*log(x) - 4))/(e^2*log (x)^2 - 8*e*log(x) + 16), x)
\[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=\int { -\frac {10 \, {\left ({\left (2 \, x - 5\right )} e^{2} \log \left (x\right )^{2} - 8 \, {\left (2 \, x - 5\right )} e \log \left (x\right ) - {\left (x - 5\right )} e + 32 \, x - 80\right )} e^{\left (\frac {1}{e \log \left (x\right ) - 4}\right )}}{e^{2} \log \left (x\right )^{2} - 8 \, e \log \left (x\right ) + 16} \,d x } \]
integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50 )*exp(1)-320*x+800)/(exp(1)^2*log(x)^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)* log(x)-4)),x, algorithm=\
integrate(-10*((2*x - 5)*e^2*log(x)^2 - 8*(2*x - 5)*e*log(x) - (x - 5)*e + 32*x - 80)*e^(1/(e*log(x) - 4))/(e^2*log(x)^2 - 8*e*log(x) + 16), x)
Timed out. \[ \int \frac {e^{\frac {1}{-4+e \log (x)}} \left (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x)\right )}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {1}{\mathrm {e}\,\ln \left (x\right )-4}}\,\left (-{\mathrm {e}}^2\,\left (20\,x-50\right )\,{\ln \left (x\right )}^2+\mathrm {e}\,\left (160\,x-400\right )\,\ln \left (x\right )-320\,x+\mathrm {e}\,\left (10\,x-50\right )+800\right )}{{\mathrm {e}}^2\,{\ln \left (x\right )}^2-8\,\mathrm {e}\,\ln \left (x\right )+16} \,d x \]
int((exp(1/(exp(1)*log(x) - 4))*(exp(1)*(10*x - 50) - 320*x - exp(2)*log(x )^2*(20*x - 50) + exp(1)*log(x)*(160*x - 400) + 800))/(exp(2)*log(x)^2 - 8 *exp(1)*log(x) + 16),x)