Integrand size = 93, antiderivative size = 32 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=x \left (1+\frac {\left (-x+25 x^2\right ) \left (5+\frac {e^x}{10+x-\log (4)}\right )}{x}\right ) \] Output:
x*((25*x^2-x)/x*(exp(x)/(10+x-2*ln(2))+5)+1)
Time = 5.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=x \left (-4+125 x+\frac {e^x (-1+25 x)}{10+x-\log (4)}\right ) \] Input:
Integrate[(-400 + 24920*x + 4996*x^2 + 250*x^3 + (80 - 4992*x - 500*x^2)*L og[4] + (-4 + 250*x)*Log[4]^2 + E^x*(-10 + 490*x + 274*x^2 + 25*x^3 + (1 - 49*x - 25*x^2)*Log[4]))/(100 + 20*x + x^2 + (-20 - 2*x)*Log[4] + Log[4]^2 ),x]
Output:
x*(-4 + 125*x + (E^x*(-1 + 25*x))/(10 + x - Log[4]))
Time = 0.53 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {7239, 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.
| \(\displaystyle \int \frac {250 x^3+4996 x^2+\left (-500 x^2-4992 x+80\right ) \log (4)+e^x \left (25 x^3+274 x^2+\left (-25 x^2-49 x+1\right ) \log (4)+490 x-10\right )+24920 x+(250 x-4) \log ^2(4)-400}{x^2+20 x+(-2 x-20) \log (4)+100+\log ^2(4)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {e^x \left (25 x^3+x^2 (274-25 \log (4))+49 x (10-\log (4))-10+\log (4)\right )}{(x+10-\log (4))^2}+250 x-4\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 125 x^2+25 e^x x-4 x-25 e^x+\frac {e^x (251-25 \log (4)) (10-\log (4))}{x+10-\log (4)}-2 e^x (113-25 \log (2))\) |
Input:
Int[(-400 + 24920*x + 4996*x^2 + 250*x^3 + (80 - 4992*x - 500*x^2)*Log[4] + (-4 + 250*x)*Log[4]^2 + E^x*(-10 + 490*x + 274*x^2 + 25*x^3 + (1 - 49*x - 25*x^2)*Log[4]))/(100 + 20*x + x^2 + (-20 - 2*x)*Log[4] + Log[4]^2),x]
Output:
-25*E^x - 4*x + 25*E^x*x + 125*x^2 - 2*E^x*(113 - 25*Log[2]) + (E^x*(251 - 25*Log[4])*(10 - Log[4]))/(10 + x - Log[4])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59
| method | result | size |
| norman | \(\frac {\left (250 \ln \left (2\right )-1246\right ) x^{2}+{\mathrm e}^{x} x -125 x^{3}-25 \,{\mathrm e}^{x} x^{2}-16 \ln \left (2\right )^{2}+160 \ln \left (2\right )-400}{2 \ln \left (2\right )-x -10}\) | \(51\) |
| parallelrisch | \(-\frac {-250 x^{2} \ln \left (2\right )+125 x^{3}+25 \,{\mathrm e}^{x} x^{2}+16 \ln \left (2\right )^{2}+400+1246 x^{2}-{\mathrm e}^{x} x -160 \ln \left (2\right )}{2 \ln \left (2\right )-x -10}\) | \(55\) |
| parts | \(125 x^{2}-4 x +\frac {100 \,{\mathrm e}^{x} \ln \left (2\right )^{2}}{10+x -2 \ln \left (2\right )}+50 \,{\mathrm e}^{x} \ln \left (2\right )+\frac {2510 \,{\mathrm e}^{x}}{10+x -2 \ln \left (2\right )}-251 \,{\mathrm e}^{x}-\frac {1002 \ln \left (2\right ) {\mathrm e}^{x}}{10+x -2 \ln \left (2\right )}+25 \,{\mathrm e}^{x} x\) | \(70\) |
| default | \(\text {Expression too large to display}\) | \(711\) |
Input:
int(((2*(-25*x^2-49*x+1)*ln(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4) *ln(2)^2+2*(-500*x^2-4992*x+80)*ln(2)+250*x^3+4996*x^2+24920*x-400)/(4*ln( 2)^2+2*(-2*x-20)*ln(2)+x^2+20*x+100),x,method=_RETURNVERBOSE)
Output:
((250*ln(2)-1246)*x^2+exp(x)*x-125*x^3-25*exp(x)*x^2-16*ln(2)^2+160*ln(2)- 400)/(2*ln(2)-x-10)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=\frac {125 \, x^{3} + 1246 \, x^{2} + {\left (25 \, x^{2} - x\right )} e^{x} - 2 \, {\left (125 \, x^{2} - 4 \, x\right )} \log \left (2\right ) - 40 \, x}{x - 2 \, \log \left (2\right ) + 10} \] Input:
integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(2 50*x-4)*log(2)^2+2*(-500*x^2-4992*x+80)*log(2)+250*x^3+4996*x^2+24920*x-40 0)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="fricas")
Output:
(125*x^3 + 1246*x^2 + (25*x^2 - x)*e^x - 2*(125*x^2 - 4*x)*log(2) - 40*x)/ (x - 2*log(2) + 10)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=125 x^{2} - 4 x + \frac {\left (25 x^{2} - x\right ) e^{x}}{x - 2 \log {\left (2 \right )} + 10} \] Input:
integrate(((2*(-25*x**2-49*x+1)*ln(2)+25*x**3+274*x**2+490*x-10)*exp(x)+4* (250*x-4)*ln(2)**2+2*(-500*x**2-4992*x+80)*ln(2)+250*x**3+4996*x**2+24920* x-400)/(4*ln(2)**2+2*(-2*x-20)*ln(2)+x**2+20*x+100),x)
Output:
125*x**2 - 4*x + (25*x**2 - x)*exp(x)/(x - 2*log(2) + 10)
\[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=\int { \frac {250 \, x^{3} + 8 \, {\left (125 \, x - 2\right )} \log \left (2\right )^{2} + 4996 \, x^{2} + {\left (25 \, x^{3} + 274 \, x^{2} - 2 \, {\left (25 \, x^{2} + 49 \, x - 1\right )} \log \left (2\right ) + 490 \, x - 10\right )} e^{x} - 8 \, {\left (125 \, x^{2} + 1248 \, x - 20\right )} \log \left (2\right ) + 24920 \, x - 400}{x^{2} - 4 \, {\left (x + 10\right )} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 20 \, x + 100} \,d x } \] Input:
integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(2 50*x-4)*log(2)^2+2*(-500*x^2-4992*x+80)*log(2)+250*x^3+4996*x^2+24920*x-40 0)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="maxima")
Output:
-1000*(2*(log(2) - 5)/(x - 2*log(2) + 10) - log(x - 2*log(2) + 10))*log(2) ^2 + 125*x^2 + 1000*x*(log(2) - 5) - 2*(log(2) - 5)*integrate(e^x/(x^2 - 4 *x*(log(2) - 5) + 4*log(2)^2 - 40*log(2) + 100), x) - 1000*(4*(log(2) - 5) *log(x - 2*log(2) + 10) + x - 4*(log(2)^2 - 10*log(2) + 25)/(x - 2*log(2) + 10))*log(2) + 9984*(2*(log(2) - 5)/(x - 2*log(2) + 10) - log(x - 2*log(2 ) + 10))*log(2) - 8*e^(-10)*exp_integral_e(2, -x + 2*log(2) - 10)*log(2)/( x - 2*log(2) + 10) + 3000*(log(2)^2 - 10*log(2) + 25)*log(x - 2*log(2) + 1 0) + 19984*(log(2) - 5)*log(x - 2*log(2) + 10) + 4996*x + (25*x^2 - x)*e^x /(x - 2*log(2) + 10) + 40*e^(-10)*exp_integral_e(2, -x + 2*log(2) - 10)/(x - 2*log(2) + 10) + 16*log(2)^2/(x - 2*log(2) + 10) - 2000*(log(2)^3 - 15* log(2)^2 + 75*log(2) - 125)/(x - 2*log(2) + 10) - 19984*(log(2)^2 - 10*log (2) + 25)/(x - 2*log(2) + 10) - 49840*(log(2) - 5)/(x - 2*log(2) + 10) - 1 60*log(2)/(x - 2*log(2) + 10) + 400/(x - 2*log(2) + 10) + 24920*log(x - 2* log(2) + 10)
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=\frac {125 \, x^{3} + 25 \, x^{2} e^{x} - 250 \, x^{2} \log \left (2\right ) + 1246 \, x^{2} - x e^{x} + 8 \, x \log \left (2\right ) - 40 \, x}{x - 2 \, \log \left (2\right ) + 10} \] Input:
integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(2 50*x-4)*log(2)^2+2*(-500*x^2-4992*x+80)*log(2)+250*x^3+4996*x^2+24920*x-40 0)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="giac")
Output:
(125*x^3 + 25*x^2*e^x - 250*x^2*log(2) + 1246*x^2 - x*e^x + 8*x*log(2) - 4 0*x)/(x - 2*log(2) + 10)
Timed out. \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=\int \frac {24920\,x+{\mathrm {e}}^x\,\left (490\,x-2\,\ln \left (2\right )\,\left (25\,x^2+49\,x-1\right )+274\,x^2+25\,x^3-10\right )-2\,\ln \left (2\right )\,\left (500\,x^2+4992\,x-80\right )+4\,{\ln \left (2\right )}^2\,\left (250\,x-4\right )+4996\,x^2+250\,x^3-400}{20\,x-2\,\ln \left (2\right )\,\left (2\,x+20\right )+4\,{\ln \left (2\right )}^2+x^2+100} \,d x \] Input:
int((24920*x + exp(x)*(490*x - 2*log(2)*(49*x + 25*x^2 - 1) + 274*x^2 + 25 *x^3 - 10) - 2*log(2)*(4992*x + 500*x^2 - 80) + 4*log(2)^2*(250*x - 4) + 4 996*x^2 + 250*x^3 - 400)/(20*x - 2*log(2)*(2*x + 20) + 4*log(2)^2 + x^2 + 100),x)
Output:
int((24920*x + exp(x)*(490*x - 2*log(2)*(49*x + 25*x^2 - 1) + 274*x^2 + 25 *x^3 - 10) - 2*log(2)*(4992*x + 500*x^2 - 80) + 4*log(2)^2*(250*x - 4) + 4 996*x^2 + 250*x^3 - 400)/(20*x - 2*log(2)*(2*x + 20) + 4*log(2)^2 + x^2 + 100), x)
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {-400+24920 x+4996 x^2+250 x^3+\left (80-4992 x-500 x^2\right ) \log (4)+(-4+250 x) \log ^2(4)+e^x \left (-10+490 x+274 x^2+25 x^3+\left (1-49 x-25 x^2\right ) \log (4)\right )}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx=\frac {x \left (-25 e^{x} x +e^{x}+250 \,\mathrm {log}\left (2\right ) x -8 \,\mathrm {log}\left (2\right )-125 x^{2}-1246 x +40\right )}{2 \,\mathrm {log}\left (2\right )-x -10} \] Input:
int(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4 )*log(2)^2+2*(-500*x^2-4992*x+80)*log(2)+250*x^3+4996*x^2+24920*x-400)/(4* log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x)
Output:
(x*( - 25*e**x*x + e**x + 250*log(2)*x - 8*log(2) - 125*x**2 - 1246*x + 40 ))/(2*log(2) - x - 10)